Subsequences of size three in an array whose sum is divisible by m | O ( M ^ 2 ) time complexity C ++ program to find count of subsequences of size three divisible by M . ; Storing frequencies of all remainders when divided by M . ; including i and j in the sum rem calculate the remainder required to make the sum divisible by M ; if the required number is less than j , we skip as we have already calculated for that value before . As j here starts with i and rem is less than j . ; if satisfies the first case . ; if satisfies the second case ; if satisfies the third case ; Driver code
#include <iostream> NEW_LINE using namespace std ; int countSubSeq ( int A [ ] , int N , int M ) { int ans = 0 ; int h [ M ] = { 0 } ; for ( int i = 0 ; i < N ; i ++ ) { A [ i ] = A [ i ] % M ; h [ A [ i ] ] ++ ; } for ( int i = 0 ; i < M ; i ++ ) { for ( int j = i ; j < M ; j ++ ) { int rem = ( M - ( i + j ) % M ) % M ; if ( rem < j ) continue ; if ( i == j && rem == j ) ans += h [ i ] * ( h [ i ] - 1 ) * ( h [ i ] - 2 ) / 6 ; else if ( i == j ) ans += h [ i ] * ( h [ i ] - 1 ) * h [ rem ] / 2 ; else if ( i == rem ) ans += h [ i ] * ( h [ i ] - 1 ) * h [ j ] / 2 ; else if ( rem == j ) ans += h [ j ] * ( h [ j ] - 1 ) * h [ i ] / 2 ; else ans = ans + h [ i ] * h [ j ] * h [ rem ] ; } } return ans ; } int main ( ) { int M = 3 ; int A [ ] = { 1 , 2 , 4 , 3 } ; int N = sizeof ( A ) / sizeof ( A [ 0 ] ) ; cout << countSubSeq ( A , N , M ) ; return 0 ; }