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Largest number less than or equal to N / 2 which is coprime to N | C ++ implementation of the above approacdh ; Function to calculate gcd of two number ; Function to check if two numbers are coprime or not ; two numbers are coprime if their gcd is 1 ; Function to find largest integer less than or equal to N / 2 and coprime with N ; Check one by one all numbers less than or equal to N / 2 ; Driver code | #include <bits/stdc++.h> NEW_LINE #define ll long long int NEW_LINE using namespace std ; ll gcd ( ll a , ll b ) { if ( b == 0 ) return a ; else return gcd ( b , a % b ) ; } bool coPrime ( ll n1 , ll n2 ) { if ( gcd ( n1 , n2 ) == 1 ) return true ; else return false ; } ll largestCoprime ( ll N ) { ll half = floor ( N / 2 ) ; while ( coPrime ( N , half ) == false ) half -- ; return half ; } int main ( ) { ll n = 50 ; cout << largestCoprime ( n ) ; return 0 ; } |
Largest number less than or equal to N / 2 which is coprime to N | C ++ implementation of the above approach ; Function to find largest integer less than or equal to N / 2 and is coprime with N ; Handle the case for N = 6 ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; long long largestCoprime ( long long N ) { if ( N == 6 ) return 1 ; else if ( N % 4 == 0 ) return ( N / 2 ) - 1 ; else if ( N % 2 == 0 ) return ( N / 2 ) - 2 ; else return ( ( N - 1 ) / 2 ) ; } int main ( ) { long long int n = 50 ; cout << largestCoprime ( n ) << endl ; return 0 ; } |
Print all safe primes below N | C ++ implementation of the approach ; Function to print first n safe primes ; Initialize all entries of integer array as 1. A value in prime [ i ] will finally be 0 if i is Not a prime , else 1 ; 0 and 1 are not primes ; If prime [ p ] is not changed , then it is a prime ; Update all multiples of p ; If i is prime ; 2 p + 1 ; If 2 p + 1 is also a prime then set prime [ 2 p + 1 ] = 2 ; i is a safe prime ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; void printSafePrimes ( int n ) { int prime [ n + 1 ] ; for ( int i = 2 ; i <= n ; i ++ ) prime [ i ] = 1 ; prime [ 0 ] = prime [ 1 ] = 0 ; for ( int p = 2 ; p * p <= n ; p ++ ) { if ( prime [ p ] == 1 ) { for ( int i = p * 2 ; i <= n ; i += p ) prime [ i ] = 0 ; } } for ( int i = 2 ; i <= n ; i ++ ) { if ( prime [ i ] != 0 ) { int temp = ( 2 * i ) + 1 ; if ( temp <= n && prime [ temp ] != 0 ) prime [ temp ] = 2 ; } } for ( int i = 5 ; i <= n ; i ++ ) if ( prime [ i ] == 2 ) cout << i << " ▁ " ; } int main ( ) { int n = 20 ; printSafePrimes ( n ) ; return 0 ; } |
Minimum multiplications with { 2 , 3 , 7 } to make two numbers equal | C ++ implementation of the approach ; Function to find powers of 2 , 3 and 7 in x ; To keep count of each divisor ; To store the result ; Count powers of 2 in x ; Count powers of 3 in x ; Count powers of 7 in x ; Remaining number which is not divisible by 2 , 3 or 7 ; Function to return the minimum number of given operations required to make a and b equal ; a = x * 2 ^ a1 * 3 ^ a2 * 7 ^ a3 va [ 0 ] = a1 va [ 1 ] = a2 va [ 2 ] = a3 va [ 3 ] = x ; Similarly for b ; If a and b cannot be made equal with the given operation . Note that va [ 3 ] and vb [ 3 ] contain remaining numbers after repeated divisions with 2 , 3 and 7. If remaining numbers are not same then we cannot make them equal . ; Minimum number of operations required ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; vector < int > Divisors ( int x ) { int c = 0 ; vector < int > v ; while ( x % 2 == 0 ) { c ++ ; x /= 2 ; } v . push_back ( c ) ; c = 0 ; while ( x % 3 == 0 ) { c ++ ; x /= 3 ; } v . push_back ( c ) ; c = 0 ; while ( x % 7 == 0 ) { c ++ ; x /= 7 ; } v . push_back ( c ) ; v . push_back ( x ) ; return v ; } int MinOperations ( int a , int b ) { vector < int > va = Divisors ( a ) ; vector < int > vb = Divisors ( b ) ; if ( va [ 3 ] != vb [ 3 ] ) return -1 ; int minOperations = abs ( va [ 0 ] - vb [ 0 ] ) + abs ( va [ 1 ] - vb [ 1 ] ) + abs ( va [ 2 ] - vb [ 2 ] ) ; return minOperations ; } int main ( ) { int a = 14 , b = 28 ; cout << MinOperations ( a , b ) ; return 0 ; } |
Product of N with its largest odd digit | C ++ program to find the product of N with its largest odd digit ; Function to return the largest odd digit in n ; If all digits are even then - 1 will be returned ; Last digit from n ; If current digit is odd and > maxOdd ; Remove last digit ; Return the maximum odd digit ; Function to return the product of n with its largest odd digit ; If there are no odd digits in n ; Product of n with its largest odd digit ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int largestOddDigit ( int n ) { int maxOdd = -1 ; while ( n > 0 ) { int digit = n % 10 ; if ( digit % 2 == 1 && digit > maxOdd ) maxOdd = digit ; n = n / 10 ; } return maxOdd ; } int getProduct ( int n ) { int maxOdd = largestOddDigit ( n ) ; if ( maxOdd == -1 ) return -1 ; return ( n * maxOdd ) ; } int main ( ) { int n = 12345 ; cout << getProduct ( n ) ; return 0 ; } |
Check if a number is perfect square without finding square root | C ++ program for above approach ; Program to find if x is a perfect square . ; Check if mid is perfect square ; Mid is small -> go right to increase mid ; Mid is large -> to left to decrease mid ; Driver Code ; Function Call | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool isPerfectSquare ( int x ) { long long left = 1 , right = x ; while ( left <= right ) { long long mid = ( left + right ) / 2 ; if ( mid * mid == x ) { return true ; } if ( mid * mid < x ) { left = mid + 1 ; } else { right = mid - 1 ; } } return false ; } int main ( ) { int x = 2500 ; if ( isPerfectSquare ( x ) ) cout << " Yes " ; else cout << " No " ; return 0 ; } |
Find k numbers which are powers of 2 and have sum N | Set 1 | CPP program to find k numbers that are power of 2 and have sum equal to N ; function to print numbers ; Count the number of set bits ; Not - possible condition ; Stores the number ; Get the set bits ; Iterate till we get K elements ; Get the topmost element ; Push the elements / 2 into priority queue ; Print all elements ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; void printNum ( int n , int k ) { int x = __builtin_popcount ( n ) ; if ( k < x k > n ) { cout << " - 1" ; return ; } priority_queue < int > pq ; int two = 1 ; while ( n ) { if ( n & 1 ) { pq . push ( two ) ; } two = two * 2 ; n = n >> 1 ; } while ( pq . size ( ) < k ) { int el = pq . top ( ) ; pq . pop ( ) ; pq . push ( el / 2 ) ; pq . push ( el / 2 ) ; } int ind = 0 ; while ( ind < k ) { cout << pq . top ( ) << " ▁ " ; pq . pop ( ) ; ind ++ ; } } int main ( ) { int n = 9 , k = 4 ; printNum ( n , k ) ; return 0 ; } |
Sum of LCM ( 1 , n ) , LCM ( 2 , n ) , LCM ( 3 , n ) , ... , LCM ( n , n ) | C ++ implementation of the approach ; Euler totient Function ; Function to return the required LCM sum ; Summation of d * ETF ( d ) where d belongs to set of divisors of n ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; #define n 1000002 NEW_LINE #define ll long long int NEW_LINE ll phi [ n + 2 ] , ans [ n + 2 ] ; void ETF ( ) { for ( int i = 1 ; i <= n ; i ++ ) { phi [ i ] = i ; } for ( int i = 2 ; i <= n ; i ++ ) { if ( phi [ i ] == i ) { phi [ i ] = i - 1 ; for ( int j = 2 * i ; j <= n ; j += i ) { phi [ j ] = ( phi [ j ] * ( i - 1 ) ) / i ; } } } } ll LcmSum ( int m ) { ETF ( ) ; for ( int i = 1 ; i <= n ; i ++ ) { for ( int j = i ; j <= n ; j += i ) { ans [ j ] += ( i * phi [ i ] ) ; } } ll answer = ans [ m ] ; answer = ( answer + 1 ) * m ; answer = answer / 2 ; return answer ; } int main ( ) { int m = 5 ; cout << LcmSum ( m ) ; return 0 ; } |
Permutations of string such that no two vowels are adjacent | CPP program to count permutations of string such that no two vowels are adjacent ; Factorial of a number ; Function to find c ( n , r ) ; Function to count permutations of string such that no two vowels are adjacent ; Finding the frequencies of the characters ; finding the no . of vowels and consonants in given word ; finding places for the vowels ; ways to fill consonants 6 ! / 2 ! ; ways to put vowels 7 C5 x 5 ! ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int factorial ( int n ) { int fact = 1 ; for ( int i = 2 ; i <= n ; i ++ ) fact = fact * i ; return fact ; } int ncr ( int n , int r ) { return factorial ( n ) / ( factorial ( r ) * factorial ( n - r ) ) ; } int countWays ( string str ) { int freq [ 26 ] = { 0 } ; int nvowels = 0 , nconsonants = 0 ; int vplaces , cways , vways ; for ( int i = 0 ; i < str . length ( ) ; i ++ ) ++ freq [ str [ i ] - ' a ' ] ; for ( int i = 0 ; i < 26 ; i ++ ) { if ( i == 0 i == 4 i == 8 i == 14 i == 20 ) nvowels += freq [ i ] ; else nconsonants += freq [ i ] ; } vplaces = nconsonants + 1 ; cways = factorial ( nconsonants ) ; for ( int i = 0 ; i < 26 ; i ++ ) { if ( i != 0 && i != 4 && i != 8 && i != 14 && i != 20 && freq [ i ] > 1 ) { cways = cways / factorial ( freq [ i ] ) ; } } vways = ncr ( vplaces , nvowels ) * factorial ( nvowels ) ; for ( int i = 0 ; i < 26 ; i ++ ) { if ( i == 0 i == 4 i == 8 i == 14 i == 20 && freq [ i ] > 1 ) { vways = vways / factorial ( freq [ i ] ) ; } } return cways * vways ; } int main ( ) { string str = " permutation " ; cout << countWays ( str ) << endl ; return 0 ; } |
Pairs from an array that satisfy the given condition | C ++ implementation of the approach ; Function to return the number of set bits in n ; Function to return the count of required pairs ; Set bits for first element of the pair ; Set bits for second element of the pair ; Set bits of the resultant number which is the XOR of both the elements of the pair ; If the condition is satisfied ; Increment the count ; Return the total count ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int setBits ( int n ) { int count = 0 ; while ( n ) { n = n & ( n - 1 ) ; count ++ ; } return count ; } int countPairs ( int a [ ] , int n ) { int count = 0 ; for ( int i = 0 ; i < n - 1 ; i ++ ) { int setbits_x = setBits ( a [ i ] ) ; for ( int j = i + 1 ; j < n ; j ++ ) { int setbits_y = setBits ( a [ j ] ) ; int setbits_xor_xy = setBits ( a [ i ] ^ a [ j ] ) ; if ( setbits_x + setbits_y == setbits_xor_xy ) count ++ ; } } return count ; } int main ( ) { int a [ ] = { 2 , 3 , 4 , 5 , 6 } ; int n = sizeof ( a ) / sizeof ( a [ 0 ] ) ; cout << countPairs ( a , n ) ; } |
Number of array elements derivable from D after performing certain operations | CPP program to find the number of array elements which can be derived by perming ( + A , - A , + B , - B ) operations on D ; Function to return gcd of a and b ; Function to Return the number of elements of arr [ ] which can be derived from D by performing ( + A , - A , + B , - B ) ; find the gcd of A and B ; counter stores the number of array elements which can be derived from D ; arr [ i ] can be derived from D only if | arr [ i ] - D | is divisible by gcd of A and B ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int gcd ( int a , int b ) { if ( a == 0 ) return b ; return gcd ( b % a , a ) ; } int findPossibleDerivables ( int arr [ ] , int n , int D , int A , int B ) { int gcdAB = gcd ( A , B ) ; int counter = 0 ; for ( int i = 0 ; i < n ; i ++ ) { if ( ( abs ( arr [ i ] - D ) % gcdAB ) == 0 ) { counter ++ ; } } return counter ; } int main ( ) { int arr [ ] = { 1 , 2 , 3 , 4 , 7 , 13 } ; int n = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; int D = 5 , A = 4 , B = 2 ; cout << findPossibleDerivables ( arr , n , D , A , B ) << " STRNEWLINE " ; int a [ ] = { 1 , 2 , 3 } ; n = sizeof ( a ) / sizeof ( a [ 0 ] ) ; D = 6 , A = 3 , B = 2 ; cout << findPossibleDerivables ( a , n , D , A , B ) << " STRNEWLINE " ; return 0 ; } |
