523. Continuous Subarray Sum ⭐

Medium

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

解題

這一題使用到前綴和的概念與mod的概念。首先，我們建立一個map來記錄目前 sum % k 的餘數，如果該數出現第二遍，代表 上一次出現該餘數的 index+1 到目前index 的值加總等於 k ，所以餘數才會再次出現。

func checkSubarraySum(nums []int, k int) bool {
    m := make(map[int]int)
    m[0] = -1
    
    sum := 0
    for i, val := range nums {
        sum += val
        if k != 0 { sum %= k }
        
        if preindex, ok := m[sum]; ok { 
            if i-preindex>1 { return true }
        } else {
            m[sum] = i
        }
    }
    
    return false
}