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Find the length of largest subarray with 0 sum
  • Difficulty Level : Medium
  • Last Updated : 25 May, 2021

Given an array of integers, find the length of the longest sub-array with a sum that equals 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: This involves the use of brute force where two nested loops are used. The outer loop is used to fix the starting position of the sub-array, and the inner loop is used for the ending position of the sub-array and if the sum of elements is equal to zero, then increase the count.
 

Algorithm:  

  1. Consider all sub-arrays one by one and check the sum of every sub-array.
  2. Run two loops: the outer loop picks the starting point i and the inner loop tries all sub-arrays starting from i.

Implementation: 



C++




/* A simple C++ program to find
largest subarray with 0 sum */
#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 Code
int main()
{
    int arr[] = { 15, -2, 2, -8, 1, 7, 10, 23 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << "Length of the longest 0 sum subarray is "
         << maxLen(arr, n);
    return 0;
}

Java




// Java code to find the largest subarray
// with 0 sum
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;
    }
 
    public static void main(String args[])
    {
        int arr[] = { 15, -2, 2, -8, 1, 7, 10, 23 };
        int n = arr.length;
        System.out.println("Length of the longest 0 sum "
                           + "subarray is " + maxLen(arr, n));
    }
}
// This code is contributed by Kamal Rawal

Python




# Python program to find the length of largest subarray with 0 sum
 
# 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
 
 
# test array
arr = [15, -2, 2, -8, 1, 7, 10, 13]
 
print "Length of the longest 0 sum subarray is % d" % maxLen(arr)

C#




// C# code to find the largest
// subarray with 0 sum
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 code
    static public void Main()
    {
        int[] arr = { 15, -2, 2, -8,
                      1, 7, 10, 23 };
        int n = arr.Length;
        Console.WriteLine("Length of the longest 0 sum "
                          + "subarray is " + maxLen(arr, n));
    }
}
 
// This code is contributed by ajit

PHP




<?php
// A simple PHP program to find
// largest subarray with 0 sum
 
// Returns length of the
// largest subarray with 0 sum
function maxLen($arr, $n)
{
    $max_len = 0; // Initialize result
 
    // 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);
echo "Length of the longest 0 " .
              "sum subarray is ",
                maxLen($arr, $n);
     
// This code is contributed by aj_36
?>

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;
    }
     
    let arr = [ 15, -2, 2, -8, 1, 7, 10, 23 ];
    let n = arr.length;
    document.write("Length of the longest 0 sum " + "subarray is " + maxLen(arr, n));
     
</script>

Output:

Length of the longest 0 sum subarray is 5

Complexity Analysis:  

  • Time Complexity: O(n^2) due to the use of nested loops.
  • Space complexity: O(1) as no extra space is used.

 
Efficient Approach: The brute force solution is calculating the sum of each and every sub-array and checking whether the sum is zero or not. Let’s now try to improve the time complexity by taking an extra space of ‘n’ length. The new array will store the sum of all the elements up to that index. The sum-index pair will be stored in a hash-map. A Hash map allows insertion and deletion of key-value pair in constant time. Therefore, the time complexity remains unaffected. So, if the same value appears twice in the array, it will be guaranteed that the particular array will be a zero-sum sub-array. 
 

Mathematical Proof:  

prefix(i) = arr[0] + arr[1] +…+ arr[i] 
prefix(j) = arr[0] + arr[1] +…+ arr[j], j>i 
ifprefix(i) == prefix(j) then prefix(j) – prefix(i) = 0 that means arr[i+1] + .. + arr[j] = 0, So a sub-array has zero sum , and the length of that sub-array is j-i+1  

Algorithm:  

  1. Create an extra space, an array of length n (prefix), a variable (sum), length (max_len), and a hash map (hm) to store the sum-index pair as a key-value pair.
  2. Move along the input array from the start to the end.
  3. For every index, update the value of sum = sum + array[i].
  4. Check every index, if the current sum is present in the hash map or not.
  5. 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.
  6. Else, put the value (sum) in the hash map, with the index as a key-value pair.
  7. Print the maximum length (max_len).

Below is a dry run of the above approach: 

Implementation: 

C++




// C++ program to find the length of largest subarray
// with 0 sum
#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 (arr[i] == 0 && max_len == 0)
            max_len = 1;
        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 Code
int main()
{
    int arr[] = { 15, -2, 2, -8, 1, 7, 10, 23 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << "Length of the longest 0 sum subarray is "
         << maxLen(arr, n);
 
    return 0;
}

Java




// A Java program to find maximum length subarray with 0 sum
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 (arr[i] == 0 && max_len == 0)
                max_len = 1;
 
            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 method
    public static void main(String arg[])
    {
        int arr[] = { 15, -2, 2, -8, 1, 7, 10, 23 };
        System.out.println("Length of the longest 0 sum subarray is "
                           + maxLen(arr));
    }
}

Python




# A python program to find maximum length subarray
# with 0 sum in o(n) time
 
# Returns the maximum length
def maxLen(arr):
     
    # NOTE: Dictonary in python in 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 arr[i] is 0 and max_len is 0:
            max_len = 1
 
        if curr_sum is 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
 
 
# test array
arr = [15, -2, 2, -8, 1, 7, 10, 13]
  
print "Length of the longest 0 sum subarray is % d" % maxLen(arr)

C#




// C# program to find maximum
// length subarray with 0 sum
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 code
    public static void Main()
    {
        int[] arr = { 15, -2, 2, -8, 1, 7, 10, 23 };
        Console.WriteLine("Length of the longest 0 sum subarray is "
                          + maxLen(arr));
    }
}
 
/* This code contributed by PrinciRaj1992 */

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 program
 
     let arr = [ 15, -2, 2, -8, 1, 7, 10, 23 ];
     document.write("Length of the longest 0 sum subarray is "
                           + maxLen(arr));
       
</script>

Output: 

Length of the longest 0 sum subarray is 5

Complexity Analysis: 

  • Time Complexity: O(n), as use of the good hashing function, will allow insertion and retrieval operations in O(1) time.
  • Space Complexity: O(n), for the use of extra space to store the prefix array and hashmap.

 

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