Find the length of largest subarray with 0 sum

Given an array of integers, find the length of the longest sub-array with sum equals 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: Please solve it on “PRACTICE” first, before moving on to the solution.

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/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[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

 ` `

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

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 */`

Output:

`Length of the longest 0 sum subarray is 5`
• Complexity Analysis:
• Time Complexity: O(n), as use of 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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