Find the sum of first N terms of the series 2 * 3 * 5 , 3 * 5 * 7 , 4 * 7 * 9 , ... | C ++ program to find sum of the first n terms of the given series ; Function to return the sum of the first n terms of the given series ; As described in the approach ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int calSum ( int n ) { return ( n * ( 2 * n * n * n + 12 * n * n + 25 * n + 21 ) ) / 2 ; } int main ( ) { int n = 3 ; cout << calSum ( n ) ; return 0 ; } |
Find elements of array using XOR of consecutive elements | C ++ program to find the array elements using XOR of consecutive elements ; Function to find the array elements using XOR of consecutive elements ; array to store the original elements ; first element a i . e elements [ 0 ] = a ; To get the next elements we have to calculate xor of previous elements with given xor of 2 consecutive elements . e . g . if a ^ b = k1 so to get b xor a both side . b = k1 ^ a ; Printing the original array elements ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; void getElements ( int a , int arr [ ] , int n ) { int elements [ n + 1 ] ; elements [ 0 ] = a ; for ( int i = 0 ; i < n ; i ++ ) { elements [ i + 1 ] = arr [ i ] ^ elements [ i ] ; } for ( int i = 0 ; i < n + 1 ; i ++ ) cout << elements [ i ] << " ▁ " ; } int main ( ) { int arr [ ] = { 13 , 2 , 6 , 1 } ; int n = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; int a = 5 ; getElements ( a , arr , n ) ; return 0 ; } |
Maximum GCD of N integers with given product | C ++ implementation of above approach ; Function to find maximum GCD of N integers with product P ; map to store prime factors of P ; prime factorization of P ; traverse all prime factors and multiply its 1 / N power to the result ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int maxGCD ( int N , int P ) { int ans = 1 ; unordered_map < int , int > prime_factors ; for ( int i = 2 ; i * i <= P ; i ++ ) { while ( P % i == 0 ) { prime_factors [ i ] ++ ; P /= i ; } } if ( P != 1 ) prime_factors [ P ] ++ ; for ( auto v : prime_factors ) ans *= pow ( v . first , v . second / N ) ; return ans ; } int main ( ) { int N = 3 , P = 24 ; cout << maxGCD ( N , P ) ; return 0 ; } |
Check if the sum of distinct digits of two integers are equal | C ++ program to check if the sum of distinct digits of two integers are equal ; Function to return the sum of distinct digits of a number ; Take last digit ; If digit has not been used before ; Set digit as used ; Remove last digit ; Function to check whether the sum of distinct digits of two numbers are equal ; Driver code | #include <iostream> NEW_LINE using namespace std ; int distinctDigitSum ( int n ) { bool used [ 10 ] ; int sum = 0 ; while ( n > 0 ) { int digit = n % 10 ; if ( ! used [ digit ] ) { used [ digit ] = true ; sum += digit ; } n = ( int ) n / 10 ; } return sum ; } string checkSum ( int m , int n ) { int sumM = distinctDigitSum ( m ) ; int sumN = distinctDigitSum ( n ) ; if ( sumM != sumN ) return " YES " ; return " NO " ; } int main ( ) { int m = 2452 , n = 9222 ; cout << ( checkSum ( m , n ) ) ; return 0 ; } |
All pairs whose xor gives unique prime | C ++ implementation of above approach ; Function that returns true if n is prime ; Corner cases ; This is checked so that we can skip middle five numbers in below loop ; Function to return the count of valid pairs ; If xor ( a [ i ] , a [ j ] ) is prime and unique ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool isPrime ( int n ) { if ( n <= 1 ) return false ; if ( n <= 3 ) return true ; if ( n % 2 == 0 n % 3 == 0 ) return false ; for ( int i = 5 ; i * i <= n ; i = i + 6 ) if ( n % i == 0 || n % ( i + 2 ) == 0 ) return false ; return true ; } int countPairs ( int a [ ] , int n ) { int count = 0 ; unordered_map < int , int > m ; for ( int i = 0 ; i < n - 1 ; i ++ ) { for ( int j = i + 1 ; j < n ; j ++ ) { if ( isPrime ( a [ i ] ^ a [ j ] ) && m [ a [ i ] ^ a [ j ] ] == 0 ) { m [ ( a [ i ] ^ a [ j ] ) ] ++ ; count ++ ; } } } return count ; } int main ( ) { int a [ ] = { 10 , 12 , 23 , 45 , 5 , 6 } ; int n = sizeof ( a ) / sizeof ( a [ 0 ] ) ; cout << countPairs ( a , n ) ; } |
Sum of element whose prime factors are present in array | C ++ program to find the sum of the elements of an array whose prime factors are present in the same array ; Stores smallest prime factor for every number ; Function to calculate SPF ( Smallest Prime Factor ) for every number till MAXN ; Marking smallest prime factor for every number to be itself . ; Separately marking spf for every even number as 2 ; If i is prime ; Marking SPF for all numbers divisible by i ; Marking spf [ j ] if it is not previously marked ; Function to return the sum of the elements of an array whose prime factors are present in the same array ; Function call to calculate smallest prime factors of all the numbers upto MAXN ; Create map for each element ; If smallest prime factor of num is present in array ; Each factor of arr [ i ] is present in the array ; Driver program ; Function call to print required answer | #include <bits/stdc++.h> NEW_LINE using namespace std ; #define MAXN 1000001 NEW_LINE int spf [ MAXN ] ; void sieve ( ) { spf [ 1 ] = 1 ; for ( int i = 2 ; i < MAXN ; i ++ ) spf [ i ] = i ; for ( int i = 4 ; i < MAXN ; i += 2 ) spf [ i ] = 2 ; for ( int i = 3 ; i * i < MAXN ; i ++ ) { if ( spf [ i ] == i ) { for ( int j = i * i ; j < MAXN ; j += i ) if ( spf [ j ] == j ) spf [ j ] = i ; } } } int sumFactors ( int arr [ ] , int n ) { sieve ( ) ; std :: map < int , int > map ; for ( int i = 0 ; i < n ; ++ i ) map [ arr [ i ] ] = 1 ; int sum = 0 ; for ( int i = 0 ; i < n ; ++ i ) { int num = arr [ i ] ; while ( num != 1 && map [ spf [ num ] ] == 1 ) { num /= spf [ num ] ; } if ( num == 1 ) sum += arr [ i ] ; } return sum ; } int main ( ) { int arr [ ] = { 5 , 11 , 55 , 25 , 100 } ; int n = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; cout << sumFactors ( arr , n ) ; return 0 ; } |
Find nth Hermite number | C ++ program to find nth Hermite number ; Utility function to calculate double factorial of a number ; Function to return nth Hermite number ; If n is even then return 0 ; If n is odd ; Calculate double factorial of ( n - 1 ) and multiply it with 2 ^ ( n / 2 ) ; If n / 2 is odd then nth Hermite number will be negative ; Return nth Hermite number ; Driver Code ; Print nth Hermite number | #include <bits/stdc++.h> NEW_LINE using namespace std ; int doubleFactorial ( int n ) { int fact = 1 ; for ( int i = 1 ; i <= n ; i = i + 2 ) { fact = fact * i ; } return fact ; } int hermiteNumber ( int n ) { if ( n % 2 == 1 ) return 0 ; else { int number = ( pow ( 2 , n / 2 ) ) * doubleFactorial ( n - 1 ) ; if ( ( n / 2 ) % 2 == 1 ) number = number * -1 ; return number ; } } int main ( ) { int n = 6 ; cout << hermiteNumber ( n ) ; return 0 ; } |
Count numbers in a range having GCD of powers of prime factors equal to 1 | C ++ implementation of the approach ; Vector to store powers greater than 3 ; Set to store perfect squares ; Set to store powers other than perfect squares ; Pushing squares ; if the values is already a perfect square means present in the set ; Run loop until some power of current number doesn 't exceed MAX ; Pushing only odd powers as even power of a number can always be expressed as a perfect square which is already present in set squares ; Inserting those sorted values of set into a vector ; Precompute the powers ; Calculate perfect squares in range using sqrtl function ; Calculate upper value of R in vector using binary search ; Calculate lower value of L in vector using binary search ; Calculate perfect powers ; Compute final answer ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; #define N 1000005 NEW_LINE #define MAX 1e18 NEW_LINE vector < long int > powers ; set < long int > squares ; set < long int > s ; void powersPrecomputation ( ) { for ( long int i = 2 ; i < N ; i ++ ) { squares . insert ( i * i ) ; if ( squares . find ( i ) != squares . end ( ) ) continue ; long int temp = i ; while ( i * i <= MAX / temp ) { temp *= ( i * i ) ; s . insert ( temp ) ; } } for ( auto x : s ) powers . push_back ( x ) ; } long int calculateAnswer ( long int L , long int R ) { powersPrecomputation ( ) ; long int perfectSquares = floor ( sqrtl ( R ) ) - floor ( sqrtl ( L - 1 ) ) ; long int high = upper_bound ( powers . begin ( ) , powers . end ( ) , R ) - powers . begin ( ) ; long int low = lower_bound ( powers . begin ( ) , powers . end ( ) , L ) - powers . begin ( ) ; long perfectPowers = perfectSquares + ( high - low ) ; long ans = ( R - L + 1 ) - perfectPowers ; return ans ; } int main ( ) { long int L = 13 , R = 20 ; cout << calculateAnswer ( L , R ) ; return 0 ; } |
Sum of integers upto N with given unit digit | C ++ implementation of the approach ; Function to return the required sum ; If the unit digit is d ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; #define ll long long int NEW_LINE ll getSum ( int n , int d ) { ll sum = 0 ; for ( int i = 1 ; i <= n ; i ++ ) { if ( i % 10 == d ) sum += i ; } return sum ; } int main ( ) { int n = 30 , d = 3 ; cout << getSum ( n , d ) ; return 0 ; } |
Split a number into 3 parts such that none of the parts is divisible by 3 | CPP program to split a number into three parts such than none of them is divisible by 3. ; Print x = 1 , y = 1 and z = N - 2 ; Otherwise , print x = 1 , y = 2 and z = N - 3 ; Driver code | #include <iostream> NEW_LINE using namespace std ; void printThreeParts ( int N ) { if ( N % 3 == 0 ) cout << " ▁ x ▁ = ▁ 1 , ▁ y ▁ = ▁ 1 , ▁ z ▁ = ▁ " << N - 2 << endl ; else cout << " ▁ x ▁ = ▁ 1 , ▁ y ▁ = ▁ 2 , ▁ z ▁ = ▁ " << N - 3 << endl ; } int main ( ) { int N = 10 ; printThreeParts ( N ) ; return 0 ; } |
Minimum absolute difference of a number and its closest prime | C ++ program to find the minimum absolute difference between a number and its closest prime ; Function to check if a number is prime or not ; Function to find the minimum absolute difference between a number and its closest prime ; Variables to store first prime above and below N ; Finding first prime number greater than N ; Finding first prime number less than N ; Variables to store the differences ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool isPrime ( int N ) { for ( int i = 2 ; i <= sqrt ( N ) ; i ++ ) { if ( N % i == 0 ) return false ; } return true ; } int getDifference ( int N ) { if ( N == 0 ) return 2 ; else if ( N == 1 ) return 1 ; else if ( isPrime ( N ) ) return 0 ; int aboveN = -1 , belowN = -1 ; int n1 ; n1 = N + 1 ; while ( true ) { if ( isPrime ( n1 ) ) { aboveN = n1 ; break ; } else n1 ++ ; } n1 = N - 1 ; while ( true ) { if ( isPrime ( n1 ) ) { belowN = n1 ; break ; } else n1 -- ; } int diff1 = aboveN - N ; int diff2 = N - belowN ; return min ( diff1 , diff2 ) ; } int main ( ) { int N = 25 ; cout << getDifference ( N ) << endl ; return 0 ; } |
Check if the sum of perfect squares in an array is divisible by x | C ++ implementation of the approach ; Function that returns true if the sum of all the perfect squares of the given array are divisible by x ; If arr [ i ] is a perfect square ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool check ( int arr [ ] , int x , int n ) { long long sum = 0 ; for ( int i = 0 ; i < n ; i ++ ) { double x = sqrt ( arr [ i ] ) ; if ( floor ( x ) == ceil ( x ) ) { sum += arr [ i ] ; } } if ( sum % x == 0 ) return true ; else return false ; } int main ( ) { int arr [ ] = { 2 , 3 , 4 , 9 , 10 } ; int n = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; int x = 13 ; if ( check ( arr , x , n ) ) { cout << " Yes " ; } else { cout << " No " ; } return 0 ; } |
