# Discuss Subarray with zero sum and given sum?

A subarray is a contiguous part of an array. An array that is inside another array

Problem Statement:

Given an array of positive and negative numbers, find if there is a subarray (of size at least one) with 0 sums.

Example:

Input: {4, 2, -3, 1, 6}
Output: true
Explanation: There is a subarray with zero-sum from index 1 to 3.

we are using the following approach:

• Traverse the array from start to end.
• From every index start another loop from i to the end of the array to get all subarrays starting from i, and keep a variable currentSum to calculate the sum of every subarray.
• For every index in inner loop update currentSum = currentSum + arr[j]
• If the currentSum is equal to the given sum then print the subarray.

Code:

``````bool hasZeroSumSubarray(int nums[], int n)
{
unordered_set<int> set;
set.insert(0);
int sum = 0;

for (int i = 0; i < n; i++)
{
sum += nums[i];
if (set.find(sum) != set.end())
return true;
else
set.insert(sum);
}
return false;
}
``````

Subarray with Given Sum

Problem Statement:

Given an array arr[] of non-negative integers and an integer sum, find a continuous subarray that adds to a given sum.

Note: There may be more than one subarray with sum as the given sum, print first such subarray.

Examples:

``````Input: arr[] = {1, 4, 20, 3, 10, 5}, sum = 33
Output: Sum found between indexes 2 and 4
Explanation: Sum of elements between indices 2 and 4 is 20 + 3 + 10 = 33
``````

Code:

``````
bool find_Subarray(int arr[], int N, int required) {
int sum = 0;
for (int i = 0; i < N; i++) {
sum = arr[i];
for (int j = i + 1; j < N + 1; j++) {
if (sum == required) {
// found subarray with given sum
return true;
} else if (sum > required) {
break;
}
sum = sum + arr[j];
}
}
return false;
}
``````