Position of rightmost different bit
Given two numbers m and n. Find the position of the rightmost different bit in the binary representation of numbers. It is guaranteed that such a bit exists.
Examples:
Input: m = 11, n = 9 Output: 2 (11)10 = (1011)2 (9)10 = (1001)2 It can be seen that 2nd bit from the right is different Input: m = 52, n = 4 Output: 5 (52)10 = (110100)2 (4)10 = (100)2, can also be written as = (000100)2 It can be seen that 5th bit from the right is different
Approach: Get the bitwise xor of m and n. Let it be xor_value = m ^ n. Now, find the position of rightmost set bit in xor_value.
Explanation: The bitwise xor operation produces a number that has set bits only at the positions where the bits of m and n differ. Thus, the position of the rightmost set bit in xor_value gives the position of the rightmost different bit.
Efficient way to find the rightmost set bit: log2(n & -n) + 1 gives us the position of the rightmost set bit. (-n) reverses all the bits from left to right till the last set bit for example: n = 16810 binary signed 2's complement of n = 00000000101010002 binary signed 2's complement of -n = 11111111010110002 ∴ (n & -n) = 00000000000010002 = 8 now, log2(n & -n) = log2(8) = 3 log2(n & -n) + 1 = 4 (position of rightmost set bit)
Below is the implementation of the above approach:
C++
// C++ implementation to find the position // of rightmost different bit #include <bits/stdc++.h> using namespace std; // Function to find the position of // rightmost set bit in 'n' // returns 0 if there is no set bit. int getRightMostSetBit( int n) { // to handle edge case when n = 0. if (n == 0) return 0; return log2(n & -n) + 1; } // Function to find the position of // rightmost different bit in the // binary representations of 'm' and 'n' // returns 0 if there is no // rightmost different bit. int posOfRightMostDiffBit( int m, int n) { // position of rightmost different // bit return getRightMostSetBit(m ^ n); } // Driver program int main() { int m = 52, n = 24; cout << "Position of rightmost different bit:" << posOfRightMostDiffBit(m, n)<<endl; return 0; } |
Java
// Java implementation to find the position // of rightmost different bit class GFG { // Function to find the position of // rightmost set bit in 'n' // return 0 if there is no set bit. static int getRightMostSetBit( int n) { if (n == 0 ) return 0 ; return ( int )((Math.log10(n & -n)) / Math.log10( 2 )) + 1 ; } // Function to find the position of // rightmost different bit in the // binary representations of 'm' and 'n' static int posOfRightMostDiffBit( int m, int n) { // position of rightmost different bit return getRightMostSetBit(m ^ n); } // Driver code public static void main(String arg[]) { int m = 52 , n = 4 ; System.out.print( "Position = " + posOfRightMostDiffBit(m, n)); } } // This code is contributed by Anant Agarwal. |
Python3
# Python implementation # to find the position # of rightmost different bit import math # Function to find the position of # rightmost set bit in 'n' def getRightMostSetBit(n): if (n = = 0 ): return 0 return math.log2(n & - n) + 1 # Function to find the position of # rightmost different bit in the # binary representations of 'm' and 'n' def posOfRightMostDiffBit(m, n): # position of rightmost different # bit return getRightMostSetBit(m ^ n) # Driver code m = 52 n = 4 print ( "position = " , int (posOfRightMostDiffBit(m, n))) # This code is contributed # by Anant Agarwal. |
C#
// C# implementation to find the position // of rightmost different bit using System; class GFG { // Function to find the position of // rightmost set bit in 'n' static int getRightMostSetBit( int n) { if (n == 0) return 0; return ( int )((Math.Log10(n & -n)) / Math.Log10(2)) + 1; } // Function to find the position of // rightmost different bit in the // binary representations of 'm' and 'n' static int posOfRightMostDiffBit( int m, int n) { // position of rightmost different bit return getRightMostSetBit(m ^ n); } // Driver code public static void Main() { int m = 52, n = 4; Console.Write( "Position = " + posOfRightMostDiffBit(m, n)); } } // This code is contributed by Smitha. |
