24. Subarray Sums Divisible by K

Given an integer array nums and an integer k, return the number of non-empty subarrays that have a sum divisible by k. A subarray is a contiguous part of an array.

Example 1:

Input: nums = [4,5,0,-2,-3,1], k = 5
Output: 7
Explanation: There are 7 subarrays with a sum divisible by k = 5:
[4, 5, 0, -2, -3, 1], [5], [5, 0], [5, 0, -2, -3], [0], [0, -2, -3], [-2, -3]

Example 2:

Input: nums = [5], k = 9
Output: 0

Solution: (Prefix sum with modulo)

// Some code
class Solution
{
public:
    int subarraysDivByK(vector<int> &nums, int k)
    {

        unordered_map<int, int> m;

        int s = 0;
        int count = 0;
        m[0]++;
        
        for (int i = 0; i < nums.size(); i++)
        {
            s = (s + nums[i])%k;
            
            if(s<0){
                s += k;
            }

            if (m.find(s) != m.end())
            {
                count += m[s];
            }

            m[s]++;
        }
        return count;
    }
};