523. Continuous Subarray Sum

LeetCode 523. Continuous Subarray Sum

Description

Given an integer array nums and an integer k, return true ifnums has a continuous subarray of size at least two whose elements sum up to a multiple of k , orfalse 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

Tags

Math, Dynamic Programming

Solution

To solve this problem, we can use prefix-sum strategy to check if each subarray whose elements sum up to a multiple of k. However, it is an O(n²) time complexity operation leading to timeout. We observe that if preSum[j] - preSum[i] == n * k, which means the sum of subarray nums(i,j] is a multiple of k, both preSum[i] and preSum[j] share the same remainder after mod k. Therefore, our goal is changed to find nums[i] and nums[j] meet the requirements that nums[i] % k == nums[j] % k and j-i ≥ 2. Obviously, we can achieve this within O(n) time complexity by applying a hash table.

Complexity

• Time complexity: $O(n)$

• Space complexity: $O(\min(n,k))$

Code

func checkSubarraySum(nums []int, k int) bool {
	rem2idx, remainder := map[int]int{0: -1}, 0
	for i, num := range nums {
		remainder = (remainder + num) % k
		if preIdx, ok := rem2idx[remainder]; ok {
			if i-preIdx > 1 {
				return true
			}
		} else {
			rem2idx[remainder] = i
		}
	}
	return false
}

Reference

1. 连续的子数组和