523. Continuous Subarray Sum

523. Continuous Subarray Sum

Medium

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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

Accepted

277,174

Submissions

1,024,226

class Solution:
    def checkSubarraySum(self, nums: List[int], k: int) -> bool:
        d=defaultdict(list)
        for i, sm in enumerate([0]+list(accumulate(nums))):
            d[sm%k]+=[i]
        
        for val in d:
            if d[val][-1]-d[val][0]>=2: return True
        return False

d中存的是余数对应的索引

之后再判断间隔就好