Subarray with given sum

PROBLEM :

Given an unsorted array of non-negative integers, find a continuous sub-array which adds to a given number.

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:
Corresponding to each test case, in a new line, print the starting and ending positions of first such occuring subarray if sum equals to subarray, else print -1.

Note: Position of 1st element of the array should be considered as 1.

Expected Time Complexity: O(n)

Constraints:
1 = T = 100
1 = N = 100
1 = an array element = 200

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 :
In first test case, sum of elements from 2nd position to 4th position is 12
In second test case, sum of elements from 1st position to 5th position is 15
     
--------------------------------------------------------------------------------
SIMPLE c++ IMPLEMENTATION :
--------------------------------------------------------------------------------
// Following solution work for Negative Nos too

#include<bits/stdc++.h>
using namespace std;

int main()
{
    int t ;
    cin>>t ;
    while(t--)
    {
        int no,target ;
   cin>>no>>target ;
 
   int arr[no] ;
   for(int i=0;i<no;i++)
       cin>>arr[i] ;
     
   unordered_map<int,int> map ;
   int sum=0 ;
   bool flag=true ;
       
        for(int i=0;i<no;i++)
        {
        sum+=arr[i] ;
       
        if(sum==target)
        {
        flag=false ;
        cout<<1<<" "<<i+1 ;
        break ;
}

if(map.find(sum-target)!=map.end())
{
flag=false ;
cout<<map[sum-target]+2<<" "<<i+1 ;
break ;
}

map[sum]=i ;
}

if(flag)
   cout<<-1 ;
cout<<endl ;
    }
return 0;
}

--------------------------------------------------------------------------------