Find the 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

We strongly recommend you to minimize your browser and try this yourself first.

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 is C++ implementation of this solution.

• C/C++
• Python

C/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) {     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; }

Python

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 is Java implementation of the above approach.

// 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));     } }

Output:

Length of the longest 0 sum subarray is 5