# Binary Subarrays With Sum in C++

Suppose an array A of 0s and 1s is given, we have to find how many non-empty subarrays have sum S? So if the input is like [1,0,1,0,1], and S = 2, then the result will be 4, as the subarrays are [1,0,1,0,1], [1,0,1,0,1], [1,0,1,0,1], [1,0,1,0,1].

To solve this, we will follow these steps −

• Define a method called atMost(), this will take array A and integer x

• if x < 0, then return 0, set j := 0 and set ret := 0

• for i in range 0 to size of A

• decrease x by A[i]

• while x < 0

• increase x by A[j], increase j by 1

• ret := ret + i – j + 1

• return ret

• From the main method do the following −

• ret := atMost(A, S) – atMost(A, S – 1)

• return ret

Let us see the following implementation to get better understanding &mius;

## Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int atMost(vector <int>& A, int x){
if(x < 0) return 0;
int j = 0;
int ret = 0;
for(int i = 0; i < A.size(); i++){
x -= A[i];
while(x < 0){
x += A[j];
j++;
}
ret += i - j + 1;
}
return ret;
}
int numSubarraysWithSum(vector<int>& A, int S) {
return atMost(A, S) - atMost(A, S - 1);
}
};
main(){
vector<int> v1 = {1,0,1,0,1};
Solution ob;
cout << (ob.numSubarraysWithSum(v1, 2));
}

### Input

[1,0,1,0,1]

## Output

4