# Find the length of largest subarray with 0 sum

Last Updated : 03 Mar, 2023

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}

Recommended Practice

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 ` `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 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 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

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## Javascript

 ``

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 ` `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 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` `(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

 ``

Output

`Length of the longest 0 sum subarray is 5`

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

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