Largest Divisor of a Number not divisible by a perfect square | C ++ Program to find the largest divisor not divisible by any perfect square greater than 1 ; Function to find the largest divisor not divisible by any perfect square greater than 1 ; set to store divisors in descending order ; If the number is divisible by i , then insert it ; Vector to store perfect squares ; Check for each divisor , if it is not divisible by any perfect square , simply return it as the answer . ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; const int MAX = 1e5 ; int findLargestDivisor ( int n ) { int m = n ; set < int , greater < int > > s ; s . insert ( 1 ) ; s . insert ( n ) ; for ( int i = 2 ; i < sqrt ( n ) + 1 ; i ++ ) { if ( n % i == 0 ) { s . insert ( n / i ) ; s . insert ( i ) ; while ( m % i == 0 ) m /= i ; } } if ( m > 1 ) s . insert ( m ) ; vector < int > vec ; for ( int i = 2 ; i <= MAX ; i ++ ) vec . push_back ( i * i ) ; for ( auto d : s ) { int divi = 0 ; for ( int j = 0 ; j < vec . size ( ) && vec [ j ] <= d ; j ++ ) { if ( d % vec [ j ] == 0 ) { divi = 1 ; break ; } } if ( ! divi ) return d ; } } int main ( ) { int n = 12 ; cout << findLargestDivisor ( n ) << endl ; n = 97 ; cout << findLargestDivisor ( n ) << endl ; return 0 ; } |
Find the Number of Maximum Product Quadruples | CPP program to find the number of Quadruples having maximum product ; Returns the number of ways to select r objects out of available n choices ; ncr = ( n * ( n - 1 ) * ( n - 2 ) * ... . . ... ( n - r + 1 ) ) / ( r * ( r - 1 ) * ... * 1 ) ; Returns the number of quadruples having maximum product ; stores the frequency of each element ; remaining_choices denotes the remaining elements to select inorder to form quadruple ; traverse the elements of the map in reverse order ; If Frequency of element < remaining choices , select all of these elements , else select only the number of elements required ; Decrement remaining_choices acc to the number of the current elements selected ; if the quadruple is formed stop the algorithm ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int NCR ( int n , int r ) { int numerator = 1 ; int denominator = 1 ; while ( r > 0 ) { numerator *= n ; denominator *= r ; n -- ; r -- ; } return ( numerator / denominator ) ; } int findWays ( int arr [ ] , int n ) { map < int , int > count ; if ( n < 4 ) return 0 ; for ( int i = 0 ; i < n ; i ++ ) { count [ arr [ i ] ] ++ ; } int remaining_choices = 4 ; int ans = 1 ; for ( auto iter = count . rbegin ( ) ; iter != count . rend ( ) ; ++ iter ) { int number = iter -> first ; int frequency = iter -> second ; int toSelect = min ( remaining_choices , frequency ) ; ans = ans * NCR ( frequency , toSelect ) ; remaining_choices -= toSelect ; if ( ! remaining_choices ) { break ; } } return ans ; } int main ( ) { int arr [ ] = { 1 , 2 , 3 , 3 , 3 , 5 } ; int n = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; int maxQuadrupleWays = findWays ( arr , n ) ; cout << maxQuadrupleWays ; return 0 ; } |
Minimum and Maximum number of pairs in m teams of n people | CPP program to find minimum and maximum no . of pairs ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; void MinimumMaximumPairs ( int n , int m ) { int max_pairs = ( ( n - m + 1 ) * ( n - m ) ) / 2 ; int min_pairs = m * ( ( ( n - m ) / m + 1 ) * ( ( n - m ) / m ) ) / 2 + ceil ( ( n - m ) / double ( m ) ) * ( ( n - m ) % m ) ; cout << " Minimum ▁ no . ▁ of ▁ pairs ▁ = ▁ " << min_pairs << " STRNEWLINE " ; cout << " Maximum ▁ no . ▁ of ▁ pairs ▁ = ▁ " << max_pairs << " STRNEWLINE " ; } int main ( ) { int n = 5 , m = 2 ; MinimumMaximumPairs ( n , m ) ; return 0 ; } |
Larger of a ^ b or b ^ a ( a raised to power b or b raised to power a ) | C ++ code for finding greater between the a ^ b and b ^ a ; Function to find the greater value ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; void findGreater ( int a , int b ) { long double x = ( long double ) a * ( long double ) ( log ( ( long double ) ( b ) ) ) ; long double y = ( long double ) b * ( long double ) ( log ( ( long double ) ( a ) ) ) ; if ( y > x ) { cout << " a ^ b ▁ is ▁ greater " << endl ; } else if ( y < x ) { cout << " b ^ a ▁ is ▁ greater " << endl ; } else { cout << " Both ▁ are ▁ equal " << endl ; } } int main ( ) { int a = 3 , b = 5 , c = 2 , d = 4 ; findGreater ( a , b ) ; findGreater ( c , d ) ; return 0 ; } |
Expressing a fraction as a natural number under modulo ' m ' | C ++ implementation of the approach ; Function to return the GCD of given numbers ; Recursive function to return ( x ^ n ) % m ; Function to return the fraction modulo mod ; ( b ^ m - 2 ) % m ; Final answer ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; #define ll long long int NEW_LINE #define m 1000000007 NEW_LINE int gcd ( int a , int b ) { if ( a == 0 ) return b ; return gcd ( b % a , a ) ; } ll modexp ( ll x , ll n ) { if ( n == 0 ) { return 1 ; } else if ( n % 2 == 0 ) { return modexp ( ( x * x ) % m , n / 2 ) ; } else { return ( x * modexp ( ( x * x ) % m , ( n - 1 ) / 2 ) % m ) ; } } ll getFractionModulo ( ll a , ll b ) { ll c = gcd ( a , b ) ; a = a / c ; b = b / c ; ll d = modexp ( b , m - 2 ) ; ll ans = ( ( a % m ) * ( d % m ) ) % m ; return ans ; } int main ( ) { ll a = 2 , b = 6 ; cout << getFractionModulo ( a , b ) << endl ; return 0 ; } |
Find sum of a number and its maximum prime factor | C ++ program to find sum of n and it 's largest prime factor ; Function to return the sum of n and it 's largest prime factor ; Initialise maxPrime to - 1. ; n must be odd at this point , thus skip the even numbers and iterate only odd numbers ; This condition is to handle the case when n is a prime number greater than 2 ; finally return the sum . ; Driver Program to check the above function . | #include <cmath> NEW_LINE #include <iostream> NEW_LINE using namespace std ; int maxPrimeFactors ( int n ) { int num = n ; int maxPrime = -1 ; while ( n % 2 == 0 ) { maxPrime = 2 ; n /= 2 ; } for ( int i = 3 ; i <= sqrt ( n ) ; i += 2 ) { while ( n % i == 0 ) { maxPrime = i ; n = n / i ; } } if ( n > 2 ) maxPrime = n ; int sum = maxPrime + num ; return sum ; } int main ( ) { int n = 19 ; cout << maxPrimeFactors ( n ) ; return 0 ; } |
Largest number less than N with digit sum greater than the digit sum of N | C ++ implementation of the approach ; Function to return the sum of the digits of n ; Loop for each digit of the number ; Function to return the greatest number less than n such that the sum of its digits is greater than the sum of the digits of n ; Starting from n - 1 ; Check until 1 ; If i satisfies the given condition ; If the condition is not satisfied ; Driver code | #include <iostream> NEW_LINE using namespace std ; int sumOfDigits ( int n ) { int res = 0 ; while ( n > 0 ) { res += n % 10 ; n /= 10 ; } return res ; } int findNumber ( int n ) { int i = n - 1 ; while ( i > 0 ) { if ( sumOfDigits ( i ) > sumOfDigits ( n ) ) return i ; i -- ; } return -1 ; } int main ( ) { int n = 824 ; cout << findNumber ( n ) ; return 0 ; } |
Find the Nth term of the series 14 , 28 , 20 , 40 , ... . . | CPP program to find the Nth term of the series 14 , 28 , 20 , 40 , ... . . ; Function to find the N - th term ; initializing the 1 st number ; loop from 2 nd term to nth term ; if i is even , double the previous number ; if i is odd , subtract 8 from previous number ; Driver Code | #include <iostream> NEW_LINE using namespace std ; int findNth ( int N ) { int b = 14 ; int i ; for ( i = 2 ; i <= N ; i ++ ) { if ( i % 2 == 0 ) b = b * 2 ; else b = b - 8 ; } return b ; } int main ( ) { int N = 6 ; cout << findNth ( N ) ; return 0 ; } |
Problem of 8 Neighbours of an element in a 2 | C ++ implementation of the approach ; Dimension of Array ; Count of 1 s ; Counting all neighbouring 1 s ; Comparing the number of neighbouring 1 s with given ranges ; Copying changes to the main matrix ; Driver code ; Function call to calculate the resultant matrix after ' K ' iterations . ; Printing Result | #include <iostream> NEW_LINE using namespace std ; #define N 4 NEW_LINE void predictMatrix ( int arr [ N ] [ N ] , int range1a , int range1b , int range0a , int range0b , int K , int b [ N ] [ N ] ) { int c = 0 ; while ( K -- ) { for ( int i = 0 ; i < N ; i ++ ) { for ( int j = 0 ; j < N ; j ++ ) { c = 0 ; if ( i > 0 && arr [ i - 1 ] [ j ] == 1 ) c ++ ; if ( j > 0 && arr [ i ] [ j - 1 ] == 1 ) c ++ ; if ( i > 0 && j > 0 && arr [ i - 1 ] [ j - 1 ] == 1 ) c ++ ; if ( i < N - 1 && arr [ i + 1 ] [ j ] == 1 ) c ++ ; if ( j < N - 1 && arr [ i ] [ j + 1 ] == 1 ) c ++ ; if ( i < N - 1 && j < N - 1 && arr [ i + 1 ] [ j + 1 ] == 1 ) c ++ ; if ( i < N - 1 && j > 0 && arr [ i + 1 ] [ j - 1 ] == 1 ) c ++ ; if ( i > 0 && j < N - 1 && arr [ i - 1 ] [ j + 1 ] == 1 ) c ++ ; if ( arr [ i ] [ j ] == 1 ) { if ( c >= range1a && c <= range1b ) b [ i ] [ j ] = 1 ; else b [ i ] [ j ] = 0 ; } if ( arr [ i ] [ j ] == 0 ) { if ( c >= range0a && c <= range0b ) b [ i ] [ j ] = 1 ; else b [ i ] [ j ] = 0 ; } } } for ( int k = 0 ; k < N ; k ++ ) for ( int m = 0 ; m < N ; m ++ ) arr [ k ] [ m ] = b [ k ] [ m ] ; } } int main ( ) { int arr [ N ] [ N ] = { 0 , 0 , 0 , 0 , 0 , 1 , 1 , 0 , 0 , 0 , 1 , 0 , 0 , 1 , 0 , 1 } ; int range1a = 2 , range1b = 2 ; int range0a = 2 , range0b = 3 ; int K = 3 , b [ N ] [ N ] = { 0 } ; predictMatrix ( arr , range1a , range1b , range0a , range0b , K , b ) ; for ( int i = 0 ; i < N ; i ++ ) { cout << endl ; for ( int j = 0 ; j < N ; j ++ ) cout << b [ i ] [ j ] << " ▁ " ; } return 0 ; } |
Number of moves required to guess a permutation . | C ++ implementation of the approach ; Function that returns the required moves ; Final move ; Driver Program to test above function | #include <bits/stdc++.h> NEW_LINE using namespace std ; int countMoves ( int n ) { int ct = 0 ; for ( int i = 1 ; i <= n ; i ++ ) ct += i * ( n - i ) ; ct += n ; return ct ; } int main ( ) { int n = 3 ; cout << countMoves ( n ) ; return 0 ; } |
Count number of triplets ( a , b , c ) such that a ^ 2 + b ^ 2 = c ^ 2 and 1 <= a <= b <= c <= n | C ++ implementation of the approach ; Function to return an Array containing all the perfect squares upto n ; While current perfect square is less than or equal to n ; Function to return the count of triplet ( a , b , c ) pairs such that a ^ 2 + b ^ 2 = c ^ 2 and 1 <= a <= b <= c <= n ; Vector of perfect squares upto n ^ 2 ; Since , a ^ 2 + b ^ 2 = c ^ 2 ; If c < a or bSquare is not a perfect square ; If triplet pair ( a , b , c ) satisfy the given condition ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; vector < int > getPerfectSquares ( int n ) { vector < int > perfectSquares ; int current = 1 , i = 1 ; while ( current <= n ) { perfectSquares . push_back ( current ) ; current = pow ( ++ i , 2 ) ; } return perfectSquares ; } int countTriplets ( int n ) { vector < int > perfectSquares = getPerfectSquares ( pow ( n , 2 ) ) ; int count = 0 ; for ( int a = 1 ; a <= n ; a ++ ) { int aSquare = pow ( a , 2 ) ; for ( int i = 0 ; i < perfectSquares . size ( ) ; i ++ ) { int cSquare = perfectSquares [ i ] ; int bSquare = abs ( cSquare - aSquare ) ; int b = sqrt ( bSquare ) ; int c = sqrt ( cSquare ) ; if ( c < a || ( find ( perfectSquares . begin ( ) , perfectSquares . end ( ) , bSquare ) == perfectSquares . end ( ) ) ) continue ; if ( ( b >= a ) && ( b <= c ) && ( aSquare + bSquare == cSquare ) ) count ++ ; } } return count ; } int main ( ) { int n = 10 ; cout << countTriplets ( n ) ; return 0 ; } |
