Given an array of positive and negative numbers, the task is to find if there is a subarray (of size at least one) with 0 sum.
Examples:
Input: {4, 2, -3, 1, 6}
Output: true
Explanation:
There is a subarray with zero sum from index 1 to 3.Input: {4, 2, 0, 1, 6}
Output: true
Explanation: The third element is zero. A single element is also a sub-array.Input: {-3, 2, 3, 1, 6}
Output: false
Subarray with 0 sum using Nested loop:
Generate every subarray and calcuate the sum of each subarray. Check if subarray sum is 0 then return true. Otherwise, if no such subarray found then return false.
Below is the implementation of the above approach:
C++
// C++ program to find if // there is a zero sum subarray #include <iostream> using namespace std; bool subArrayExists( int arr[], int n) { for ( int i = 0; i < n; i++) { // starting point of the sub arrray and // sum is initialized with first element of subarray // a[i] int sum = arr[i]; if (sum == 0) return true ; for ( int j = i + 1; j < n; j++) { // we are finding the sum till jth index // starting from ith index sum += arr[j]; if (sum == 0) return true ; } } return false ; } // Driver's code int main() { int arr[] = { -3, 2, 3, 1, 6 }; int N = sizeof (arr) / sizeof (arr[0]); // Function call if (subArrayExists(arr, N)) cout << "Found a subarray with 0 sum" ; else cout << "No Such Sub Array Exists!" ; return 0; } // This code is contributed by Tapesh(tapeshdua420) |
Java
// Java program to find if // there is a zero sum subarray import java.util.Arrays; public class Main { public static boolean subArrayExists( int arr[], int n) { for ( int i = 0 ; i < n; i++) { int sum = arr[i]; if (sum == 0 ) return true ; for ( int j = i + 1 ; j < n; j++) { sum += arr[j]; if (sum == 0 ) return true ; } } return false ; } // Driver's code public static void main(String[] args) { int arr[] = { - 3 , 2 , 3 , 1 , 6 }; int N = arr.length; // Function call if (subArrayExists(arr, N)) System.out.println( "Found a subarray with 0 sum" ); else System.out.println( "No Such Sub Array Exists!" ); } } // This code is contributed by Utkarsh Kumar |
Python3
def subArrayExists(arr, n): for i in range (n): # Starting point of the subarray and # sum is initialized with the first element of subarray sum = arr[i] if sum = = 0 : return True for j in range (i + 1 , n): # We are finding the sum till the jth index # starting from the ith index sum + = arr[j] if sum = = 0 : return True return False # Driver's code if __name__ = = "__main__" : arr = [ - 3 , 2 , 3 , 1 , 6 ] N = len (arr) # Function call if subArrayExists(arr, N): print ( "Found a subarray with 0 sum" ) else : print ( "No Such Sub Array Exists!" ) |
C#
// C# program to find if // there is a zero sum subarray using System; public class GFG { public static bool subArrayExists( int [] arr, int n) { for ( int i = 0; i < n; i++) { int sum = arr[i]; if (sum == 0) return true ; for ( int j = i + 1; j < n; j++) { sum += arr[j]; if (sum == 0) return true ; } } return false ; } // Driver's code public static void Main() { int [] arr = { -3, 2, 3, 1, 6 }; int N = arr.Length; // Function call if (subArrayExists(arr, N)) Console.WriteLine( "Found a subarray with 0 sum" ); else Console.WriteLine( "No Such Sub Array Exists!" ); } } // This code is contributed by Pushpesh Raj |
Javascript
function subArrayExists(arr) { const n = arr.length; for (let i = 0; i < n; i++) { // Starting point of the subarray, and the sum is initialized with the first element of the subarray. let sum = arr[i]; if (sum === 0) return true ; for (let j = i + 1; j < n; j++) { // We are finding the sum until the jth index starting from the ith index. sum += arr[j]; if (sum === 0) return true ; } } return false ; } const arr = [-3, 2, 3, 1, 6]; // Function call if (subArrayExists(arr)) console.log( "Found a subarray with 0 sum" ); else console.log( "No such subarray exists!" ); |
No Such Sub Array Exists!
Time Complexity: O(N2)
Auxiliary Space: O(1)
Subarray with 0 sum using Hashing:
The idea is to iterate through the array and for every element arr[i], calculate the sum of elements from 0 to i (this can simply be done as sum += arr[i]). If the current sum has been seen before, then there must be a zero-sum subarray. Hashing is used to store the sum values so that sum can be stored quickly and find out whether the current sum is seen before or not.
Follow the given steps to solve the problem:
- Declare a variable sum, to store the sum of prefix elements
- Traverse the array and at each index, add the element into the sum and check if this sum exists earlier. If the sum exists, then return true
- Also, insert every prefix sum into a map, so that later on it can be found whether the current sum is seen before or not
- At the end return false, as no such subarray is found
Illustration:
arr[] = {1, 4, -2, -2, 5, -4, 3}
Consider all prefix sums, one can notice that there is a subarray with 0 sum when :
- Either a prefix sum repeats
- Or, prefix sum becomes 0.
