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和可被 K 整除的子数组

Subarray Sums Divisible by K

专题
Algorithmic Programming / 算法编程
难度
L3
来源
Citadel

题目详情

问题：和可被 K 整除的子数组

考察：数组、动态规划

来源：Citadel

链接：https://www.jointaro.com/interviews/questions/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
解析

思路：若两个前缀和模 k 的余数相同，它们之间的子数组和可被 k 整除。遍历数组维护余数计数，当前余数已有多少次就新增多少个答案。

复杂度：时间 O(n)，空间 O(k)。