Count Numbers with N digits which consists of even number of 0 ’ s | C ++ program to count numbers with N digits which consists of odd number of 0 's ; Function to count Numbers with N digits which consists of odd number of 0 's ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int countNumbers ( int N ) { return ( pow ( 10 , N ) - 1 ) - ( pow ( 10 , N ) - pow ( 8 , N ) ) / 2 ; } int main ( ) { int n = 2 ; cout << countNumbers ( n ) << endl ; return 0 ; } |
Find determinant of matrix generated by array rotation | C ++ program for finding determinant of generated matrix ; Function to calculate determinant ; Driver code | #include <bits/stdc++.h> NEW_LINE #define N 3 NEW_LINE using namespace std ; int calcDeterminant ( int arr [ ] ) { int determinant = 0 ; for ( int i = 0 ; i < N ; i ++ ) { determinant += pow ( arr [ i ] , 3 ) ; } determinant -= 3 * arr [ 0 ] * arr [ 1 ] * arr [ 2 ] ; return determinant ; } int main ( ) { int arr [ ] = { 4 , 5 , 3 } ; cout << calcDeterminant ( arr ) ; return 0 ; } |
Minimum elements to be added in a range so that count of elements is divisible by K | C ++ implementation of the approach ; Total elements in the range ; If total elements are already divisible by k ; Value that must be added to count in order to make it divisible by k ; Driver Program to test above function | #include <bits/stdc++.h> NEW_LINE using namespace std ; int minimumMoves ( int k , int l , int r ) { int count = r - l + 1 ; if ( count % k == 0 ) return 0 ; return ( k - ( count % k ) ) ; } int main ( ) { int k = 3 , l = 10 , r = 10 ; cout << minimumMoves ( k , l , r ) ; return 0 ; } |
Sum of all even numbers in range L and R | C ++ program to print the sum of all even numbers in range L and R ; Function to return the sum of all natural numbers ; Function to return sum of even numbers in range L and R ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int sumNatural ( int n ) { int sum = ( n * ( n + 1 ) ) ; return sum ; } int sumEven ( int l , int r ) { return sumNatural ( r / 2 ) - sumNatural ( ( l - 1 ) / 2 ) ; } int main ( ) { int l = 2 , r = 5 ; cout << " Sum ▁ of ▁ Natural ▁ numbers ▁ from ▁ L ▁ to ▁ R ▁ is ▁ " << sumEven ( l , r ) ; return 0 ; } |
Check if N is divisible by a number which is composed of the digits from the set { A , B } | C ++ program to find if number N is divisible by a number that contains only a and b as it 's digits ; Function to check whether n is divisible by a number whose digits are either a or b ; base condition ; recursive call ; Check for all numbers beginning with ' a ' or ' b ' ; Driver program | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool isDivisibleRec ( int x , int a , int b , int n ) { if ( x > n ) return false ; if ( n % x == 0 ) return true ; return ( isDivisibleRec ( x * 10 + a , a , b , n ) || isDivisibleRec ( x * 10 + b , a , b , n ) ) ; } bool isDivisible ( int a , int b , int n ) { return isDivisibleRec ( a , a , b , n ) || isDivisibleRec ( b , a , b , n ) ; } int main ( ) { int a = 3 , b = 5 , n = 53 ; if ( isDivisible ( a , b , n ) ) cout << " Yes " ; else cout << " No " ; return 0 ; } |
Minimum number of moves required to reach the destination by the king in a chess board | C ++ program to Find the minimum number of moves required to reach the destination by the king in a chess board ; function to Find the minimum number of moves required to reach the destination by the king in a chess board ; minimum number of steps ; while the king is not in the same row or column as the destination ; Go up ; Go down ; Go left ; Go right ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; void MinSteps ( int SourceX , int SourceY , int DestX , int DestY ) { cout << max ( abs ( SourceX - DestX ) , abs ( SourceY - DestY ) ) << endl ; while ( ( SourceX != DestX ) || ( SourceY != DestY ) ) { if ( SourceX < DestX ) { cout << ' U ' ; SourceX ++ ; } if ( SourceX > DestX ) { cout << ' D ' ; SourceX -- ; } if ( SourceY > DestY ) { cout << ' L ' ; SourceY -- ; } if ( SourceY < DestY ) { cout << ' R ' ; SourceY ++ ; } cout << endl ; } } int main ( ) { int sourceX = 4 , sourceY = 4 ; int destinationX = 7 , destinationY = 0 ; MinSteps ( sourceX , sourceY , destinationX , destinationY ) ; return 0 ; } |
Count of pairs in an array whose sum is a perfect square | CPP implementation of the approach ; Function to return an ArrayList containing all the perfect squares upto n ; while current perfect square is less than or equal to n ; Function to print the sum of maximum two elements from the array ; Function to return the count of numbers that when added with n give a perfect square ; temp > n is checked so that pairs ( x , y ) and ( y , x ) don 't get counted twice ; Function to count the pairs whose sum is a perfect square ; Sum of the maximum two elements from the array ; List of perfect squares upto max ; Contains all the array elements ; Add count of the elements that when added with arr [ i ] give a perfect square ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; vector < int > getPerfectSquares ( int n ) { vector < int > perfectSquares ; int current = 1 , i = 1 ; while ( current <= n ) { perfectSquares . push_back ( current ) ; current = static_cast < int > ( pow ( ++ i , 2 ) ) ; } return perfectSquares ; } int maxPairSum ( vector < int > & arr ) { int n = arr . size ( ) ; int max , secondMax ; if ( arr [ 0 ] > arr [ 1 ] ) { max = arr [ 0 ] ; secondMax = arr [ 1 ] ; } else { max = arr [ 1 ] ; secondMax = arr [ 0 ] ; } for ( int i = 2 ; i < n ; i ++ ) { if ( arr [ i ] > max ) { secondMax = max ; max = arr [ i ] ; } else if ( arr [ i ] > secondMax ) { secondMax = arr [ i ] ; } } return ( max + secondMax ) ; } int countPairsWith ( int n , vector < int > & perfectSquares , unordered_set < int > & nums ) { int count = 0 ; for ( int i = 0 ; i < perfectSquares . size ( ) ; i ++ ) { int temp = perfectSquares [ i ] - n ; if ( temp > n && find ( nums . begin ( ) , nums . end ( ) , temp ) != nums . end ( ) ) { count ++ ; } } return count ; } int countPairs ( vector < int > & arr ) { int i , n = arr . size ( ) ; int max = maxPairSum ( arr ) ; vector < int > perfectSquares = getPerfectSquares ( max ) ; unordered_set < int > nums ; for ( i = 0 ; i < n ; i ++ ) { nums . insert ( arr [ i ] ) ; } int count = 0 ; for ( i = 0 ; i < n ; i ++ ) { count += countPairsWith ( arr [ i ] , perfectSquares , nums ) ; } return count ; } int main ( ) { vector < int > arr = { 2 , 3 , 6 , 9 , 10 , 20 } ; cout << countPairs ( arr ) << endl ; return 0 ; } |
Element equal to the sum of all the remaining elements | C ++ implementation of the above approach ; Function to find the element ; sum is use to store sum of all elements of array ; iterate over all elements ; Driver code | #include <bits/stdc++.h> NEW_LINE #define ll long long int NEW_LINE using namespace std ; int findEle ( int arr [ ] , int n ) { ll sum = 0 ; for ( int i = 0 ; i < n ; i ++ ) sum += arr [ i ] ; for ( int i = 0 ; i < n ; i ++ ) if ( arr [ i ] == sum - arr [ i ] ) return arr [ i ] ; return -1 ; } int main ( ) { int arr [ ] = { 1 , 2 , 3 , 6 } ; int n = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; cout << findEle ( arr , n ) ; return 0 ; } |
Sum of all natural numbers in range L to R | C ++ program to print the sum of all numbers in range L and R ; Function to return the sum of all natural numbers ; Function to return the sum of all numbers in range L and R ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int sumNatural ( int n ) { int sum = ( n * ( n + 1 ) ) / 2 ; return sum ; } int suminRange ( int l , int r ) { return sumNatural ( r ) - sumNatural ( l - 1 ) ; } int main ( ) { int l = 2 , r = 5 ; cout << " Sum ▁ of ▁ Natural ▁ numbers ▁ from ▁ L ▁ to ▁ R ▁ is ▁ " << suminRange ( l , r ) ; return 0 ; } |
Check if a large number is divisible by 75 or not | C ++ implementation to check the number is divisible by 75 or not ; check divisibleBy3 ; to store sum of Digit ; traversing through each digit ; summing up Digit ; check if sumOfDigit is divisibleBy3 ; check divisibleBy25 ; if a single digit number ; length of the number ; taking the last two digit ; checking if the lastTwo digit is divisibleBy25 ; Function to check divisibleBy75 or not ; check if divisibleBy3 and divisibleBy25 ; Drivers code ; divisible ; if divisibleBy75 | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool divisibleBy3 ( string number ) { int sumOfDigit = 0 ; for ( int i = 0 ; i < number . length ( ) ; i ++ ) sumOfDigit += number [ i ] - '0' ; if ( sumOfDigit % 3 == 0 ) return true ; return false ; } bool divisibleBy25 ( string number ) { if ( number . length ( ) < 2 ) return false ; int length = number . length ( ) ; int lastTwo = ( number [ length - 2 ] - '0' ) * 10 + ( number [ length - 1 ] - '0' ) ; if ( lastTwo % 25 == 0 ) return true ; return false ; } bool divisibleBy75 ( string number ) { if ( divisibleBy3 ( number ) && divisibleBy25 ( number ) ) return true ; return false ; } int main ( ) { string number = "754586672150" ; bool divisible = divisibleBy75 ( number ) ; if ( divisible ) cout << " Yes " ; else cout << " No " ; return 0 ; } |
Find the other number when LCM and HCF given | CPP program to find other number from given HCF and LCM ; Function that will calculates the zeroes at the end ; Driver code ; Calling function . | #include <bits/stdc++.h> NEW_LINE using namespace std ; int otherNumber ( int A , int Lcm , int Hcf ) { return ( Lcm * Hcf ) / A ; } int main ( ) { int A = 8 , Lcm = 8 , Hcf = 1 ; int result = otherNumber ( A , Lcm , Hcf ) ; cout << " B ▁ = ▁ " << result ; return 0 ; } |
Overall percentage change from successive changes | C ++ implementation of above approach ; Calculate successive change of 1 st 2 change ; Calculate successive change for rest of the value ; Driver code ; Calling function | #include <bits/stdc++.h> NEW_LINE using namespace std ; float successiveChange ( int arr [ ] , int N ) { float var1 , var2 , result = 0 ; var1 = arr [ 0 ] ; var2 = arr [ 1 ] ; result = var1 + var2 + ( float ( var1 * var2 ) / 100 ) ; for ( int i = 2 ; i < N ; i ++ ) result = result + arr [ i ] + ( float ( result * arr [ i ] ) / 100 ) ; return result ; } int main ( ) { int arr [ ] = { 10 , 20 , 30 , 10 } ; int N = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; float result = successiveChange ( arr , N ) ; cout << " Percentage ▁ change ▁ is ▁ = ▁ " << result << " ▁ % " ; return 0 ; } |
Minimum numbers ( smaller than or equal to N ) with sum S | C ++ program to find the minimum numbers required to get to S ; Function to find the minimum numbers required to get to S ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int minimumNumbers ( int n , int s ) { if ( s % n ) return s / n + 1 ; else return s / n ; } int main ( ) { int n = 5 ; int s = 11 ; cout << minimumNumbers ( n , s ) ; return 0 ; } |
Sum of multiples of A and B less than N | CPP program to find the sum of all multiples of A and B below N ; Function to find sum of AP series ; Number of terms ; Function to find the sum of all multiples of A and B below N ; Since , we need the sum of multiples less than N ; common factors of A and B ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; #define ll long long int NEW_LINE ll sumAP ( ll n , ll d ) { n /= d ; return ( n ) * ( 1 + n ) * d / 2 ; } ll sumMultiples ( ll A , ll B , ll n ) { n -- ; ll common = ( A * B ) / __gcd ( A , B ) ; return sumAP ( n , A ) + sumAP ( n , B ) - sumAP ( n , common ) ; } int main ( ) { ll n = 100 , A = 5 , B = 10 ; cout << " Sum ▁ = ▁ " << sumMultiples ( A , B , n ) ; return 0 ; } |
