Min in Rotated Sorted Array

/* find min element in rotated sorted array 
Sample Input 1:
2
7 
4 6 9 11 1 2 2
5 
4 5 2 2 4
Sample Output 1:
1
2
Explanation Of Sample Input 1:
For the first test case,

The minimum value present in ‘ARR’ is 1. Hence, the answer is 1.

For the second test case:

The minimum value present in ‘ARR’ is 1. Hence, the answer is 2.
Sample Input 2:
2
8 
9 9 9 10 2 3 6 6 
6 
4 6 7 9 2 4
Sample Output 2:
2
2
*/

Approach:
   1st: Traverse the list and update the min element with current if required. T.C. O(N).
   
   2nd OPtimized: As Array is sorted but rotated so straightforward binary search can't be applied. But little modification to  binary can be done.
                  As at any point/index, there are 2 possibility, current element is bigger than left or lesser. If bigger than you can be on either side
                  of pivot, so to check the side, we can compare it with the left element of the current parted array. otherwise 2nd case is the found point/
                  pivot point.
                  
#include <bits/stdc++.h> 
int help(vector<int>&arr, int l, int r){
    if(l==r) return arr[l];
    if(l+1==r) return min(arr[l],arr[r]);
    int m = (l+r)/2;
    if(arr[m]<arr[m-1]) return arr[m];
    if(arr[m]<arr[l]) return help(arr, l, m);
    else return help(arr, m, r);
}
int ninjaChallenge(int n, vector<int> &arr)
{
    return help(arr,0,n-1);
}

// T.C. : O(LogN)
// S.C. : O(1)