Subarray with given sum.cpp

/*
Given an unsorted array A of size N that contains only non-negative integers, find a continuous sub-array which adds to a given number S.

 

Example 1:

Input:
N = 5, S = 12
A[] = {1,2,3,7,5}
Output: 2 4
Explanation: The sum of elements 
from 2nd position to 4th position 
is 12.
 

Example 2:

Input:
N = 10, S = 15
A[] = {1,2,3,4,5,6,7,8,9,10}
Output: 1 5
Explanation: The sum of elements 
from 1st position to 5th position
is 15.
 

Your Task:
You don't need to read input or print anything. The task is to complete the function subarraySum() which takes arr, N and S as input parameters and returns a list containing the starting and ending positions of the first such occurring subarray from the left where sum equals to S. The two indexes in the list should be according to 1-based indexing. If no such subarray is found, return an array consisting only one element that is -1.

 

Expected Time Complexity: O(N)
Expected Auxiliary Space: O(1)

 

Constraints:
1 <= N <= 105
1 <= Ai <= 109
*/

/* A simple program to print subarray
with sum as given sum */
#include <bits/stdc++.h>
using namespace std;

/* Returns true if the there is a subarray
of arr[] with sum equal to 'sum' otherwise
returns false. Also, prints the result */
int subArraySum(int arr[], int n, int sum)
{
	int curr_sum, i, j;

	// Pick a starting point
	for (i = 0; i < n; i++) {
		curr_sum = arr[i];

		// try all subarrays starting with 'i'
		for (j = i + 1; j <= n; j++) {
			if (curr_sum == sum) {
				cout << "Sum found between indexes "
					<< i << " and " << j - 1;
				return 1;
			}
			if (curr_sum > sum || j == n)
				break;
			curr_sum = curr_sum + arr[j];
		}
	}

	cout << "No subarray found";
	return 0;
}

// Driver Code
int main()
{
	int arr[] = { 15, 2, 4, 8, 9, 5, 10, 23 };
	int n = sizeof(arr) / sizeof(arr[0]);
	int sum = 23;
	subArraySum(arr, n, sum);
	return 0;
}