# 974.Subarray-Sums-Divisible-by-K

## 974. Subarray Sums Divisible by K

## 题目地址

<https://leetcode.com/problems/subarray-sums-divisible-by-k/>

## 题目描述

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

## 代码

### Approach #1

Time Complexity: O(N) Space Complexity: O(K)

if sum\[0, i] % K == sum\[0, j] % K, sum\[i + 1, j] is divisible by by K.

```java
class Solution {
  public int subarraysDivByK(int[] A, int K) {
    int[] map = new int[K];
map[0] = 1;
    int count = 0, sum = 0;
    for(int a : A) {
        sum = (sum + a) % K;
        if(sum < 0) sum += K;  // Because -1 % 5 = -1, but we need the positive mod 4
        count += map[sum];
        map[sum]++;
    }
    return count;
  }
}
```