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Position of rightmost different bit in C++
In this problem, we are given two numbers N and M. Our task is to find the index of the rightmost different bit in the binary representation of the number.
Let’s take an example to understand the problem,
Input − N = 12 , M = 9
Output − 2
Explanation − (12)2 = 1100 and (10)2 = 1010.
The second bit from right is a different bit.
To solve this problem, we will have to find all the different bits of the numbers. To find all different bits we will find xor of M and N. Then we will find the rightmost bit of M^N.
This seems a bit complex to understand let’s solve an example using this method.
N = 12 , M = 9 N^M = 0110.
A rightmost set bit here is at index 2.
Example
Program to show the implementation of our solution,
#include <iostream>
#include <math.h>
using namespace std;
int rightSetBit(int N) {
int bitIndex = log2(N & -N)+1;
return bitIndex;
}
void rightDiffBit(int m, int n) {
int diffBit = rightSetBit(m^n);
cout<<diffBit;
}
int main() {
int N = 12, M = 10;
cout<<"Postion of first right different bit of the number "<<N<<" & "<<M<<" is ";
rightDiffBit(N, M);
return 0;
}Output
Postion of first right different bit of the number 12 & 10 is 2
Updated on: 17-Apr-2020
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