Count number of subsets with sum equal to k

This is a variation of the subset sum problem, which is NP-Hard - so there is no known polynomial solution to it. (In fact, the subset sum problem says it is hard to find if there is even one subset that sums to the given sum).

Possible approaches to solve it are brute force (check all possible subsets), or if the set contains relatively small integers, you can use the pseudo-polynomial dynamic programming technique:

f(i,0) = 1    (i >= 0) //succesful base clause
f(0,j) = 0    (j != 0) //non succesful base clause
f(i,j) = f(i-1,j) + f(i-1,j-arr[i])  //step

Applying dynamic programming to the above recursive formula gives you O(k*n) time and space solution.

Invoke with f(n,k) [assuming 1 based index for arrays].

Following is memoized Dynamic Programming code to print the count of the number of subsets with a given sum. The repeating values of DP are stores in "tmp" array. To attain a DP solution first always start with a recursive solution to the problem and then store the repeating value in a tmp array to arrive at a memoized solution.

#include <bits/stdc++.h>

using namespace std; 

int tmp[1001][1001];  


int subset_count(int* arr, int sum, int n) 
{ ` if(sum==0)
        return 1;
    if(n==0)
        return 0;
    if(tmp[n][sum]!=-1)
        return tmp[n][sum];
    else{
        if(arr[n-1]>sum)
            return tmp[n][sum]=subset_count(arr,sum, n-1);
        else{
            return tmp[n][required_sum]=subset_count(arr,sum, n- 1)+subset_count(arr,sum-arr[n-1], n-1);`
        }
    }
} 

// Driver code 

int main() 
{ ` memset(tmp,-1,sizeof(tmp));
    int arr[] = { 2, 3, 5, 6, 8, 10 }; 
    int n = sizeof(arr) / sizeof(int); 
    int sum = 10; `


    cout << subset_count(arr,sum, n); 

    return 0; 
}