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[10/26] 523. Continuous Subarray Sum
본문
Medium
3953391Add to ListShareGiven 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 <= 1050 <= nums[i] <= 1090 <= sum(nums[i]) <= 231 - 11 <= k <= 231 - 1
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Submissions
1,275,263
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Runtime: 1064 ms, faster than 91.09% of Python3 online submissions for Continuous Subarray Sum.
Memory Usage: 34 MB, less than 10.30% of Python3 online submissions for Continuous Subarray Sum.
Memory Usage: 34 MB, less than 10.30% of Python3 online submissions for Continuous Subarray Sum.
class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
dic = {0:-1}
modsum = 0
for idx, num in enumerate(nums):
modsum = (modsum + num)%k
if modsum not in dic:
dic[modsum] = idx
elif idx - dic[modsum] >= 2:
return True
return False