Prefix sums for above array are: 1, 5, 3, 1, 6, 2, 5
Since prefix sum 1 repeats, we have a subarray with 0 sum.
Below is the Implementation of the above approach:
C++
// C++ program to find if // there is a zero sum subarray #include <bits/stdc++.h> using namespace std; bool subArrayExists( int arr[], int N) { unordered_set< int > sumSet; // Traverse through array // and store prefix sums int sum = 0; for ( int i = 0; i < N; i++) { sum += arr[i]; // If prefix sum is 0 or // it is already present if (sum == 0 || sumSet.find(sum) != sumSet.end()) return true ; sumSet.insert(sum); } return false ; } // Driver's code int main() { int arr[] = {-3, 2, 3, 1, 6}; int N = sizeof (arr) / sizeof (arr[0]); // Function call if (subArrayExists(arr, N)) cout << "Found a subarray with 0 sum" ; else cout << "No Such Sub Array Exists!" ; return 0; } |
Java
// Java program to find // if there is a zero sum subarray import java.util.HashSet; import java.util.Set; class ZeroSumSubarray { // Returns true if arr[] // has a subarray with sero sum static Boolean subArrayExists( int arr[]) { // Creates an empty hashset hs Set<Integer> hs = new HashSet<Integer>(); // Initialize sum of elements int sum = 0 ; // Traverse through the given array for ( int i = 0 ; i < arr.length; i++) { // Add current element to sum sum += arr[i]; // Return true in following cases // a) Current element is 0 // b) sum of elements from 0 to i is 0 // c) sum is already present in hash set if (arr[i] == 0 || sum == 0 || hs.contains(sum)) return true ; // Add sum to hash set hs.add(sum); } // We reach here only when there is // no subarray with 0 sum return false ; } // Driver's code public static void main(String arg[]) { int arr[] = {- 3 , 2 , 3 , 1 , 6 }; // Function call if (subArrayExists(arr)) System.out.println( "Found a subarray with 0 sum" ); else System.out.println( "No Such Sub Array Exists!" ); } } |
Python3
# python3 program to find if # there is a zero sum subarray def subArrayExists(arr, N): # traverse through array # and store prefix sums n_sum = 0 s = set () for i in range (N): n_sum + = arr[i] # If prefix sum is 0 or # it is already present if n_sum = = 0 or n_sum in s: return True s.add(n_sum) return False # Driver's code if __name__ = = '__main__' : arr = [ - 3 , 2 , 3 , 1 , 6 ] N = len (arr) # Function call if subArrayExists(arr, N) = = True : print ( "Found a subarray with 0 sum" ) else : print ( "No Such sub array exits!" ) # This code is contributed by Shrikant13 |
C#
// C# program to find if there // is a zero sum subarray using System; using System.Collections.Generic; class GFG { // Returns true if arr[] has // a subarray with sero sum static bool SubArrayExists( int [] arr) { // Creates an empty HashSet hM HashSet< int > hs = new HashSet< int >(); // Initialize sum of elements int sum = 0; // Traverse through the given array for ( int i = 0; i < arr.Length; i++) { // Add current element to sum sum += arr[i]; // Return true in following cases // a) Current element is 0 // b) sum of elements from 0 to i is 0 // c) sum is already present in hash set if (arr[i] == 0 || sum == 0 || hs.Contains(sum)) return true ; // Add sum to hash set hs.Add(sum); } // Reach here only when there is // no subarray with 0 sum return false ; } // Driver's code public static void Main() { int [] arr = {-3, 2, 3, 1, 6}; // Function call if (SubArrayExists(arr)) Console.WriteLine( "Found a subarray with 0 sum" ); else Console.WriteLine( "No Such Sub Array Exists!" ); } } |
Javascript
// A Javascript program to // find if there is a zero sum subarray const subArrayExists = (arr) => { const sumSet = new Set(); // Traverse through array // and store prefix sums let sum = 0; for (let i = 0 ; i < arr.length ; i++) { sum += arr[i]; // If prefix sum is 0 // or it is already present if (sum === 0 || sumSet.has(sum)) return true ; sumSet.add(sum); } return false ; } // Driver code const arr = [-3, 2, 3, 1, 6]; if (subArrayExists(arr)) console.log( "Found a subarray with 0 sum" ); else console.log( "No Such Sub Array Exists!" ); |
No Such Sub Array Exists!
Time Complexity: O(N) under the assumption that a good hashing function is used, that allows insertion and retrieval operations in O(1) time.
Auxiliary Space: O(N) Here extra space is required for hashing
This article is contributed by Chirag Gupta. Please write comments if you find anything incorrect, or if you want to share more information about the topic discussed above
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