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# Subarray Sums Divisible by K

Given an array `A` of integers, return the number of (contiguous, non-empty) subarrays that have a sum divisible by `K`.

**Example 1:**

```
Input: A = [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]
```

**Note:**

1. `1 <= A.length <= 30000`
2. `-10000 <= A[i] <= 10000`
3. `2 <= K <= 10000`

```java
class Solution {
    public int subarraysDivByK(int[] A, int K) {
        // Map of Sum -> Frequency
        HashMap<Integer, Integer> map = new HashMap<>();
        map.put(0, 1);
        int count = 0, sum = 0;
        for (int x : A) {
            sum = (sum + x);
            int remainder = sum % K;
            if (remainder < 0)
                remainder += K; // Because -1 % 5 = -1, but we need the positive mod 4
            count += map.getOrDefault(remainder, 0);
            map.put(remainder, map.getOrDefault(remainder, 0) + 1);
        }
        return count;
    }
}
```


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