Check if a prime number can be expressed as sum of two Prime Numbers | C ++ program to check if a prime number can be expressed as sum of two Prime Numbers ; Function to check whether a number is prime or not ; Function to check if a prime number can be expressed as sum of two Prime Numbers ; if the number is prime , and number - 2 is also prime ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool isPrime ( int n ) { if ( n <= 1 ) return false ; for ( int i = 2 ; i <= sqrt ( n ) ; i ++ ) { if ( n % i == 0 ) return false ; } return true ; } bool isPossible ( int N ) { if ( isPrime ( N ) && isPrime ( N - 2 ) ) return true ; else return false ; } int main ( ) { int n = 13 ; if ( isPossible ( n ) ) cout << " Yes " ; else cout << " No " ; return 0 ; } |
Count pairs from two arrays whose modulo operation yields K | C ++ implementation of above approach ; Function to return the total pairs of elements whose modulo yield K ; set is used to avoid duplicate pairs ; check which element is greater and proceed according to it ; check if modulo is equal to K ; return size of the set ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int totalPairs ( int arr1 [ ] , int arr2 [ ] , int K , int n , int m ) { set < pair < int , int > > s ; for ( int i = 0 ; i < n ; i ++ ) { for ( int j = 0 ; j < m ; j ++ ) { if ( arr1 [ i ] > arr2 [ j ] ) { if ( arr1 [ i ] % arr2 [ j ] == K ) s . insert ( make_pair ( arr1 [ i ] , arr2 [ j ] ) ) ; } else { if ( arr2 [ j ] % arr1 [ i ] == K ) s . insert ( make_pair ( arr2 [ j ] , arr1 [ i ] ) ) ; } } } return s . size ( ) ; } int main ( ) { int arr1 [ ] = { 8 , 3 , 7 , 50 } ; int arr2 [ ] = { 5 , 1 , 10 , 4 } ; int K = 3 ; int n = sizeof ( arr1 ) / sizeof ( arr1 [ 0 ] ) ; int m = sizeof ( arr2 ) / sizeof ( arr2 [ 0 ] ) ; cout << totalPairs ( arr1 , arr2 , K , n , m ) ; return 0 ; } |
Largest sub | C ++ program to find the length of the largest sub - array of an array every element of whose is a perfect square ; function to return the length of the largest sub - array of an array every element of whose is a perfect square ; if both a and b are equal then arr [ i ] is a perfect square ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int contiguousPerfectSquare ( int arr [ ] , int n ) { int a ; float b ; int current_length = 0 ; int max_length = 0 ; for ( int i = 0 ; i < n ; i ++ ) { b = sqrt ( arr [ i ] ) ; a = b ; if ( a == b ) current_length ++ ; else current_length = 0 ; max_length = max ( max_length , current_length ) ; } return max_length ; } int main ( ) { int arr [ ] = { 9 , 75 , 4 , 64 , 121 , 25 } ; int n = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; cout << contiguousPerfectSquare ( arr , n ) ; return 0 ; } |
Count pairs of numbers from 1 to N with Product divisible by their Sum | C ++ program to count pairs of numbers from 1 to N with Product divisible by their Sum ; Function to count pairs ; variable to store count ; Generate all possible pairs such that 1 <= x < y < n ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int countPairs ( int n ) { int count = 0 ; for ( int x = 1 ; x < n ; x ++ ) { for ( int y = x + 1 ; y <= n ; y ++ ) { if ( ( y * x ) % ( y + x ) == 0 ) count ++ ; } } return count ; } int main ( ) { int n = 15 ; cout << countPairs ( n ) ; return 0 ; } |
Find the index of the left pointer after possible moves in the array | C ++ program to find the index of the left pointer ; Function that returns the index of the left pointer ; there 's only one element in the array ; initially both are at end ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int getIndex ( int a [ ] , int n ) { if ( n == 1 ) return 0 ; int ptrL = 0 , ptrR = n - 1 , sumL = a [ 0 ] , sumR = a [ n - 1 ] ; while ( ptrR - ptrL > 1 ) { if ( sumL < sumR ) { ptrL ++ ; sumL += a [ ptrL ] ; } else if ( sumL > sumR ) { ptrR -- ; sumR += a [ ptrR ] ; } else { break ; } } return ptrL ; } int main ( ) { int a [ ] = { 2 , 7 , 9 , 8 , 7 } ; int n = sizeof ( a ) / sizeof ( a [ 0 ] ) ; cout << getIndex ( a , n ) ; return 0 ; } |
Find the position of the last removed element from the array | C ++ program to find the position of the last removed element from the array ; Function to find the original position of the element which will be removed last ; take ceil of every number ; Since position is index + 1 ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int getPosition ( int a [ ] , int n , int m ) { for ( int i = 0 ; i < n ; i ++ ) { a [ i ] = ( a [ i ] / m + ( a [ i ] % m != 0 ) ) ; } int ans = -1 , max = -1 ; for ( int i = n - 1 ; i >= 0 ; i -- ) { if ( max < a [ i ] ) { max = a [ i ] ; ans = i ; } } return ans + 1 ; } int main ( ) { int a [ ] = { 2 , 5 , 4 } ; int n = sizeof ( a ) / sizeof ( a [ 0 ] ) ; int m = 2 ; cout << getPosition ( a , n , m ) ; return 0 ; } |
Find the value of f ( n ) / f ( r ) * f ( n | CPP to find the value of f ( n ) / f ( r ) * f ( n - r ) ; Function to find value of given F ( n ) ; iterate over n ; calculate result ; return the result ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int calcFunction ( int n , int r ) { int finalDenominator = 1 ; int mx = max ( r , n - r ) ; for ( int i = mx + 1 ; i <= n ; i ++ ) { int denominator = ( int ) pow ( i , i ) ; int numerator = ( int ) pow ( i - mx , i - mx ) ; finalDenominator = ( finalDenominator * denominator ) / numerator ; } return finalDenominator ; } int main ( ) { int n = 6 , r = 2 ; cout << "1 / " << calcFunction ( n , r ) << endl ; return 0 ; } |
Check if the array has an element which is equal to sum of all the remaining elements | C ++ program to Check if the array has an element which is equal to sum of all the remaining elements ; Function to check if such element exists or not ; Storing frequency in map ; Stores the sum ; Traverse the array and count the array elements ; Only possible if sum is even ; If half sum is available ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool isExists ( int a [ ] , int n ) { unordered_map < int , int > freq ; int sum = 0 ; for ( int i = 0 ; i < n ; i ++ ) { freq [ a [ i ] ] ++ ; sum += a [ i ] ; } if ( sum % 2 == 0 ) { if ( freq [ sum / 2 ] ) return true ; } return false ; } int main ( ) { int a [ ] = { 5 , 1 , 2 , 2 } ; int n = sizeof ( a ) / sizeof ( a [ 0 ] ) ; if ( isExists ( a , n ) ) cout << " Yes " ; else cout << " No " ; return 0 ; } |
Print all numbers less than N with at | C ++ program to print all the numbers less than N which have at most 2 unique digits ; Function to generate all possible numbers ; If the number is less than n ; If the number exceeds ; Check if it is not the same number ; Function to print all numbers ; All combination of digits ; Print all numbers ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; set < int > st ; void generateNumbers ( int n , int num , int a , int b ) { if ( num > 0 && num < n ) st . insert ( num ) ; if ( num >= n ) return ; if ( num * 10 + a > num ) generateNumbers ( n , num * 10 + a , a , b ) ; generateNumbers ( n , num * 10 + b , a , b ) ; } void printNumbers ( int n ) { for ( int i = 0 ; i <= 9 ; i ++ ) for ( int j = i + 1 ; j <= 9 ; j ++ ) generateNumbers ( n , 0 , i , j ) ; cout << " The ▁ numbers ▁ are : ▁ " ; while ( ! st . empty ( ) ) { cout << * st . begin ( ) << " ▁ " ; st . erase ( st . begin ( ) ) ; } } int main ( ) { int n = 12 ; printNumbers ( n ) ; return 0 ; } |
Smallest prime number missing in an array | C ++ implementation of the above approach ; this store all prime number upto 10 ^ 5 this function find all prime ; use sieve to find prime ; if number is prime then store it in prime vector ; Function to find the smallest prime missing ; first of all find all prime ; now store all number as index of freq arr so that we can improve searching ; now check for each prime ; Driver code ; find smallest prime which is not present | #include <bits/stdc++.h> NEW_LINE using namespace std ; #define ll long long int NEW_LINE vector < ll > findPrime ( int MAX ) { bool pm [ MAX + 1 ] ; memset ( pm , true , sizeof ( pm ) ) ; pm [ 0 ] = pm [ 1 ] = false ; for ( int i = 2 ; i <= MAX ; i ++ ) if ( pm [ i ] ) for ( int j = 2 * i ; j <= MAX ; j += i ) pm [ j ] = false ; vector < ll > prime ; for ( int i = 0 ; i <= MAX ; i ++ ) if ( pm [ i ] ) prime . push_back ( i ) ; return prime ; } int findSmallest ( int arr [ ] , int n ) { int MAX = * max_element ( arr , arr + n ) ; vector < ll > prime = findPrime ( MAX ) ; unordered_set < int > s ; for ( int i = 0 ; i < n ; i ++ ) s . insert ( arr [ i ] ) ; int ans = -1 ; for ( int i = 0 ; i < prime . size ( ) ; i ++ ) if ( s . find ( prime [ i ] ) == s . end ( ) ) { ans = prime [ i ] ; break ; } return ans ; } int main ( ) { int arr [ ] = { 3 , 0 , 1 , 2 , 7 } ; int n = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; if ( findSmallest ( arr , n ) == -1 ) cout << " No ▁ prime ▁ number ▁ missing " ; else cout << findSmallest ( arr , n ) ; return 0 ; } |
Find the number after successive division | C ++ program to implement above approach ; Function to find the number ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int findNum ( int div [ ] , int rem [ ] , int N ) { int num = rem [ N - 1 ] ; for ( int i = N - 2 ; i >= 0 ; i -- ) { num = num * div [ i ] + rem [ i ] ; } return num ; } int main ( ) { int div [ ] = { 8 , 3 } ; int rem [ ] = { 2 , 2 } ; int N = sizeof ( div ) / sizeof ( div [ 0 ] ) ; cout << findNum ( div , rem , N ) ; return 0 ; } |
Ways to form n / 2 pairs such that difference of pairs is minimum | C ++ implementation of the above approach ; Using mod because the number of ways might be very large ; ways is serving the same purpose as discussed ; pairing up zero people requires one way . ; map count stores count of s . ; sort according to value ; Iterating backwards . ; Checking if current count is odd . ; if current count = 5 , multiply ans by ways [ 4 ] . ; left out person will be selected in current_count ways ; left out person will pair with previous person in previous_count ways ; if previous count is odd , * then multiply answer by ways [ prev_count - 1 ] . * since one has already been reserved , * remaining will be even . * reduce prev_count = 0 , since we don 't need it now. ; if prev count is even , one will be reserved , * therefore decrement by 1. * In the next iteration , prev_count will become odd * and it will be handled in the same way . ; if current count is even , * then simply multiply ways [ current_count ] * to answer . ; multiply answer by ways [ first__count ] since that is left out , after iterating the array . ; Driver code | #include <bits/stdc++.h> NEW_LINE #define mp make_pair NEW_LINE #define pb push_back NEW_LINE #define S second NEW_LINE #define ll long long NEW_LINE using namespace std ; const int mod = 1000000007 ; const int MAX = 100000 ; ll ways [ MAX + 1 ] ; void preCompute ( ) { ways [ 0 ] = 1LL ; ways [ 2 ] = 1LL ; for ( int i = 4 ; i <= MAX ; i += 2 ) { ways [ i ] = ( 1LL * ( i - 1 ) * ways [ i - 2 ] ) % mod ; } } void countWays ( int * arr , int n ) { map < int , int > count ; for ( int i = 0 ; i < n ; i ++ ) count [ arr [ i ] ] ++ ; vector < pair < int , int > > count_vector ; map < int , int > :: iterator it ; for ( it = count . begin ( ) ; it != count . end ( ) ; it ++ ) { count_vector . pb ( mp ( it -> first , it -> second ) ) ; } sort ( count_vector . begin ( ) , count_vector . end ( ) ) ; ll ans = 1 ; for ( int i = count_vector . size ( ) - 1 ; i > 0 ; i -- ) { int current_count = count_vector [ i ] . S ; int prev_count = count_vector [ i - 1 ] . S ; if ( current_count & 1 ) { ans = ( ans * ways [ current_count - 1 ] ) % mod ; ans = ( ans * current_count ) % mod ; ans = ( ans * prev_count ) % mod ; if ( prev_count & 1 ) { ans = ( ans * ways [ prev_count - 1 ] ) % mod ; count_vector [ i - 1 ] . S = 0 ; } else { count_vector [ i - 1 ] . S -- ; } } else { ans = ( ans * ways [ current_count ] ) % mod ; } } ans = ( ans * ways [ count_vector [ 0 ] . S ] ) % mod ; cout << ans << " STRNEWLINE " ; } int main ( ) { preCompute ( ) ; int arr [ ] = { 2 , 3 , 3 , 3 , 3 , 4 , 4 , 4 , 4 , 4 } ; int n = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; countWays ( arr , n ) ; return 0 ; } |
