974 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

Constraints:

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

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class Solution:
    def subarraysDivByK(self, nums: List[int], k: int) -> int:
        # sum[i] - sum[j] = k * x
        # (sum[i] - sum[j]) % k  == (k * x) % k == 0
        # sum[i] % k == sum[j] % k
        res = 0
        prefix = 0
        count = [1] + [0] * k
        for num in nums:
            prefix = (prefix + num) % k
            res += count[prefix]
            count[prefix] += 1
        return res