Find Subarrays With Equal Sum LeetCode Solution

Problem – Find Subarrays With Equal Sum LeetCode Solution

Given a 0-indexed integer array nums, determine whether there exist two subarrays of length 2 with equal sum. Note that the two subarrays must begin at different indices.

Return true if these subarrays exist, and false otherwise.

A subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

Input: nums = [4,2,4]
Output: true
Explanation: The subarrays with elements [4,2] and [2,4] have the same sum of 6.

Example 2:

Input: nums = [1,2,3,4,5]
Output: false
Explanation: No two subarrays of size 2 have the same sum.

Example 3:

Input: nums = [0,0,0]
Output: true
Explanation: The subarrays [nums[0],nums[1]] and [nums[1],nums[2]] have the same sum of 0. 
Note that even though the subarrays have the same content, the two subarrays are considered different because they are in different positions in the original array.

Constraints:

• 2 <= nums.length <= 1000
• -109 <= nums[i] <= 109

Find Subarrays With Equal Sum LeetCode Solution in C++

class Solution {
public:
	bool findSubarrays(vector<int>& nums) {
		int n=nums.size();
		int count=0;
		for(int i=0;i<n-1;i++){
			for(int j=i+1;j<n-1;j++){
				if(nums[j]+nums[j+1]==nums[i]+nums[i+1]) count++;
			}
		}
		return count>=1;
	}
};

Find Subarrays With Equal Sum LeetCode Solution in Python

class Solution:
    def findSubarrays(self, nums: List[int]) -> bool:
        sums = set()
        
        for i in range(len(nums)-1):
            t = nums[i]+nums[i+1]
            if t in sums:
                return True
            else:
                sums.add(t)
            
        return False

Find Subarrays With Equal Sum LeetCode Solution in Java

class Solution {
    public boolean findSubarrays(int[] nums) {
        
        if(nums.length<2)
        return false;
        
        HashSet<Integer> set=new HashSet<>(); //  To Track previous sum calculated 
        
        for(int i=0;i<nums.length-1;i++){
            
            if(!set.add(nums[i]+nums[i+1]))  //   Check if sum already exist 
            return true;
            
        }
        
        return false;    
        
    }
}

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