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<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; } |
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 ?> |
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 <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 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
// 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) |
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.
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