Skip to content
Related Articles

Related Articles

Find the length of largest subarray with 0 sum

Improve Article
Save Article
  • Difficulty Level : Medium
  • Last Updated : 17 Oct, 2022
Improve Article
Save Article

Given an array arr[] of length N, find the length of the longest sub-array with a sum equal to 0.

Examples:

Input: arr[] = {15, -2, 2, -8, 1, 7, 10, 23}
Output: 5
Explanation: The longest sub-array with elements summing up-to 0 is {-2, 2, -8, 1, 7}

Input: arr[] = {1, 2, 3}
Output: 0
Explanation: There is no subarray with 0 sum

Input:  arr[] = {1, 0, 3}
Output:  1
Explanation: The longest sub-array with elements summing up-to 0 is {0}

Naive Approach: Follow the steps below to solve the problem using this approach:

  • Consider all sub-arrays one by one and check the sum of every sub-array.
  • If the sum of the current subarray is equal to zero then update the maximum length accordingly

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Returns length of the largest
// subarray with 0 sum
int maxLen(int arr[], int N)
{
    // Initialize result
    int max_len = 0;
 
    // Pick a starting point
    for (int i = 0; i < N; i++) {
 
        // Initialize currr_sum for
        // every starting point
        int curr_sum = 0;
 
        // Try all subarrays starting with 'i'
        for (int j = i; j < N; j++) {
            curr_sum += arr[j];
 
            // If curr_sum becomes 0,
            // then update max_len
            // if required
            if (curr_sum == 0)
                max_len = max(max_len, j - i + 1);
        }
    }
    return max_len;
}
 
// Driver's Code
int main()
{
    int arr[] = {15, -2, 2, -8, 1, 7, 10, 23};
    int N = sizeof(arr) / sizeof(arr[0]);
   
  // Function call
    cout << "Length of the longest 0 sum subarray is "
         << maxLen(arr, N);
    return 0;
}

Java




// Java code for the above approach
 
class GFG {
   
    // Returns length of the largest subarray
    // with 0 sum
    static int maxLen(int arr[], int N)
    {
        int max_len = 0;
 
        // Pick a starting point
        for (int i = 0; i < N; i++) {
           
            // Initialize curr_sum for every
            // starting point
            int curr_sum = 0;
 
            // try all subarrays starting with 'i'
            for (int j = i; j < N; j++) {
                curr_sum += arr[j];
 
                // If curr_sum becomes 0, then update
                // max_len
                if (curr_sum == 0)
                    max_len = Math.max(max_len, j - i + 1);
            }
        }
        return max_len;
    }
 
  // Driver's code
    public static void main(String args[])
    {
        int arr[] = {15, -2, 2, -8, 1, 7, 10, 23};
        int N = arr.length;
       
      // Function call
        System.out.println("Length of the longest 0 sum "
                           + "subarray is " + maxLen(arr, N));
    }
}

Python3




# Python program for the above approach
 
# returns the length
def maxLen(arr):
     
    # initialize result
    max_len = 0
 
    # pick a starting point
    for i in range(len(arr)):
         
        # initialize sum for every starting point
        curr_sum = 0
         
        # try all subarrays starting with 'i'
        for j in range(i, len(arr)):
         
            curr_sum += arr[j]
 
            # if curr_sum becomes 0, then update max_len
            if curr_sum == 0:
                max_len = max(max_len, j-i + 1)
 
    return max_len
 
# Driver's code
if __name__ == "__main__":
# test array
    arr = [15, -2, 2, -8, 1, 7, 10, 13]
     
    # Function call
    print ("Length of the longest 0 sum subarray is % d" % maxLen(arr))

C#




// C# code for the above approach
using System;
 
class GFG {
     
    // Returns length of the
    // largest subarray with 0 sum
    static int maxLen(int[] arr, int N)
    {
        int max_len = 0;
 
        // Pick a starting point
        for (int i = 0; i < N; i++) {
             
            // Initialize curr_sum
            // for every starting point
            int curr_sum = 0;
 
            // try all subarrays
            // starting with 'i'
            for (int j = i; j < N; j++) {
                curr_sum += arr[j];
 
                // If curr_sum becomes 0,
                // then update max_len
                if (curr_sum == 0)
                    max_len = Math.Max(max_len,
                                       j - i + 1);
            }
        }
        return max_len;
    }
 
    // Driver's code
    static public void Main()
    {
        int[] arr = {15, -2, 2, -8,
                      1, 7, 10, 23};
        int N = arr.Length;
         
        // Function call
        Console.WriteLine("Length of the longest 0 sum "
                          + "subarray is " + maxLen(arr, N));
    }
}

PHP




<?php
// PHP program for the above approach
 
// Returns length of the
// largest subarray with 0 sum
function maxLen($arr, $N)
{
    // Initialize result
    $max_len = 0;
 
