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523. Continuous Subarray Sum

Description

Given an integer array nums and an integer k, return true if nums has a good subarray or false otherwise.

A good subarray is a subarray where:

  • its length is at least two, and
  • the sum of the elements of the subarray is a multiple of k.

Note that:

  • A subarray is a contiguous part of the array.
  • 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

  • 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;
        }
    }
    
  • 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;
        }
    };
    
  • 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
    
    
  • 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
    }
    
  • function checkSubarraySum(nums: number[], k: number): boolean {
        const n = nums.length;
        let prefixSum = 0;
        const map = new Map([[0, 0]]);
    
        for (let i = 0; i < n; i++) {
            prefixSum += nums[i];
            const remainder = prefixSum % k;
    
            if (!map.has(remainder)) {
                map.set(remainder, i + 1);
            } else if (i - map.get(remainder)! > 0) return true;
        }
    
        return false;
    }
    
    

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