Subarray-with-given-sum

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

Input:
The first line of input contains an integer T denoting the number of test cases. Then T test cases follow. 
Each test case consists of two lines. 
The first line of each test case is N and S, where N is the size of array and S is the sum. 
The second line of each test case contains N space separated integers denoting the array elements.

Output:
For each testcase, in a new line, print the starting and ending positions(1 indexing) of first such occuring subarray 
from the left if sum equals to subarray, else print -1.

Constraints:
1 <= T <= 100
1 <= N <= 107
1 <= Ai <= 1010

Example:
Input:
2
5 12
1 2 3 7 5
10 15
1 2 3 4 5 6 7 8 9 10
Output:
2 4
1 5

Explanation : 
Testcase1: sum of elements from 2nd position to 4th position is 12
Testcase2: sum of elements from 1st position to 5th position is 15*/

#include<iostream>
using namespace std;	

pair<int, int> solve(int *arr, int arraySize, int sum) {
	int counter = 0, fsum = 0, end = 0;

	for (int i = 0; i < arraySize; i++) {
		if (fsum < sum) {
			fsum += arr[i];
			counter++;
			end = i + 1;
		}
		
		else if (fsum == sum) 
			return pair<int, int>(i - counter + 1, i);
		
		else {
			i -= counter;
			fsum = 0;
			counter = 0;
		}
	}

	if (fsum == sum)
		return pair<int, int>(end - counter + 1, end);
	else
		return pair<int, int>(-1, -1);
}

void main() {

	int testCases, arraySize, sum;
	cout << "Number of test cases: ";
	cin >> testCases;

	for (int i = 0; i < testCases; i++) {
		cout << "Array size: ";
		cin >> arraySize;
		cout << "Sum: ";
		cin >> sum;

		int *arr = new int[arraySize];
		cout << "Elements of array: " << endl;
		
		for (int j = 0; j < arraySize; j++) {
			cin >> arr[j];
		}
		pair<int, int> par = solve(arr, arraySize, sum);
		cout << "Solution: " << endl;
		cout << par.first << " " << par.second << endl;
		delete[] arr; arr = nullptr;
	}

	system("pause>0");
}