    // Pick a starting point
    for ($i = 0; $i < $N; $i++)
    {
        // Initialize currr_sum
        // for every starting point
        $curr_sum = 0;
 
        // try all subarrays
        // starting with 'i'
        for ($j = $i; $j < $N; $j++)
        {
            $curr_sum += $arr[$j];
 
            // If curr_sum becomes 0,
            // then update max_len
            // if required
            if ($curr_sum == 0)
            $max_len = max($max_len,
                           $j - $i + 1);
        }
    }
    return $max_len;
}
 
// Driver Code
$arr = array(15, -2, 2, -8,
              1, 7, 10, 23);
$N = sizeof($arr);
 
// Function call
echo "Length of the longest 0 " .
              "sum subarray is ",
                maxLen($arr, $N);
     
?>

Javascript




<script>
    // Javascript code to find the largest
    // subarray with 0 sum
     
    // Returns length of the
    // largest subarray with 0 sum
    function maxLen(arr, N)
    {
        let max_len = 0;
   
        // Pick a starting point
        for (let i = 0; i < N; i++) {
            // Initialize curr_sum
            // for every starting point
            let curr_sum = 0;
   
            // try all subarrays
            // starting with 'i'
            for (let j = i; j < N; j++) {
                curr_sum += arr[j];
   
                // If curr_sum becomes 0,
                // then update max_len
                if (curr_sum == 0)
                    max_len = Math.max(max_len, j - i + 1);
            }
        }
        return max_len;
    }
     
    // Driver's code
    let arr = [15, -2, 2, -8, 1, 7, 10, 23];
    let N = arr.length;
     
    // Function call
    document.write("Length of the longest 0 sum " + "subarray is " + maxLen(arr, N));
     
</script>

Output

Length of the longest 0 sum subarray is 5

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

Find the length of the largest subarray with 0 sum using hashmap:

Follow the below idea to solve the problem using this approach: 

Let us say prefixsum of array till index i is represented as Si .
Now consider two indices i and j (j > i) such that Si = Sj .

So, 
Si = arr[0] + arr[1] + . . . + arr[i]
Sj = arr[0] + arr[1] + . . . + arr[i] + arr[i+1] + . . . + arr[j]

Now if we subtract Si from Sj .
Sj – Si = (arr[0] + arr[1] + . . . + arr[i] + arr[i+1] + . . . + arr[j]) – (arr[0] + arr[1] + . . . + arr[i])
0 = (arr[0] – arr[0]) + (arr[1] – arr[1]) + . . . + (arr[i] – arr[i]) + arr[i+1] + arr[i+2] + . . . + arr[j]
0 = arr[i+1] + arr[i+2] + . . . + arr[j]

So we can see if there are two indices i and j (j > i) for which the prefix sum are same then the subarray from i+1 to j has sum = 0.

We can use hashmap to store the prefix sum, and if we reach any index for which there is already a prefix with same sum, we will find a subarray with sum as 0. Compare the length of that subarray with the current longest subarray and update the maximum value accordingly.

Follow the steps mentioned below to implement the approach:

  • Create a variable (sum), length (max_len), and a hash map (hm) to store the sum-index pair as a key-value pair.
  • Traverse the input array and for every index, 
    • Update the value of sum = sum + array[i].
    • Check every index, if the current sum is present in the hash map or not.
    • If present, update the value of max_len to a maximum difference of two indices (current index and index in the hash-map) and max_len.
    • Else, put the value (sum) in the hash map, with the index as a key-value pair.
  • Print the maximum length (max_len).

Below is a dry run of the above approach: 

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Returns Length of the required subarray
int maxLen(int arr[], int N)
{
    // Map to store the previous sums
    unordered_map<int, int> presum;
 
    int sum = 0; // Initialize the sum of elements
    int max_len = 0; // Initialize result
 
    // Traverse through the given array
    for (int i = 0; i < N; i++) {
 
        // Add current element to sum
        sum += arr[i];
        if (sum == 0)
            max_len = i + 1;
 
        // Look for this sum in Hash table
        if (presum.find(sum) != presum.end()) {
 
            // If this sum is seen before, then update
            // max_len
            max_len = max(max_len, i - presum[sum]);
        }
        else {
            // Else insert this sum with index
            // in hash table
            presum[sum] = i;
        }
    }
 
    return max_len;
}
 
// Driver's Code
int main()
{
    int arr[] = { 15, -2, 2, -8, 1, 7, 10, 23 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function call
    cout << "Length of the longest 0 sum subarray is "
         << maxLen(arr, N);
 
    return 0;
}

Java




// Java program for the above approach
 
import java.util.HashMap;
 
class MaxLenZeroSumSub {
 
    // Returns length of the maximum length
    // subarray with 0 sum
    static int maxLen(int arr[])
    {
        // Creates an empty hashMap hM
        HashMap<Integer, Integer> hM
            = new HashMap<Integer, Integer>();
 
        int sum = 0; // Initialize sum of elements
        int max_len = 0; // Initialize result
 
