README_EN.md

523. Continuous Subarray Sum

中文文档

Description

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

Solutions

Python3

class Solution:
    def checkSubarraySum(self, nums: List[int], k: int) -> bool:
        s = 0
        mp = {0: -1}
        for i, v in enumerate(nums):
            s += v
            r = s % k
            if r in mp and i - mp[r] >= 2:
                return True
            if r not in mp:
                mp[r] = i
        return False

Java

class Solution {
    public boolean checkSubarraySum(int[] nums, int k) {
        Map<Integer, Integer> mp = new HashMap<>();
        mp.put(0, -1);
        int s = 0;
        for (int i = 0; i < nums.length; ++i) {
            s += nums[i];
            int r = s % k;
            if (mp.containsKey(r) && i - mp.get(r) >= 2) {
                return true;
            }
            if (!mp.containsKey(r)) {
                mp.put(r, i);
            }
        }
        return false;
    }
}

C++

class Solution {
public:
    bool checkSubarraySum(vector<int>& nums, int k) {
        unordered_map<int, int> mp;
        mp[0] = -1;
        int s = 0;
        for (int i = 0; i < nums.size(); ++i)
        {
            s += nums[i];
            int r = s % k;
            if (mp.count(r) && i - mp[r] >= 2) return true;
            if (!mp.count(r)) mp[r] = i;
        }
        return false;
    }
};

Go

func checkSubarraySum(nums []int, k int) bool {
	mp := map[int]int{0: -1}
	s := 0
	for i, v := range nums {
		s += v
		r := s % k
		if j, ok := mp[r]; ok && i-j >= 2 {
			return true
		}
		if _, ok := mp[r]; !ok {
			mp[r] = i
		}
	}
	return false
}

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