Find if there is a subarray with 0 sum - GeeksforGeeks

11 Feb, 2022
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
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++
Java
Python3
C#
Javascript
// 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 codeint 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;}
// A Java program to find// if there is a zero sum subarrayimport 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 code    public static void main(String arg[])    {        int arr[] = { -3, 2, 3, 1, 6 };        if (subArrayExists(arr))            System.out.println(                "Found a subarray with 0 sum");        else            System.out.println("No Such Sub Array Exists!");    }}
# A python 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 codearr = [-3, 2, 3, 1, 6]n = len(arr)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
// A C# program to find if there// is a zero sum subarrayusing 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);        }         // We reach here only when there is        // no subarray with 0 sum        return false;    }     // Main Method    public static void Main()    {        int[] arr = { -3, 2, 3, 1, 6 };        if (SubArrayExists(arr))            Console.WriteLine(                "Found a subarray with 0 sum");        else            Console.WriteLine("No Such Sub Array Exists!");    }}
// 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!");
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. 
Exercise: Extend the above program to print starting and ending indexes of all subarrays with 0 sum.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 
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