find the different rightmost bit - xor

Question

I am stuck on this problem. Thanks for your assistance.
You're given two integers, n and m. Find position of the rightmost bit in which they differ in their binary representations (it is guaranteed that such a bit exists), counting from right to left.
Return the value of 2position_of_the_found_bit (0-based).
Example
For n = 11 and m = 13, the output should be
differentRightmostBit(n, m) = 2.
11 (subscript 10) = 1011 (subscript 2), 13 (subscript) 10 = 1101 (subscript 2), the rightmost bit in which they differ is the bit at position 1 (0-based) from the right in the binary representations.
So the answer is 2 to the 1st power = 2.

Answer 1

After playing around with the bitwise operators I got it !! the answer is (n ^ m) & -(n ^ m)
I could have easily done this in ruby without using the bitwise operators by converting them to a binary string and finding the first non - match starting on the right side and returning (2 ** position) but it was required to be a one liner using the bitwise operators that was the tricky part.
I give credit to Ryan for pointing me in the right direction. Thanks Ryan !!

Answer 2

Since all integers are stored in a 32 bit register hence we need to check value in all these 32 registers now since at 32th register we require 1 from right so 33-32=1 hence at last 33-a[k-1] for the answer. Here i just stored the positions where the binary value is unequal in array 'a' and printed the last element in the array would even work for -ve integers.
#include <iostream>
using namespace std;
int main()
{
int n,m,k=0,y=0;
cin>>n>>m;
int a[32];
for(int i=31;i>=0;i--)
{
y=y+1;
if(n & (1<<i))
{if( !(m & (1<<i)) )
{
a[k]=y;
k++;
}
}else
if ( !(n & (1<<i)) )
{if (m & (1<<i))
{
a[k]=y;
k++;
}
}
}
cout<<33-a[k-1];
return 0;
}

Answer 3

The answer of your question is the below code written in java.The input of this code is in following format:
Input:
First line of input contains a single integer T which denotes the number of test cases. T test cases follows. First line of each test case contains two space separated integers M and N.
Example:
Input:
2
11 9
52 4
Output:
2
5
In binary "11" is "1011"
and "9" is "1001" and as you can see the 2nd bit(from R.H.S) is different and that should be our answer.
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
//sourabh agrawal
class GFG {
public static void main(String[] args)throws IOException{
final BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine().trim());
StringBuilder sb = new StringBuilder();
while(t-->0){
StringTokenizer st = new StringTokenizer(br.readLine().trim());
int a = Integer.parseInt(st.nextToken().trim());
int b = Integer.parseInt(st.nextToken().trim());
int c = (a^b)&-(a^b); //this is the key to solve the question
String ans = Integer.toBinaryString(c);
int size = ans.length();
int position = 1;
while(size-->0){
if(ans.charAt(size)=='0'){
position++;
}else{
break;
}
}
sb.append(position).append("\n");
}
System.out.println(sb);
}
}

Answer 4

One More way:
int RightMostDiffBit(int M,int N){
int count = 1;
for(int i=0;i<32;i++){
if((M & (count << i)) ^ (N & (count << i)))
return i+1;
}
return -1;
}

Answer 5

int main()
{
int a,b,i=0, tempa,tempb;
printf("enter values for elements a and b\n");
scanf("%d %d", &a,&b);
while(i<=31)
{
tempa= a&(1<<i);
tempb= b&(1<<i);
if((tempa^tempb))
{
printf("pos of different bit is %d\n",i);
break;
}
i++;
}
return 0;
}