Program to find the profit or loss when CP of N items is equal to SP of M items | C ++ implementation of above formula ; Function to calculate Profit or loss ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; void profitLoss ( int N , int M ) { if ( N == M ) cout << " No ▁ Profit ▁ nor ▁ Loss " ; else { float result = 0.0 ; result = float ( abs ( N - M ) ) / M ; if ( N - M < 0 ) cout << " Loss ▁ = ▁ - " << result * 100 << " % " ; else cout << " Profit ▁ = ▁ " << result * 100 << " % " ; } } int main ( ) { int N = 8 , M = 9 ; profitLoss ( N , M ) ; return 0 ; } |
Aronson 's Sequence | C ++ program to generate the first n terms of Aronson 's sequence ; Returns the given number in words ; Get number of digits in given number ; Base cases ; The following arrays contain one digit ( both cardinal and ordinal forms ) , two digit ( < 20 , ordinal forms ) numbers , and multiples ( ordinal forms ) and powers of 10. ; If single digit number ; here len can be 3 or 4 ; last two digits ; Handle all powers of 10 ; Handle two digit numbers < 20 ; Function to print the first n terms of Aronson 's sequence ; check if character is alphabet or not ; convert number to words in ordinal format and append ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; string convert_to_words ( string num ) { int len = num . length ( ) ; if ( len == 0 len > 4 ) { return " " ; } string single_digits_temp [ ] = { " " , " one " , " two " , " three " , " four " , " five " , " six " , " seven " , " eight " , " nine " } ; string single_digits [ ] = { " " , " first " , " second " , " third " , " fourth " , " fifth " , " sixth " , " seventh " , " eighth " , " ninth " } ; string two_digits [ ] = { " " , " tenth " , " eleventh " , " twelfth " , " thirteenth " , " fourteenth " , " fifteenth " , " sixteenth " , " seventeenth " , " eighteenth " , " nineteenth " } ; string tens_multiple [ ] = { " " , " tenth " , " twentieth " , " thirtieth " , " fortieth " , " fiftieth " , " sixtieth " , " seventieth " , " eightieth " , " ninetieth " } ; string tens_power [ ] = { " hundred " , " thousand " } ; string word = " " ; if ( len == 1 ) { word += single_digits [ num [ 0 ] - '0' ] ; return word ; } int i = 0 , ctr = 0 ; string s = " ▁ " ; while ( i < len ) { if ( len >= 3 ) { if ( num [ i ] != '0' ) { word += single_digits_temp [ num [ i ] - '0' ] + " ▁ " ; word += tens_power [ len - 3 ] + " ▁ " ; ctr ++ ; } len -- ; num . erase ( 0 , 1 ) ; } else { if ( ctr != 0 ) { s = " ▁ and ▁ " ; word . erase ( word . length ( ) - 1 ) ; } if ( num [ i + 1 ] == '0' ) if ( num [ i ] == '0' ) word = word + " th " ; else word += s + tens_multiple [ num [ i ] - '0' ] ; else if ( num [ i ] == '1' ) word += s + two_digits [ num [ i + 1 ] - '0' + 1 ] ; else { if ( num [ i ] != '0' ) word += s + tens_multiple [ num [ i ] - '0' ] . substr ( 0 , tens_multiple [ num [ i ] - '0' ] . length ( ) - 4 ) + " y ▁ " ; else word += s ; word += single_digits [ num [ i + 1 ] - '0' ] ; } i += 2 ; } if ( i == len ) { if ( word [ 0 ] == ' ▁ ' ) word . erase ( 0 , 1 ) ; } } return word ; } void Aronsons_sequence ( int n ) { string str = " T ▁ is ▁ the ▁ " ; int ind = 0 ; for ( int i = 0 ; i < str . length ( ) ; i ++ ) { if ( isalpha ( str [ i ] ) ) { ind += 1 ; } if ( str [ i ] == ' t ' or str [ i ] == ' T ' ) { n -= 1 ; str += convert_to_words ( to_string ( ind ) ) + " , ▁ " ; cout << ind << " , ▁ " ; } if ( n == 0 ) break ; } } int main ( void ) { int n = 6 ; Aronsons_sequence ( n ) ; return 0 ; } |
Find all good indices in the given Array | C ++ program to find all good indices in the given array ; Function to find all good indices in the given array ; hash to store frequency of each element ; Storing frequency of each element and calculating sum simultaneously ; check if array is good after removing i - th index element ; print good indices ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; void niceIndices ( int A [ ] , int n ) { int sum = 0 ; map < int , int > m ; for ( int i = 0 ; i < n ; ++ i ) { m [ A [ i ] ] ++ ; sum += A [ i ] ; } for ( int i = 0 ; i < n ; ++ i ) { int k = sum - A [ i ] ; if ( k % 2 == 0 ) { k = k >> 1 ; if ( m . find ( k ) != m . end ( ) ) { if ( ( A [ i ] == k && m [ k ] > 1 ) || ( A [ i ] != k ) ) cout << ( i + 1 ) << " ▁ " ; } } } } int main ( ) { int A [ ] = { 8 , 3 , 5 , 2 } ; int n = sizeof ( A ) / sizeof ( A [ 0 ] ) ; niceIndices ( A , n ) ; return 0 ; } |
Count pieces of circle after N cuts | C ++ program to find number of pieces of circle after N cuts ; Function to find number of pieces of circle after N cuts ; Driver program | #include <bits/stdc++.h> NEW_LINE using namespace std ; int countPieces ( int N ) { return 2 * N ; } int main ( ) { int N = 100 ; cout << countPieces ( N ) ; return 0 ; } |
Sum of all the multiples of 3 and 7 below N | C ++ program to find the sum of all multiples of 3 and 7 below N ; Function to find sum of AP series ; Number of terms ; Function to find the sum of all multiples of 3 and 7 below N ; Since , we need the sum of multiples less than N ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; long long sumAP ( long long n , long long d ) { n /= d ; return ( n ) * ( 1 + n ) * d / 2 ; } long long sumMultiples ( long long n ) { n -- ; return sumAP ( n , 3 ) + sumAP ( n , 7 ) - sumAP ( n , 21 ) ; } int main ( ) { long long n = 24 ; cout << sumMultiples ( n ) ; return 0 ; } |
Check whether product of digits at even places is divisible by sum of digits at odd place of a number | C ++ implementation of the above approach ; below function checks whether product of digits at even places is divisible by sum of digits at odd places ; if size is even ; if size is odd ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool productSumDivisible ( int n , int size ) { int sum = 0 , product = 1 ; while ( n > 0 ) { if ( size % 2 == 0 ) { product *= n % 10 ; } else { sum += n % 10 ; } n = n / 10 ; size -- ; } if ( product % sum == 0 ) return true ; return false ; } int main ( ) { int n = 1234 ; int len = 4 ; if ( productSumDivisible ( n , len ) ) cout << " TRUE " ; else cout << " FALSE " ; return 0 ; } |
GCD of a number raised to some power and another number | CPP program to find GCD of a ^ n and b . ; Returns GCD of a ^ n and b . ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; typedef long long int ll ; ll gcd ( ll a , ll b ) { if ( a == 0 ) return b ; return gcd ( b % a , a ) ; } ll powGCD ( ll a , ll n , ll b ) { for ( int i = 0 ; i < n ; i ++ ) a = a * a ; return gcd ( a , b ) ; } int main ( ) { ll a = 10 , b = 5 , n = 2 ; cout << powGCD ( a , n , b ) ; return 0 ; } |
Position after taking N steps to the right and left in an alternate manner | C ++ program to find the last coordinate where it ends his journey ; Function to return the last destination ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int lastCoordinate ( int n , int a , int b ) { return ( ( n + 1 ) / 2 ) * a - ( n / 2 ) * b ; } int main ( ) { int n = 3 , a = 5 , b = 2 ; cout << lastCoordinate ( n , a , b ) ; return 0 ; } |
Smallest number greater than or equal to N divisible by K | C ++ implementation of the above approach ; Function to find the smallest number greater than or equal to N that is divisible by k ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int findNum ( int N , int K ) { int rem = ( N + K ) % K ; if ( rem == 0 ) return N ; else return N + K - rem ; } int main ( ) { int N = 45 , K = 6 ; cout << " Smallest ▁ number ▁ greater ▁ than ▁ or ▁ equal ▁ to ▁ " << N << " that is divisible by " ▁ < < ▁ K ▁ < < ▁ " is " return 0 ; } |
Sum and Product of digits in a number that divide the number | C ++ implementation of the above approach ; Print the sum and product of digits that divides the number . ; Fetching each digit of the number ; Checking if digit is greater than 0 and can divides n . ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; void countDigit ( int n ) { int temp = n , sum = 0 , product = 1 ; while ( temp != 0 ) { int d = temp % 10 ; temp /= 10 ; if ( d > 0 && n % d == 0 ) { sum += d ; product *= d ; } } cout << " Sum ▁ = ▁ " << sum ; cout << " Product = " } int main ( ) { int n = 1012 ; countDigit ( n ) ; return 0 ; } |
Largest number smaller than or equal to N divisible by K | C ++ implementation of the above approach ; Function to find the largest number smaller than or equal to N that is divisible by k ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int findNum ( int N , int K ) { int rem = N % K ; if ( rem == 0 ) return N ; else return N - rem ; } int main ( ) { int N = 45 , K = 6 ; cout << " Largest ▁ number ▁ smaller ▁ than ▁ or ▁ equal ▁ to ▁ " << N << " that is divisible by " < < ▁ K ▁ < < ▁ " is " return 0 ; } |
Check if any permutation of a number is divisible by 3 and is Palindromic | C ++ program to check if any permutation of a number is divisible by 3 and is Palindromic ; Function to check if any permutation of a number is divisible by 3 and is Palindromic ; Hash array to store frequency of digits of n ; traverse the digits of integer and store their frequency ; Calculate the sum of digits simultaneously ; Check if number is not divisible by 3 ; If more than one digits have odd frequency , palindromic permutation not possible ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool isDivisiblePalindrome ( int n ) { int hash [ 10 ] = { 0 } ; int digitSum = 0 ; while ( n ) { digitSum += n % 10 ; hash [ n % 10 ] ++ ; n /= 10 ; } if ( digitSum % 3 != 0 ) return false ; int oddCount = 0 ; for ( int i = 0 ; i < 10 ; i ++ ) { if ( hash [ i ] % 2 != 0 ) oddCount ++ ; } if ( oddCount > 1 ) return false ; else return true ; } int main ( ) { int n = 34734 ; isDivisiblePalindrome ( n ) ? cout << " True " : cout << " False " ; return 0 ; } |
Minimum time to return array to its original state after given modifications | C ++ implementation of above approach ; Function to return lcm of two numbers ; Make a graph ; Add edge ; Count reachable nodes from every node . ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int lcm ( int a , int b ) { return ( a * b ) / ( __gcd ( a , b ) ) ; } int dfs ( int src , vector < int > adj [ ] , vector < bool > & visited ) { visited [ src ] = true ; int count = 1 ; for ( int i = 0 ; i < adj [ src ] . size ( ) ; i ++ ) if ( ! visited [ adj [ src ] [ i ] ] ) count += dfs ( adj [ src ] [ i ] , adj , visited ) ; return count ; } int findMinTime ( int arr [ ] , int P [ ] , int n ) { vector < int > adj [ n + 1 ] ; for ( int i = 0 ; i < n ; i ++ ) { adj [ arr [ i ] ] . push_back ( P [ i ] ) ; } vector < bool > visited ( n + 1 ) ; int ans = 1 ; for ( int i = 0 ; i < n ; i ++ ) { if ( ! visited [ i ] ) { ans = lcm ( ans , dfs ( i , adj , visited ) ) ; } } return ans ; } int main ( ) { int arr [ ] = { 1 , 2 , 3 } ; int P [ ] = { 3 , 2 , 1 } ; int n = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; cout << findMinTime ( arr , P , n ) ; return 0 ; } |
