# Find the length of largest subarray with 0 sum

Given an array of integers, find length of the largest subarray with sum equals to 0.

Examples :

```Input: arr[] = {15, -2, 2, -8, 1, 7, 10, 23};
Output: 5
The largest subarray with 0 sum is -2, 2, -8, 1, 7

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

Input: arr[] = {1, 0, 3}
Output: 1
```

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. Time complexity of this method is O(n2).

Below are implementations of this solution.

## C/C++

 `/* A simple C++ program to find largest subarray with 0 sum */` `#include ` `using` `namespace` `std; ` ` `  `// Returns length of the largest subarray with 0 sum ` `int` `maxLen(``int` `arr[], ``int` `n) ` `{ ` `    ``int` `max_len = 0; ``// Initialize result ` ` `  `    ``// 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 program to test above function ` `int` `main() ` `{ ` `    ``int` `arr[] = {15, -2, 2, -8, 1, 7, 10, 23}; ` `    ``int` `n = ``sizeof``(arr)/``sizeof``(arr); ` `    ``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

 ` `

Output :

`Length of the longest 0 sum subarray is 5`

We can Use Hashing to solve this problem in O(n) time. 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.

Following are implementations of the above approach.

## C++

 `// C++ program to find the length of largest subarray  ` `// with 0 sum ` `#include ` `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

## 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 hM = ``new` `HashMap(); ` ` `  `        ``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) `

Output :

`Length of the longest 0 sum subarray is 5`

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.

This article is attributed to GeeksforGeeks.org

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