        // Traverse through the given array
        for (int i = 0; i < arr.length; i++) {
            // Add current element to sum
            sum += arr[i];
 
            if (sum == 0)
                max_len = i + 1;
 
            // Look this sum in hash table
            Integer prev_i = hM.get(sum);
 
            // If this sum is seen before, then update
            // max_len if required
            if (prev_i != null)
                max_len = Math.max(max_len, i - prev_i);
            else // Else put this sum in hash table
                hM.put(sum, i);
        }
 
        return max_len;
    }
 
    // Drive's code
    public static void main(String arg[])
    {
        int arr[] = { 15, -2, 2, -8, 1, 7, 10, 23 };
 
        // Function call
        System.out.println(
            "Length of the longest 0 sum subarray is "
            + maxLen(arr));
    }
}

Python3




# Python program for the above approach
 
# Returns the maximum length
 
 
def maxLen(arr):
 
    # NOTE: Dictionary in python is
    # implemented as Hash Maps
    # Create an empty hash map (dictionary)
    hash_map = {}
 
    # Initialize result
    max_len = 0
 
    # Initialize sum of elements
    curr_sum = 0
 
    # Traverse through the given array
    for i in range(len(arr)):
 
        # Add the current element to the sum
        curr_sum += arr[i]
 
        if curr_sum == 0:
            max_len = i + 1
 
        # NOTE: 'in' operation in dictionary
        # to search key takes O(1). Look if
        # current sum is seen before
        if curr_sum in hash_map:
            max_len = max(max_len, i - hash_map[curr_sum])
        else:
 
            # else put this sum in dictionary
            hash_map[curr_sum] = i
 
    return max_len
 
 
# Driver's code
if __name__ == "__main__":
 
    # test array
    arr = [15, -2, 2, -8, 1, 7, 10, 13]
 
    # Function call
    print("Length of the longest 0 sum subarray is % d" % maxLen(arr))

C#




// C# program for the above approach
 
using System;
using System.Collections.Generic;
 
public class MaxLenZeroSumSub {
 
    // Returns length of the maximum
    // length subarray with 0 sum
    static int maxLen(int[] arr)
    {
        // Creates an empty hashMap hM
        Dictionary<int, int> hM
            = new Dictionary<int, int>();
 
        int sum = 0; // Initialize sum of elements
        int max_len = 0; // Initialize result
 
        // Traverse through the given array
        for (int i = 0; i < arr.GetLength(0); i++) {
 
            // Add current element to sum
            sum += arr[i];
 
            if (arr[i] == 0 && max_len == 0)
                max_len = 1;
 
            if (sum == 0)
                max_len = i + 1;
 
            // Look this sum in hash table
            int prev_i = 0;
            if (hM.ContainsKey(sum)) {
                prev_i = hM[sum];
            }
 
            // If this sum is seen before, then update
            // max_len if required
            if (hM.ContainsKey(sum))
                max_len = Math.Max(max_len, i - prev_i);
            else {
                // Else put this sum in hash table
                if (hM.ContainsKey(sum))
                    hM.Remove(sum);
 
                hM.Add(sum, i);
            }
        }
 
        return max_len;
    }
 
    // Driver's code
    public static void Main()
    {
        int[] arr = { 15, -2, 2, -8, 1, 7, 10, 23 };
 
        // Function call
        Console.WriteLine(
            "Length of the longest 0 sum subarray is "
            + maxLen(arr));
    }
}

Javascript




<script>
 
// Javascript program to find maximum length subarray with 0 sum
 
    // Returns length of the maximum length subarray with 0 sum
    function maxLen(arr)
    {
        // Creates an empty hashMap hM
        let hM = new Map();
  
        let sum = 0; // Initialize sum of elements
        let max_len = 0; // Initialize result
  
        // Traverse through the given array
        for (let i = 0; i < arr.length; i++) {
             
            // Add current element to sum
            sum += arr[i];
  
            if (arr[i] == 0 && max_len == 0)
                max_len = 1;
  
            if (sum == 0)
                max_len = i + 1;
  
            // Look this sum in hash table
            let prev_i = hM.get(sum);
  
            // If this sum is seen before, then update max_len
            // if required
            if (prev_i != null)
                max_len = Math.max(max_len, i - prev_i);
                 
            else // Else put this sum in hash table
                hM.set(sum, i);
        }
  
        return max_len;
    }
 
 
// Driver's program
 
     let arr = [15, -2, 2, -8, 1, 7, 10, 23];
     // Function call
     document.write("Length of the longest 0 sum subarray is "
                           + maxLen(arr));
       
</script>

Output

Length of the longest 0 sum subarray is 5

Time Complexity: O(N)
Auxiliary Space: O(N)


My Personal Notes arrow_drop_up
Related Articles

Start Your Coding Journey Now!