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: The third element is zero. A single element is also a sub-array.

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

Naive approach:

consider all subarrays one by one and check the sum of every subarray. Run two loops: the outer loop picks a starting point i and the inner loop tries all subarrays starting from i 

C++

// C++ program to find if
// there is a zero sum subarray
#include <iostream>
using namespace std;

bool subArrayExists(int arr[], int n)
{
    for (int i = 0; i < n; i++) {
        int sum = arr[i];
        if (sum == 0)
            return true;
        for (int j = i + 1; j < n; j++) {
            sum += arr[j];
            if (sum == 0)
                return true;
        }
    }
    return false;
}

// Driver's code
int main()
{
    int arr[] = { -3, 2, 3, 1, 6 };
    int N = sizeof(arr) / sizeof(arr[0]);

    // Function call
    if (subArrayExists(arr, N))
        cout << "Found a subarray with 0 sum";
    else
        cout << "No Such Sub Array Exists!";
    return 0;
}

// This code is contributed by Tapesh(tapeshdua420)

Output

No Such Sub Array Exists!

Time Complexity: O(N²)
Auxiliary Space: O(1)

Find if there is a subarray with 0 sum using 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 sum can be stored quickly and find out whether the current sum is seen before or not.

Follow the given steps to solve the problem:

• Declare a variable sum, to store the sum of prefix elements
• Traverse the array and at each index, add the element into the sum and check if this sum exists earlier. If the sum exists, then return true
• Also, insert every prefix sum into a map, so that later on it can be found whether the current sum is seen before or not
• At the end return false, as no such subarray is found

Illustration:

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

Consider all prefix sums, one can notice that there is a subarray with 0 sum when :

• Either a prefix sum repeats or
• 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. 

Below is the Implementation of the above approach:

C++

// 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's code
int main()
{
    int arr[] = {-3, 2, 3, 1, 6};
    int N = sizeof(arr) / sizeof(arr[0]);

    // Function call
    if (subArrayExists(arr, N))
        cout << "Found a subarray with 0 sum";
    else
        cout << "No Such Sub Array Exists!";
    return 0;
}

Java

// Java program to find
// if there is a zero sum subarray

import java.util.HashSet;
import java.util.Set;

class ZeroSumSubarray {

    // Returns true if arr[]
    // has a subarray with sero sum
    static Boolean subArrayExists(int arr[])
    {
        // Creates an empty hashset hs
        Set<Integer> hs = new HashSet<Integer>();

        // Initialize sum of elements
        int sum = 0;

        // Traverse through the given array
        for (int i = 0; i < arr.length; i++) {
            // Add current element to sum
            sum += arr[i];

            // Return true in following cases
            // a) Current element is 0
            // b) sum of elements from 0 to i is 0
            // c) sum is already present in hash set
            if (arr[i] == 0 || sum == 0 || hs.contains(sum))
                return true;

            // Add sum to hash set
            hs.add(sum);
        }

        // We reach here only when there is
        // no subarray with 0 sum
        return false;
    }

    // Driver's code
    public static void main(String arg[])
    {
        int arr[] = {-3, 2, 3, 1, 6};

        // Function call
        if (subArrayExists(arr))
            System.out.println(
                "Found a subarray with 0 sum");
        else
            System.out.println("No Such Sub Array Exists!");
    }
}

Python3

# python3 program to find if
# there is a zero sum subarray


def subArrayExists(arr, N):
    # traverse through array
    # and store prefix sums
    n_sum = 0
    s = set()

    for i in range(N):
        n_sum += arr[i]

        # If prefix sum is 0 or
        # it is already present
        if n_sum == 0 or n_sum in s:
            return True
        s.add(n_sum)

    return False


# Driver's code
if __name__ == '__main__':
    arr = [-3, 2, 3, 1, 6]
    N = len(arr)

    # Function call
    if subArrayExists(arr, N) == True:
        print("Found a sunbarray with 0 sum")
    else:
        print("No Such sub array exits!")

# This code is contributed by Shrikant13

C#

// C# program to find if there
// is a zero sum subarray

using System;
using System.Collections.Generic;

class GFG {

    // Returns true if arr[] has
    // a subarray with sero sum
    static bool SubArrayExists(int[] arr)
    {
        // Creates an empty HashSet hM
        HashSet<int> hs = new HashSet<int>();
        // Initialize sum of elements
        int sum = 0;

        // Traverse through the given array
        for (int i = 0; i < arr.Length; i++) {
            // Add current element to sum
            sum += arr[i];

            // Return true in following cases
            // a) Current element is 0
            // b) sum of elements from 0 to i is 0
            // c) sum is already present in hash set
            if (arr[i] == 0 || sum == 0 || hs.Contains(sum))
                return true;

            // Add sum to hash set
            hs.Add(sum);
        }

        // Reach here only when there is
        // no subarray with 0 sum
        return false;
    }

    // Driver's code
    public static void Main()
    {
        int[] arr = {-3, 2, 3, 1, 6};

        // Function call
        if (SubArrayExists(arr))
            Console.WriteLine(
                "Found a subarray with 0 sum");
        else
            Console.WriteLine("No Such Sub Array Exists!");
    }
}

Javascript

// A Javascript program to
//  find if there is a zero sum subarray

const subArrayExists = (arr) => {
    const sumSet = new Set();

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

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

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

// Driver code

const arr =  [-3, 2, 3, 1, 6];
if (subArrayExists(arr))
    console.log("Found a subarray with 0 sum");
else
    console.log("No Such Sub Array Exists!");

Output

No Such Sub Array Exists!

Time Complexity: O(N) under the assumption that a good hashing function is used, that allows insertion and retrieval operations in O(1) time. 
Auxiliary Space: O(N) Here extra space is required for hashing

This article is contributed by Chirag Gupta. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above