Last updated on October 5th, 2024 at 04:51 pm

Here, We see Continuous Subarray Sum LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

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Continuous Subarray Sum LeetCode Solution

Problem Statement

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

1. Continuous Subarray Sum LeetCode Solution C++

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

2. Continuous Subarray Sum LeetCode Solution Java

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

3. Continuous Subarray Sum Solution JavaScript

var checkSubarraySum = function (nums, k) {
	let sum = 0
	let prefix = 0;
	const hash = new Set();
	for (let i = 0; i < nums.length; i++) {
		sum += nums[i]
		if (k != 0) sum %= k
		if(hash.has(sum)) return true
		hash.add(prefix);
		prefix = sum;
	}
	return false
};

4. Continuous Subarray Sum Solution Python

class Solution(object):
    def checkSubarraySum(self, nums, k):
        d , s = {0:-1} , 0
        for i, n in enumerate(nums):
            if k != 0:
                s = (s + n) % k
            else:
                s += n
            if s not in d:
                d[s] = i
            else:
                if i - d[s] >= 2:
                    return True
        return False