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++
#include <iostream>
using namespace std;
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;
}
int main()
{
int arr[] = { -3, 2, 3, 1, 6 };
int N = sizeof(arr) / sizeof(arr[0]);
if (subArrayExists(arr, N))
cout << "Found a subarray with 0 sum";
else
cout << "No Such Sub Array Exists!";
return 0;
}
|
Java
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;
}
public static void main(String[] args)
{
int arr[] = { -3, 2, 3, 1, 6 };
int N = arr.length;
if (subArrayExists(arr, N))
System.out.println("Found a subarray with 0 sum");
else
System.out.println("No Such Sub Array Exists!");
}
}
|
Python3
def subArrayExists(arr, n):
for i in range(n):
sum = arr[i]
if sum == 0:
return True
for j in range(i + 1, n):
sum += arr[j]
if sum == 0:
return True
return False
if __name__ == "__main__":
arr = [-3, 2, 3, 1, 6]
N = len(arr)
if subArrayExists(arr, N):
print("Found a subarray with 0 sum")
else:
print("No Such Sub Array Exists!")
|
C#
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;
}
public static void Main()
{
int[] arr = { -3, 2, 3, 1, 6 };
int N = arr.Length;
if (subArrayExists(arr, N))
Console.WriteLine("Found a subarray with 0 sum");
else
Console.WriteLine("No Such Sub Array Exists!");
}
}
|
OutputNo 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++
#include <bits/stdc++.h>
using namespace std;
bool subArrayExists(int arr[], int N)
{
unordered_set<int> sumSet;
int sum = 0;
for (int i = 0; i < N; i++) {
sum += arr[i];
if (sum == 0 || sumSet.find(sum) != sumSet.end())
return true;
sumSet.insert(sum);
}
return false;
}
int main()
{
int arr[] = {-3, 2, 3, 1, 6};
int N = sizeof(arr) / sizeof(arr[0]);
if (subArrayExists(arr, N))
cout << "Found a subarray with 0 sum";
else
cout << "No Such Sub Array Exists!";
return 0;
}
|
Java
import java.util.HashSet;
import java.util.Set;
class ZeroSumSubarray {
static Boolean subArrayExists(int arr[])
{
Set<Integer> hs = new HashSet<Integer>();
int sum = 0;
for (int i = 0; i < arr.length; i++) {
sum += arr[i];
if (arr[i] == 0 || sum == 0 || hs.contains(sum))
return true;
hs.add(sum);
}
return false;
}
public static void main(String arg[])
{
int arr[] = {-3, 2, 3, 1, 6};
if (subArrayExists(arr))
System.out.println(
"Found a subarray with 0 sum");
else
System.out.println("No Such Sub Array Exists!");
}
}
|
Python3
def subArrayExists(arr, N):
n_sum = 0
s = set()
for i in range(N):
n_sum += arr[i]
if n_sum == 0 or n_sum in s:
return True
s.add(n_sum)
return False
if __name__ == '__main__':
arr = [-3, 2, 3, 1, 6]
N = len(arr)
if subArrayExists(arr, N) == True:
print("Found a subarray with 0 sum")
else:
print("No Such sub array exits!")
|
C#
using System;
using System.Collections.Generic;
class GFG {
static bool SubArrayExists(int[] arr)
{
HashSet<int> hs = new HashSet<int>();
int sum = 0;
for (int i = 0; i < arr.Length; i++) {
sum += arr[i];
if (arr[i] == 0 || sum == 0 || hs.Contains(sum))
return true;
hs.Add(sum);
}
return false;
}
public static void Main()
{
int[] arr = {-3, 2, 3, 1, 6};
if (SubArrayExists(arr))
Console.WriteLine(
"Found a subarray with 0 sum");
else
Console.WriteLine("No Such Sub Array Exists!");
}
}
|
Javascript
const subArrayExists = (arr) => {
const sumSet = new Set();
let sum = 0;
for (let i = 0 ; i < arr.length ; i++)
{
sum += arr[i];
if (sum === 0 || sumSet.has(sum))
return true;
sumSet.add(sum);
}
return false;
}
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!");
|
OutputNo 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