Find an array of size N that satisfies the given conditions | C ++ implementation of the approach ; Utility function to print the contents of an array ; Function to generate and print the required array ; Initially all the positions are empty ; To store the count of positions i such that arr [ i ] = s ; To store the final array elements ; Set arr [ i ] = s and the gap between them is exactly 2 so in for loop we use i += 2 ; Mark the i 'th position as visited  as we put arr[i] = s ; Increment the count ; Finding the next odd number after s ; If the i 'th position is not visited  it means we did not put any value  at position i so we put 1 now ; Print the final array ; Driver code
#include <bits/stdc++.h> NEW_LINE using namespace std ; void printArr ( int arr [ ] , int n ) { for ( int i = 0 ; i < n ; i ++ ) cout << arr [ i ] << " ▁ " ; } void findArray ( int n , int k , int s ) { int vis [ n ] = { 0 } ; int cnt = 0 ; int arr [ n ] ; for ( int i = 0 ; i < n && cnt < k ; i += 2 ) { arr [ i ] = s ; vis [ i ] = 1 ; cnt ++ ; } int val = s ; if ( s % 2 == 0 ) val ++ ; else val = val + 2 ; for ( int i = 0 ; i < n ; i ++ ) { if ( vis [ i ] == 0 ) { arr [ i ] = val ; } } printArr ( arr , n ) ; } int main ( ) { int n = 8 , k = 3 , s = 12 ; findArray ( n , k , s ) ; return 0 ; }