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}
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:
- Consider all sub-arrays one by one and check the sum of every sub-array.
- Run two loops: the outer loop picks the starting point i and the inner loop tries all sub-arrays starting from i.
Implementation:
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) { // Initialize result int max_len = 0; // 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 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; } |
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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 |
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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) |
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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 |
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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 ?> |
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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:
- 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
- Move along the input array from starting to the end
- For every index update the value of sum = sum + array[i]
- Check for every index, if the current sum is present in the hash map or not
- 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
- 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:
Implementation:
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 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; } |
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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)); } } |
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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) |
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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 */ |
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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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