Check whether product of digits at even places of a number is divisible by K | C ++ implementation of the above approach ; below function checks whether product of digits at even places is divisible by K ; if position is even ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool productDivisible ( int n , int k ) { int product = 1 , position = 1 ; while ( n > 0 ) { if ( position % 2 == 0 ) product *= n % 10 ; n = n / 10 ; position ++ ; } if ( product % k == 0 ) return true ; return false ; } int main ( ) { int n = 321922 ; int k = 3 ; if ( productDivisible ( n , k ) ) cout << " YES " ; else cout << " NO " ; return 0 ; } |
Permutations of n things taken r at a time with k things together | CPP program to find the number of permutations of n different things taken r at a time with k things grouped together ; Function to find factorial of a number ; Function to calculate p ( n , r ) ; Function to find the number of permutations of n different things taken r at a time with k things grouped together ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int factorial ( int n ) { int fact = 1 ; for ( int i = 2 ; i <= n ; i ++ ) fact = fact * i ; return fact ; } int npr ( int n , int r ) { int pnr = factorial ( n ) / factorial ( n - r ) ; return pnr ; } int countPermutations ( int n , int r , int k ) { return factorial ( k ) * ( r - k + 1 ) * npr ( n - k , r - k ) ; } int main ( ) { int n = 8 ; int r = 5 ; int k = 2 ; cout << countPermutations ( n , r , k ) ; return 0 ; } |
Greatest Integer Function | CPP program to illustrate greatest integer Function ; Function to calculate the GIF value of a number ; GIF is the floor of a number ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int GIF ( float n ) { return floor ( n ) ; } int main ( ) { int n = 2.3 ; cout << GIF ( n ) ; return 0 ; } |
Number of triangles formed from a set of points on three lines | CPP program to find the possible number of triangles that can be formed from set of points on three lines ; Returns factorial of a number ; calculate c ( n , r ) ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int factorial ( int n ) { int fact = 1 ; for ( int i = 2 ; i <= n ; i ++ ) fact = fact * i ; return fact ; } int ncr ( int n , int r ) { return factorial ( n ) / ( factorial ( r ) * factorial ( n - r ) ) ; } int main ( ) { int m = 3 , n = 4 , k = 5 ; int totalTriangles = ncr ( m + n + k , 3 ) - ncr ( m , 3 ) - ncr ( n , 3 ) - ncr ( k , 3 ) ; cout << totalTriangles << endl ; } |
Check whether sum of digits at odd places of a number is divisible by K | C ++ implementation of the approach ; function that checks the divisibility of the sum of the digits at odd places of the given number ; if position is odd ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool SumDivisible ( int n , int k ) { int sum = 0 , position = 1 ; while ( n > 0 ) { if ( position % 2 == 1 ) sum += n % 10 ; n = n / 10 ; position ++ ; } if ( sum % k == 0 ) return true ; return false ; } int main ( ) { int n = 592452 ; int k = 3 ; if ( SumDivisible ( n , k ) ) cout << " YES " ; else cout << " NO " ; return 0 ; } |
Check whether sum of digits at odd places of a number is divisible by K | C ++ implementation of the above approach ; Converting integer to string ; Traversing the string ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool sumDivisible ( int n , int k ) { int sum = 0 ; string num = to_string ( n ) ; int i ; for ( i = 0 ; i < num . size ( ) ; i ++ ) { if ( i % 2 != 0 ) { sum = sum + ( num [ i ] - '0' ) ; } } if ( sum % k == 0 ) { return true ; } else { return false ; } } int main ( ) { int n = 592452 ; int k = 3 ; if ( sumDivisible ( n , k ) ) { cout << ( " YES " ) ; } else { cout << ( " NO " ) ; } return 0 ; } |
Check if a triangle of positive area is possible with the given angles | C ++ program to check if a triangle of positive area is possible with the given angles ; Checking if the sum of three angles is 180 and none of the angles is zero ; Checking if sum of any two angles is greater than equal to the third one ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; string isTriangleExists ( int a , int b , int c ) { if ( a != 0 && b != 0 && c != 0 && ( a + b + c ) == 180 ) if ( ( a + b ) >= c || ( b + c ) >= a || ( a + c ) >= b ) return " YES " ; else return " NO " ; else return " NO " ; } int main ( ) { int a = 50 , b = 60 , c = 70 ; cout << isTriangleExists ( a , b , c ) << endl ; return 0 ; } |
Find maximum value of x such that n ! % ( k ^ x ) = 0 | C ++ program to maximize the value of x such that n ! % ( k ^ x ) = 0 ; Function to maximize the value of x such that n ! % ( k ^ x ) = 0 ; Find square root of k and add 1 to it ; Run the loop from 2 to m and k should be greater than 1 ; optimize the value of k ; Minimum store ; Driver Code | #include <iostream> NEW_LINE #include <math.h> NEW_LINE using namespace std ; class GfG { public : int findX ( int n , int k ) { int r = n , v , u ; int m = sqrt ( k ) + 1 ; for ( int i = 2 ; i <= m && k > 1 ; i ++ ) { if ( i == m ) { i = k ; } for ( u = v = 0 ; k % i == 0 ; v ++ ) { k /= i ; } if ( v > 0 ) { int t = n ; while ( t > 0 ) { t /= i ; u += t ; } r = min ( r , u / v ) ; } } return r ; } } ; int main ( ) { GfG g ; int n = 5 ; int k = 2 ; cout << g . findX ( n , k ) ; } |
Ways of selecting men and women from a group to make a team | C ++ implementation of the approach ; Returns factorial of the number ; Function to calculate ncr ; Function to calculate the total possible ways ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int fact ( int n ) { int fact = 1 ; for ( int i = 2 ; i <= n ; i ++ ) fact *= i ; return fact ; } int ncr ( int n , int r ) { int ncr = fact ( n ) / ( fact ( r ) * fact ( n - r ) ) ; return ncr ; } int ways ( int m , int w , int n , int k ) { int ans = 0 ; while ( m >= k ) { ans += ncr ( m , k ) * ncr ( w , n - k ) ; k += 1 ; } return ans ; } int main ( ) { int m , w , n , k ; m = 7 ; w = 6 ; n = 5 ; k = 3 ; cout << ways ( m , w , n , k ) ; } |
Product of every K ’ th prime number in an array | C ++ implementation of the approach ; Create a boolean array " prime [ 0 . . n ] " and initialize all the entries as true . A value in prime [ i ] will finally be false if i is Not a prime , else true . ; 0 and 1 are not prime numbers ; If prime [ p ] is not changed , then it is a prime ; Update all multiples of p ; compute the answer ; count of primes ; product of the primes ; traverse the array ; if the number is a prime ; increase the count ; if it is the K 'th prime ; Driver code ; create the sieve | #include <bits/stdc++.h> NEW_LINE using namespace std ; #define MAX 1000000 NEW_LINE bool prime [ MAX + 1 ] ; void SieveOfEratosthenes ( ) { memset ( prime , true , sizeof ( prime ) ) ; prime [ 1 ] = false ; prime [ 0 ] = false ; for ( int p = 2 ; p * p <= MAX ; p ++ ) { if ( prime [ p ] == true ) { for ( int i = p * 2 ; i <= MAX ; i += p ) prime [ i ] = false ; } } } void productOfKthPrimes ( int arr [ ] , int n , int k ) { int c = 0 ; long long int product = 1 ; for ( int i = 0 ; i < n ; i ++ ) { if ( prime [ arr [ i ] ] ) { c ++ ; if ( c % k == 0 ) { product *= arr [ i ] ; c = 0 ; } } } cout << product << endl ; } int main ( ) { SieveOfEratosthenes ( ) ; int n = 5 , k = 2 ; int arr [ n ] = { 2 , 3 , 5 , 7 , 11 } ; productOfKthPrimes ( arr , n , k ) ; return 0 ; } |
Sum of greatest odd divisor of numbers in given range | C ++ program to find sum of greatest odd divisor of numbers in given range ; Function to return sum of first n odd numbers ; Recursive function to return sum of greatest odd divisor of numbers in range [ 1 , n ] ; if ( n % 2 == 1 ) { Odd n ; else { Even n ; Function to return sum of greatest odd divisor of numbers in range [ a , b ] ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int square ( int n ) { return n * n ; } int sum ( int n ) { if ( n == 0 ) return 0 ; return square ( ( n + 1 ) / 2 ) + sum ( n / 2 ) ; } return square ( n / 2 ) + sum ( n / 2 ) ; } } int oddDivSum ( int a , int b ) { return sum ( b ) - sum ( a - 1 ) ; } int main ( ) { int a = 3 , b = 9 ; cout << oddDivSum ( a , b ) << endl ; return 0 ; } |
Minimum numbers needed to express every integer below N as a sum | CPP program to find count of integers needed to express all numbers from 1 to N . ; function to count length of binary expression of n ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int countBits ( int n ) { int count = 0 ; while ( n ) { count ++ ; n >>= 1 ; } return count ; } int main ( ) { int n = 32 ; cout << " Minimum ▁ value ▁ of ▁ K ▁ is ▁ = ▁ " << countBits ( n ) << endl ; return 0 ; } |
Check if a number is an Achilles number or not | Program to check if the given number is an Achilles Number ; function to check if the number is powerful number ; First divide the number repeatedly by 2 ; If only 2 ^ 1 divides n ( not higher powers ) , then return false ; if n is not a power of 2 then this loop will execute repeat above process ; Find highest power of " factor " that divides n ; If only factor ^ 1 divides n ( not higher powers ) , then return false ; n must be 1 now if it is not a prime number . Since prime numbers are not powerful , we return false if n is not 1. ; Utility function to check if number is a perfect power or not ; Function to check Achilles Number ; Driver Program | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool isPowerful ( int n ) { while ( n % 2 == 0 ) { int power = 0 ; while ( n % 2 == 0 ) { n /= 2 ; power ++ ; } if ( power == 1 ) return false ; } for ( int factor = 3 ; factor <= sqrt ( n ) ; factor += 2 ) { int power = 0 ; while ( n % factor == 0 ) { n = n / factor ; power ++ ; } if ( power == 1 ) return false ; } return ( n == 1 ) ; } bool isPower ( int a ) { if ( a == 1 ) return true ; for ( int i = 2 ; i * i <= a ; i ++ ) { double val = log ( a ) / log ( i ) ; if ( ( val - ( int ) val ) < 0.00000001 ) return true ; } return false ; } bool isAchillesNumber ( int n ) { if ( isPowerful ( n ) && ! isPower ( n ) ) return true ; else return false ; } int main ( ) { int n = 72 ; if ( isAchillesNumber ( n ) ) cout << " YES " << endl ; else cout << " NO " << endl ; n = 36 ; if ( isAchillesNumber ( n ) ) cout << " YES " << endl ; else cout << " NO " << endl ; return 0 ; } |
Program to find count of numbers having odd number of divisors in given range | C ++ program to find count of numbers having odd number of divisors in given range ; Function to count numbers having odd number of divisors in range [ A , B ] ; variable to odd divisor count ; iterate from a to b and count their number of divisors ; variable to divisor count ; if count of divisor is odd then increase res by 1 ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int OddDivCount ( int a , int b ) { int res = 0 ; for ( int i = a ; i <= b ; ++ i ) { int divCount = 0 ; for ( int j = 1 ; j <= i ; ++ j ) { if ( i % j == 0 ) { ++ divCount ; } } if ( divCount % 2 ) { ++ res ; } } return res ; } int main ( ) { int a = 1 , b = 10 ; cout << OddDivCount ( a , b ) << endl ; return 0 ; } |
Largest factor of a given number which is a perfect square | C ++ program to find the largest factor of a number which is also a perfect square ; Function to find the largest factor of a given number which is a perfect square ; Initialise the answer to 1 ; Finding the prime factors till sqrt ( num ) ; Frequency of the prime factor in the factorisation initialised to 0 ; If the frequency is odd then multiply i frequency - 1 times to the answer ; Else if it is even , multiply it frequency times ; Driver Code | #include <cmath> NEW_LINE #include <iostream> NEW_LINE using namespace std ; int largestSquareFactor ( int num ) { int answer = 1 ; for ( int i = 2 ; i < sqrt ( num ) ; ++ i ) { int cnt = 0 ; int j = i ; while ( num % j == 0 ) { cnt ++ ; j *= i ; } if ( cnt & 1 ) { cnt -- ; answer *= pow ( i , cnt ) ; } else { answer *= pow ( i , cnt ) ; } } return answer ; } int main ( ) { int N = 420 ; cout << largestSquareFactor ( N ) ; return 0 ; } |
