C++ Program to Find if there is a subarray with 0 sum

Given an array of positive and negative numbers, find if there is a subarray (of size at-least one) with 0 sum.

Examples :

Input: {4, 2, -3, 1, 6}
Output: true 
Explanation:
There is a subarray with zero sum from index 1 to 3.

Input: {4, 2, 0, 1, 6}
Output: true 
Explanation:
There is a subarray with zero sum from index 2 to 2.

Input: {-3, 2, 3, 1, 6}
Output: false

A simple solution is to consider all subarrays one by one and check the sum of every subarray. We can run two loops: the outer loop picks a starting point i and the inner loop tries all subarrays starting from i (See this for implementation). The time complexity of this method is O(n²).
We can also use hashing. The idea is to iterate through the array and for every element arr[i], calculate the sum of elements from 0 to i (this can simply be done as sum += arr[i]). If the current sum has been seen before, then there is a zero-sum array. Hashing is used to store the sum values so that we can quickly store sum and find out whether the current sum is seen before or not.
Example :

arr[] = {1, 4, -2, -2, 5, -4, 3}

If we consider all prefix sums, we can
notice that there is a subarray with 0
sum when :
1) Either a prefix sum repeats or
2) Or prefix sum becomes 0.

Prefix sums for above array are:
1, 5, 3, 1, 6, 2, 5

Since prefix sum 1 repeats, we have a subarray
with 0 sum. 

Following is implementation of the above approach. 

C++

// A C++ program to find if
// there is a zero sum subarray
#include <bits/stdc++.h>
using namespace std;

bool subArrayExists(int arr[], int n)
{
    unordered_set<int> sumSet;

    // Traverse through array
    // and store prefix sums
    int sum = 0;
    for (int i = 0; i < n; i++)
    {
        sum += arr[i];

        // If prefix sum is 0 or
        // it is already present
        if (sum == 0
            || sumSet.find(sum)
            != sumSet.end())
            return true;

        sumSet.insert(sum);
    }
    return false;
}

// Driver code
int main()
{
    int arr[] = { -3, 2, 3, 1, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
    if (subArrayExists(arr, n))
        cout << "Found a subarray with 0 sum";
    else
        cout << "No Such Sub Array Exists!";
    return 0;
}

Output

No Such Sub Array Exists!

Time Complexity of this solution can be considered as O(n) under the assumption that we have good hashing function that allows insertion and retrieval operations in O(1) time. 
Space Complexity: O(n) .Here we required extra space for unordered_set to insert array elements.

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