Find if there is a subarray with 0 sum

3.1

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 
There is a subarray with zero sum from index 1 to 3.

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

Input: {-3, 2, 3, 1, 6}
Output: false
There is no subarray with zero sum.

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). 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 sum of elements form 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_map<int,bool> sumMap;

    // Traverse throught 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 || sumMap[sum] == true)
            return true;

        sumMap[sum] = true;
    }
    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;
}

Java

// A Java program to find if there is a zero sum subarray
import java.util.HashMap;
 
class ZeroSumSubarray {
    
    // Returns true if arr[] has a subarray with sero sum
    static Boolean printZeroSumSubarray(int arr[])
    {
        // Creates an empty hashMap hM
        HashMap<Integer, Integer> hM = 
                          new HashMap<Integer, 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 map
            if (arr[i] == 0 || sum == 0 || hM.get(sum) != null)                            
               return true;
            
            // Add sum to hash map
            hM.put(sum, i);
        }    
        
        // We reach here only when there is no subarray with 0 sum
        return false;
    }        
    
    public static void main(String arg[])
    {
        int arr[] = {4, 2, -3, 1, 6};
        if (printZeroSumSubarray(arr))
            System.out.println("Found a subarray with 0 sum");
        else
            System.out.println("No Subarray with 0 sum");            
    }            
}


Output:
Found a subarray with 0 sum

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.

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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