Find the Nth term of the series 2 + 6 + 13 + 23 + . . . | CPP program to find Nth term of the series 2 + 6 + 13 + 23 + 36 + ... ; calculate Nth term of given series ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int Nth_Term ( int n ) { return ( 3 * pow ( n , 2 ) - n + 2 ) / ( 2 ) ; } int main ( ) { int N = 5 ; cout << Nth_Term ( N ) << endl ; } |
All possible numbers of N digits and base B without leading zeros | C ++ implementation of the approach ; function to count all permutations ; count of all permutations ; count of permutations with leading zeros ; Return the permutations without leading zeros ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; void countPermutations ( int N , int B ) { int x = pow ( B , N ) ; int y = pow ( B , N - 1 ) ; cout << x - y << " STRNEWLINE " ; } int main ( ) { int N = 6 ; int B = 4 ; countPermutations ( N , B ) ; return 0 ; } |
Absolute difference between the Product of Non | C ++ program to find the Absolute Difference between the Product of Non - Prime numbers and Prime numbers of an Array ; Function to find the difference between the product of non - primes and the product of primes of an array . ; Find maximum value in the array ; USE SIEVE TO FIND ALL PRIME NUMBERS LESS THAN OR EQUAL TO max_val Create a boolean array " prime [ 0 . . n ] " . A value in prime [ i ] will finally be false if i is Not a prime , else true . ; Remaining part of SIEVE ; If prime [ p ] is not changed , then it is a prime ; Update all multiples of p ; Store the product of primes in P1 and the product of non primes in P2 ; the number is prime ; the number is non - prime ; Return the absolute difference ; Driver Code ; Find the absolute difference | #include <bits/stdc++.h> NEW_LINE using namespace std ; int calculateDifference ( int arr [ ] , int n ) { int max_val = * max_element ( arr , arr + n ) ; vector < bool > prime ( max_val + 1 , true ) ; prime [ 0 ] = false ; prime [ 1 ] = false ; for ( int p = 2 ; p * p <= max_val ; p ++ ) { if ( prime [ p ] == true ) { for ( int i = p * 2 ; i <= max_val ; i += p ) prime [ i ] = false ; } } int P1 = 1 , P2 = 1 ; for ( int i = 0 ; i < n ; i ++ ) { if ( prime [ arr [ i ] ] ) { P1 *= arr [ i ] ; } else if ( arr [ i ] != 1 ) { P2 *= arr [ i ] ; } } return abs ( P2 - P1 ) ; } int main ( ) { int arr [ ] = { 1 , 3 , 5 , 10 , 15 , 7 } ; int n = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; cout << calculateDifference ( arr , n ) ; return 0 ; } |
Maximum count of equal numbers in an array after performing given operations | CPP program to find the maximum number of equal numbers in an array ; Function to find the maximum number of equal numbers in an array ; to store sum of elements ; if sum of numbers is not divisible by n ; Driver Code ; size of an array | #include <bits/stdc++.h> NEW_LINE using namespace std ; int EqualNumbers ( int a [ ] , int n ) { int sum = 0 ; for ( int i = 0 ; i < n ; i ++ ) sum += a [ i ] ; if ( sum % n ) return n - 1 ; return n ; } int main ( ) { int a [ ] = { 1 , 4 , 1 } ; int n = sizeof ( a ) / sizeof ( a [ 0 ] ) ; cout << EqualNumbers ( a , n ) ; return 0 ; } |
Count number of ordered pairs with Even and Odd Sums | C ++ implementation of the above approach ; function to count odd sum pair ; if number is even ; if number is odd ; count of ordered pairs ; function to count even sum pair ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int count_odd_pair ( int n , int a [ ] ) { int odd = 0 , even = 0 ; for ( int i = 0 ; i < n ; i ++ ) { if ( a [ i ] % 2 == 0 ) even ++ ; else odd ++ ; } int ans = odd * even * 2 ; return ans ; } int count_even_pair ( int odd_sum_pairs , int n ) { int total_pairs = ( n * ( n - 1 ) ) ; int ans = total_pairs - odd_sum_pairs ; return ans ; } int main ( ) { int n = 6 ; int a [ ] = { 2 , 4 , 5 , 9 , 1 , 8 } ; int odd_sum_pairs = count_odd_pair ( n , a ) ; int even_sum_pairs = count_even_pair ( odd_sum_pairs , n ) ; cout << " Even ▁ Sum ▁ Pairs ▁ = ▁ " << even_sum_pairs << endl ; cout << " Odd ▁ Sum ▁ Pairs = ▁ " << odd_sum_pairs << endl ; return 0 ; } |
Steps required to visit M points in order on a circular ring of N points | C ++ implementation of the approach ; Function to count the steps required ; Start at 1 ; Initialize steps ; If nxt is greater than cur ; Now we are at a [ i ] ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int findSteps ( int n , int m , int a [ ] ) { int cur = 1 ; int steps = 0 ; for ( int i = 0 ; i < m ; i ++ ) { if ( a [ i ] >= cur ) steps += ( a [ i ] - cur ) ; else steps += ( n - cur + a [ i ] ) ; cur = a [ i ] ; } return steps ; } int main ( ) { int n = 3 , m = 3 ; int a [ ] = { 2 , 1 , 2 } ; cout << findSteps ( n , m , a ) ; } |
Program to Convert Hexadecimal Number to Binary | C ++ program to convert Hexadecimal number to Binary ; function to convert Hexadecimal to Binary Number ; driver code ; Get the Hexadecimal number ; Convert HexaDecimal to Binary | #include <bits/stdc++.h> NEW_LINE using namespace std ; void HexToBin ( string hexdec ) { long int i = 0 ; while ( hexdec [ i ] ) { switch ( hexdec [ i ] ) { case '0' : cout << "0000" ; break ; case '1' : cout << "0001" ; break ; case '2' : cout << "0010" ; break ; case '3' : cout << "0011" ; break ; case '4' : cout << "0100" ; break ; case '5' : cout << "0101" ; break ; case '6' : cout << "0110" ; break ; case '7' : cout << "0111" ; break ; case '8' : cout << "1000" ; break ; case '9' : cout << "1001" ; break ; case ' A ' : case ' a ' : cout << "1010" ; break ; case ' B ' : case ' b ' : cout << "1011" ; break ; case ' C ' : case ' c ' : cout << "1100" ; break ; case ' D ' : case ' d ' : cout << "1101" ; break ; case ' E ' : case ' e ' : cout << "1110" ; break ; case ' F ' : case ' f ' : cout << "1111" ; break ; default : cout << " Invalid hexadecimal digit " << hexdec [ i ] ; } i ++ ; } } int main ( ) { char hexdec [ 100 ] = "1AC5" ; cout << " Equivalent Binary value is : " HexToBin ( hexdec ) ; return 0 ; } |
Count unordered pairs ( i , j ) such that product of a [ i ] and a [ j ] is power of two | C ++ program to Count unordered pairs ( i , j ) in array such that product of a [ i ] and a [ j ] can be expressed as power of two ; Function to check if x is power of 2 ; First x in the below expression is for the case when x is 0 ; Function to Count unordered pairs ; is a number can be expressed as power of two ; count total number of unordered pairs ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; bool isPowerOfTwo ( int x ) { return x && ( ! ( x & ( x - 1 ) ) ) ; } void Count_pairs ( int a [ ] , int n ) { int count = 0 ; for ( int i = 0 ; i < n ; i ++ ) { if ( isPowerOfTwo ( a [ i ] ) ) count ++ ; } int ans = ( count * ( count - 1 ) ) / 2 ; cout << ans << " STRNEWLINE " ; } int main ( ) { int a [ ] = { 2 , 5 , 8 , 16 , 128 } ; int n = sizeof ( a ) / sizeof ( a [ 0 ] ) ; Count_pairs ( a , n ) ; return 0 ; } |
Ways of dividing a group into two halves such that two elements are in different groups | CPP Program to count Number of ways in which two Beautiful girls are in different group ; This function will return the factorial of a given number ; This function will calculate nCr of given n and r ; This function will Calculate number of ways ; Driver Code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int factorial ( int n ) { int result = 1 ; for ( int i = 1 ; i <= n ; i ++ ) result = result * i ; return result ; } int nCr ( int n , int r ) { return factorial ( n ) / ( factorial ( r ) * factorial ( n - r ) ) ; } int calculate_result ( int n ) { int result = 2 * nCr ( ( n - 2 ) , ( n / 2 - 1 ) ) ; return result ; } int main ( void ) { int a = 2 , b = 4 ; cout << calculate_result ( 2 * a ) << endl ; cout << calculate_result ( 2 * b ) << endl ; return 0 ; } |
Print values of ' a ' in equation ( a + b ) <= n and a + b is divisible by x | CPP program to Find values of a , in equation ( a + b ) <= n and a + b is divisible by x . ; function to Find values of a , in equation ( a + b ) <= n and a + b is divisible by x . ; least possible which is divisible by x ; run a loop to get required answer ; increase value by x ; answer is possible ; Driver code ; function call | #include <bits/stdc++.h> NEW_LINE using namespace std ; void PossibleValues ( int b , int x , int n ) { int leastdivisible = ( b / x + 1 ) * x ; int flag = 1 ; while ( leastdivisible <= n ) { if ( leastdivisible - b >= 1 ) { cout << leastdivisible - b << " ▁ " ; leastdivisible += x ; flag = 0 ; } else break ; } if ( flag ) cout << -1 ; } int main ( ) { int b = 10 , x = 6 , n = 40 ; PossibleValues ( b , x , n ) ; return 0 ; } |
Check if all sub | C ++ program to check if all sub - numbers have distinct Digit product ; Function to calculate product of digits between given indexes ; Function to check if all sub - numbers have distinct Digit product ; Length of number N ; Digit array ; set to maintain digit products ; Finding all possible subarrays ; Driver code | #include <bits/stdc++.h> NEW_LINE using namespace std ; int digitProduct ( int digits [ ] , int start , int end ) { int pro = 1 ; for ( int i = start ; i <= end ; i ++ ) { pro *= digits [ i ] ; } return pro ; } bool isDistinct ( int N ) { string s = to_string ( N ) ; int len = s . length ( ) ; int digits [ len ] ; unordered_set < int > products ; for ( int i = 0 ; i < len ; i ++ ) { digits [ i ] = s [ i ] - '0' ; } for ( int i = 0 ; i < len ; i ++ ) { for ( int j = i ; j < len ; j ++ ) { int val = digitProduct ( digits , i , j ) ; if ( products . find ( val ) != products . end ( ) ) return false ; else products . insert ( val ) ; } } return true ; } int main ( ) { int N = 324 ; if ( isDistinct ( N ) ) cout << " YES " ; else cout << " NO " ; return 0 ; } |
Ulam Number Sequence | Cpp code to print nth Ulam number ; Array to store Ulam Number ; function to compute ulam Number ; Set to search specific Ulam number efficiently ; push First 2 two term of the sequence in the array and set for further calculation ; loop to generate Ulam number ; traverse the array and check if i can be represented as sum of two distinct element of the array ; Check if i - arr [ j ] exist in the array or not using set If yes , Then i can be represented as sum of arr [ j ] + ( i - arr [ j ] ) ; if Count is greater than 2 break the loop ; If count is 2 that means i can be represented as sum of two distinct terms of the sequence ; i is ulam number ; Driver code ; pre compute Ulam Number sequence ; print nth Ulam number | #include <bits/stdc++.h> NEW_LINE using namespace std ; #define MAX 10000 NEW_LINE vector < int > arr ; void ulam ( ) { unordered_set < int > s ; arr . push_back ( 1 ) ; s . insert ( 1 ) ; arr . push_back ( 2 ) ; s . insert ( 2 ) ; for ( int i = 3 ; i < MAX ; i ++ ) { int count = 0 ; for ( int j = 0 ; j < arr . size ( ) ; j ++ ) { if ( s . find ( i - arr [ j ] ) != s . end ( ) && arr [ j ] != ( i - arr [ j ] ) ) count ++ ; if ( count > 2 ) break ; } if ( count == 2 ) { arr . push_back ( i ) ; s . insert ( i ) ; } } } int main ( ) { ulam ( ) ; int n = 9 ; cout << arr [ n - 1 ] ; return 0 ; } |
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