10. Subarray with 0 sum

The problem can be found at the following link: Question Link

My Approach

To check if there is a subarray with a sum of 0, I maintain a running sum while traversing the array. I use an unordered set to store the cumulative sum at each step. If, at any point, I find that the sum already exists in the set, it means there is a subarray with a sum of 0. This is because the difference between the current sum and the sum stored in the set would be 0.

Explain with Example

Consider the array [4, 2, -3, 1, 6]. The running sum would be [4, 6, 3, 4, 10]. At the third step, the sum becomes 4, and we check if 4 is already in the set. Since it is, we know there is a subarray with a sum of 0 (from index 1 to 3: 2 - 3 + 1).

Time and Auxiliary Space Complexity

• Time Complexity : O(n), where n is the size of the array.

• Auxiliary Space Complexity : O(n), where n is the size of the array.

Code (C++)

class Solution {
public:
    bool subArrayExists(int arr[], int n) {
        int sum = 0;
        unordered_set<int> s;
        s.insert(0);
        
        for (int i = 0; i < n; i++) {
            sum += arr[i];
            if (s.find(sum) != s.end())
                return true;
            s.insert(sum);
        }
        
        return false;
    }
};

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