Find if there is a subarray with 0 sum
- Difficulty Level : Medium
- Last Updated : 11 Feb, 2022
Given an array of positive and negative numbers, 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: trueExplanation :
The third element is zero. A single element is also a sub-array.
Input: {-3, 2, 3, 1, 6}
Output: falseA 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 (See this for implementation). The time complexity of this method is O(n2).
We can also use 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 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.
Example :arr[] = {1, 4, -2, -2, 5, -4, 3} If we consider all prefix sums, we can notice that there is a subarray with 0 sum when : 1) Either a prefix sum repeats or 2) 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.Following is implementation of the above approach.
C++
// A 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 code
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
// A 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 code
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
# A python 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 code
arr
=
[
-
3
,
2
,
3
,
1
,
6
]
n
=
len
(arr)
if
subArrayExists(arr, n)
=
=
True
:
(
"Found a sunbarray with 0 sum"
)
else
:
(
"No Such sub array exits!"
)
# This code is contributed by Shrikant13
C#
// A 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);
}
// We reach here only when there is
// no subarray with 0 sum
return
false
;
}
// Main Method
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
// 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!"
);
OutputNo Such Sub Array Exists!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.
Space Complexity: O(n) .Here we required extra space for unordered_set to insert array elements.
Exercise:
Extend the above program to print starting and ending indexes of all subarrays with 0 sum.
This article is contributed by Chirag Gupta. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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