# 974. Subarray Sums Divisible by K

## Description

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


Constraints:

• 1 <= nums.length <= 3 * 104
• -104 <= nums[i] <= 104
• 2 <= k <= 104

## Solutions

• class Solution {
public int subarraysDivByK(int[] nums, int k) {
Map<Integer, Integer> cnt = new HashMap<>();
cnt.put(0, 1);
int ans = 0, s = 0;
for (int x : nums) {
s = ((s + x) % k + k) % k;
ans += cnt.getOrDefault(s, 0);
cnt.merge(s, 1, Integer::sum);
}
return ans;
}
}

• class Solution {
public:
int subarraysDivByK(vector<int>& nums, int k) {
unordered_map<int, int> cnt{ {0, 1} };
int ans = 0, s = 0;
for (int& x : nums) {
s = ((s + x) % k + k) % k;
ans += cnt[s]++;
}
return ans;
}
};

• class Solution:
def subarraysDivByK(self, nums: List[int], k: int) -> int:
cnt = Counter({0: 1})
ans = s = 0
for x in nums:
s = (s + x) % k
ans += cnt[s]
cnt[s] += 1
return ans


• func subarraysDivByK(nums []int, k int) (ans int) {
cnt := map[int]int{0: 1}
s := 0
for _, x := range nums {
s = ((s+x)%k + k) % k
ans += cnt[s]
cnt[s]++
}
return
}

• function subarraysDivByK(nums: number[], k: number): number {
const counter = new Map();
counter.set(0, 1);
let s = 0,
ans = 0;
for (const num of nums) {
s += num;
const t = ((s % k) + k) % k;
ans += counter.get(t) || 0;
counter.set(t, (counter.get(t) || 0) + 1);
}
return ans;
}