523. Continuous Subarray Sum

Medium (Trung bình) Python 🔗 Xem trên LeetCode

📋 Đề Bài

Given an integer array nums and an integer k, return true if nums has a continuous subarray of size at least two whose elements sum up to a multiple of k, or false otherwise.

An integer x is a multiple of k if there exists an integer n such that x = n * k. 0 is always a multiple of k.

 

Example 1:

Input: nums = [23,2,4,6,7], k = 6
Output: true
Explanation: [2, 4] is a continuous subarray of size 2 whose elements sum up to 6.

Example 2:

Input: nums = [23,2,6,4,7], k = 6
Output: true
Explanation: [23, 2, 6, 4, 7] is an continuous subarray of size 5 whose elements sum up to 42.
42 is a multiple of 6 because 42 = 7 * 6 and 7 is an integer.

Example 3:

Input: nums = [23,2,6,4,7], k = 13
Output: false

 

Constraints:

  • 1 <= nums.length <= 105
  • 0 <= nums[i] <= 109
  • 0 <= sum(nums[i]) <= 231 - 1
  • 1 <= k <= 231 - 1

💻 Lời Giải

Python 0523-continuous-subarray-sum.py 
class Solution:
    def checkSubarraySum(self, nums: List[int], k: int) -> bool:
        seen, cur = {0: -1}, 0
        for i, a in enumerate(nums):
            cur = (cur + a) % abs(k) if k else cur + a
            if i - seen.setdefault(cur, i) > 1: return True
        return False