0974._Subarray_Sums_Divisible_by_K.md

974. Subarray Sums Divisible by K

难度: Medium

刷题内容

原题连接

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

内容描述

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

解题方案

思路 1 - 时间复杂度: O(N)- 空间复杂度: O(N)******

一看就是前缀和，处理比较巧妙，两个index的前缀和相等说明中间那个区间都是K的倍数

class Solution:
    def subarraysDivByK(self, A, K):
        """
        :type A: List[int]
        :type K: int
        :rtype: int
        """
        prefix = [0]
        for num in A:
            prefix.append((prefix[-1] + num) % K)
        count = collections.Counter(prefix)
        return sum(v * (v-1) // 2 for v in count.values())