Find the element that appears once in sorted array

PROBLEM :

Given a sorted array of integers, every element appears twice except for one. Find that single one in linear time complexity and without using extra memory.

Input:
The first line of input consists number of the test cases. The description of T test cases is as follows:
The first line of each test case contains the size of the array, and the second line has the elements of the array.

Output:
In each separate line print the number that appears only once in the array.

Constraints:
1 = T = 70
1 = N = 100
0 = A[i] = 100000

Example:
Input:
1
11
1 1 2 2 3 3 4 50 50 65 65

Output:
4

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SIMPLE c++ IMPLEMENTATION :( O(n) Solution using Bitwise)
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#include<iostream>
using namespace std;

int main()
{
int t ;
cin>>t ;
while(t--)
{
   int no ;
   cin>>no ;
 
   int arr[no] ;
   for(int i=0;i<no;i++)
       cin>>arr[i] ;
     
   int XOR=0 ;
   for(int i=0;i<no;i++)
       XOR=XOR^arr[i] ;
     
   cout<<XOR<<endl ;
}
return 0;
}

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SIMPLE c++ IMPLEMENTATION :( O(Logn) Solution using Binary Search)
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#include<iostream>
using namespace std;
int FindElement(int [],int ,int ) ;
int main()
{
int t ;
cin>>t ;
while(t--)
{
   int no ;
   cin>>no ;
 
   int arr[no] ;
   for(int i=0;i<no;i++)
       cin>>arr[i] ;
     
   cout<<FindElement(arr,0,no-1)<<endl ;
}
return 0;
}

int FindElement(int arr[],int start,int end)
{
    if(start>end)
        return -1 ;
       
    if(start==end)
        return arr[start] ;
       
    int mid=(start+end)/2 ;
   
    if(mid%2==0)
    {
        if(arr[mid]==arr[mid+1])
            return FindElement(arr,mid+2,end) ;
        else
            return FindElement(arr,start,mid) ;
    }
    else
    {
        if(arr[mid]==arr[mid-1])
            return FindElement(arr,mid+1,end) ;
        else
            return FindElement(arr,start,mid-1) ;
    }
}

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