[leetcode] 974. Subarray Sums Divisible by K

Description

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. 1 <= A.length <= 30000
2. -10000 <= A[i] <= 10000
3. 2 <= K <= 10000

分析

题目的意思是：这道题统计连续子串能够被K整除的个数，解法还是挺奇妙的，首先给P数组放一个0，然后遍历A进行求和，P[i+1] = A[0] + A[1] + … + A[i]. 然后再进行数的频率统计，最后统计的结果按照v*(v-1)/2求和就行了，不要问我为什么这样做就行了，我也不知道。但是我验证了一下结果却是是对的。

代码

class Solution:
    def subarraysDivByK(self, A: List[int], K: int) -> int:
        n=len(A)
        P=[0]
        for x in A:
            P.append((P[-1]+x)%K)
        count=collections.Counter(P)
        cnt=0
        for v in count.values():
            cnt+=v*(v-1)/2
        return int(cnt)

参考文献

[LeetCode] Approach 1: Prefix Sums and Counting