# Continuous Subarray Sum in C++

Suppose we have a list of non-negative numbers and a target integer k, we have to write a function to check whether the array has a continuous subarray of size at least 2 that sums up to a multiple of k, sums up to n*k where n is also an integer. So if the input is like [23,2,4,6,7], and k = 6, then the result will be True, as [2,4] is a continuous subarray of size 2 and sums up to 6.

To solve this, we will follow these steps −

• Make a map m, set m[0] := -1 and sum := 0, n := size of nums array
• for i in range 0 to n – 1
• sum := sum + nums[i]
• if k is non zero, then sum := sum mod k
• if m has sum, and i – m[sum] >= 2, then return true
• if m does not has sum, set m[sum] := i
• return false

Let us see the following implementation to get better understanding −

## Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
bool checkSubarraySum(vector<int>& nums, int k) {
unordered_map<int, int> m;
m[0] = -1;
int sum = 0;
int n = nums.size();
for(int i = 0; i < n; i++){
sum += nums[i];
if(k)
sum %= k;
if(m.count(sum) && i - m[sum] >= 2){
return true;
}
if(!m.count(sum)) m[sum] = i;
}
return false;
}
};
main(){
vector<int> v = {23,2,4,6,7};
Solution ob;
cout << (ob.checkSubarraySum(v, 6));
}

## Input

[23,2,4,6,7]
6

## Output

1

Updated on: 02-May-2020

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