Toggle The Middle - Solution

Problem : Toggle The Middle

You are given a number N. You need to toggle the middle bit of N in binary form and print the result in decimal form.

    If number of bits in binary form of N are odd then toggle the middle bit (Like 111 to 101).
    If number of bits in binary form of N are even then toggle both the middle bits (Like 1111 to 1001)

    Note: Toggling a bit means converting a 0 to 1 and vice versa.

Hint

Input:
The first line of input contains T denoting the number of testcases. T testcases follow. Each testcase contains a single number N.

Output:
For each testcase, in a new line, print the decimal form after toggling the middle bit of N.

Constraints:
1 <= T <= 100
1 <= N <= 106

Examples:
Input:
5
1
2
3
4
5
Output:
0
1
0
6
7

Examples:
Testcase3: N=3. Binary is 11. Toggle the middle bits: 00. 00 in decimal is 0
Testcase5: N=5. Binary is 101. Toggle the middle bit: 111. 111 in decimal is 7

Solution:

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

int main() {
    //code
    int T;
    cin>>T;
    
    while(T--){
        int N;
        cin>>N;
        
        if(N==1){
            cout<<"0"<<endl;
        } else if(N==0){
            cout<<"1"<<endl;
        } else {
            int noOfBits = floor(log(N)/log(2)) + 1;
            
            if(!(noOfBits % 2)){
                int mid = noOfBits/2;
                int before_mid = mid -1;
                
                N = (N ^ (1 <<(mid)));
                N = (N ^ (1 <<(before_mid)));
                
                cout<<N<<endl;
            } else {
                int mid = noOfBits/2;
                N = (N^ (1 <<(mid)));
                cout<<N<<endl;
            }
        }
    }
    return 0;
}