---
title: '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
icon: dot
topics:
  - { name: "Array", slug: "array" }
  - { name: "Hash Table", slug: "hash-table" }
  - { name: "Prefix Sum", slug: "prefix-sum" }
issue: "https://github.com/prdlk/leetcode/issues/40"
sidebar:
  label: 'Subarray Sums Divisible by K'
  badge: 'Medium'
---

### 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 * 10^4`
- `-10^4 <= nums[i] <= 10^4`
- `2 <= k <= 10^4`

## Solution

```py
class Solution:
    def subarraysDivByK(self, nums: list[int], k: int) -> int:
        remainder_counts = {0: 1}
        cur_sum = 0
        count = 0

        for n in nums:
            cur_sum += n
            remainder = cur_sum % k

            if remainder in remainder_counts:
                count += remainder_counts[remainder]

            remainder_counts[remainder] = remainder_counts.get(remainder, 0) + 1

        return count
```