PHP
<?php // PHP implementation to // find the position of // rightmost different bit // Function to find the position // of rightmost set bit in 'n' function getRightMostSetBit( $n ) { if ( $n == 0) return 0; return log( $n & - $n , (2)) + 1; } // Function to find the position of // rightmost different bit in the // binary representations of 'm' // and 'n' function posOfRightMostDiffBit( $m , $n ) { // position of rightmost // different bit return getRightMostSetBit( $m ^ $n ); } // Driver Code $m = 52; $n = 4; echo posOfRightMostDiffBit( $m , $n ); // This code is contributed by Ajit ?> |
Javascript
<script> // JavaScript implementation to find the position // of rightmost different bit // Function to find the position of // rightmost set bit in 'n' // returns 0 if there is no set bit. function getRightMostSetBit(n) { // to handle edge case when n = 0. if (n == 0) return 0; return Math.log2(n & -n) + 1; } // Function to find the position of // rightmost different bit in the // binary representations of 'm' and 'n' // returns 0 if there is no // rightmost different bit. function posOfRightMostDiffBit(m, n) { // position of rightmost different // bit return getRightMostSetBit(m ^ n); } // Driver program let m = 52, n = 24; document.write( "Position of rightmost different bit:" + posOfRightMostDiffBit(m, n) + "<br>" ); // This code is contributed by Surbhi Tyagi. </script> |
Position of rightmost different bit:3
Using ffs() function
C++
// C++ implementation to find the // position of rightmost different // bit in two number. #include <bits/stdc++.h> using namespace std; // function to find rightmost different // bit in two numbers. int posOfRightMostDiffBit( int m, int n) { return ffs(m ^ n); } // Driver code int main() { int m = 52, n = 4; cout << "Position = " << posOfRightMostDiffBit(m, n); return 0; } |
Java
// Java implementation to find the // position of rightmost different // bit in two number. import java.util.*; class GFG{ // function to find rightmost // different bit in two numbers. static int posOfRightMostDiffBit( int m, int n) { return ( int )Math.floor( Math.log10( Math.pow(m ^ n, 2 )))+ 2 ; } // Driver code public static void main(String[] args) { int m = 52 , n = 4 ; System.out.println( "Position = " + posOfRightMostDiffBit(m, n)); } } // This code is contributed by gauravrajput1 |
Python3
# Python3 implementation to find the # position of rightmost different # bit in two number. from math import floor, log10 # Function to find rightmost different # bit in two numbers. def posOfRightMostDiffBit(m, n): return floor(log10( pow (m ^ n, 2 ))) + 2 # Driver code if __name__ = = '__main__' : m, n = 52 , 4 print ( "Position = " , posOfRightMostDiffBit(m, n)) # This code is contributed by mohit kumar 29 |
C#
// C# implementation to find the // position of rightmost different // bit in two number. using System; class GFG { // function to find rightmost // different bit in two numbers. static int posOfRightMostDiffBit( int m, int n) { return ( int )Math.Floor(Math.Log10( Math.Pow(m ^ n, 2))) + 2; } // Driver code public static void Main(String[] args) { int m = 52, n = 4; Console.Write( "Position = " + posOfRightMostDiffBit(m, n)); } } // This code is contributed by shivanisinghss2110 |
PHP
<?php // PHP implementation to find the // position of rightmost different // bit in two number. // function to find rightmost // different bit in two numbers. function posOfRightMostDiffBit( $m , $n ) { $t = floor (log( $m ^ $n , 2)); return $t ; } // Driver code $m = 52; $n = 4; echo "Position = " , posOfRightMostDiffBit( $m , $n ); // This code is contributed by ajit ?> |
Javascript
<script> // Javascript implementation to find the // position of rightmost different // bit in two number. // function to find rightmost // different bit in two numbers. function posOfRightMostDiffBit(m, n) { return parseInt(Math.floor (Math.log10(Math.pow(m ^ n, 2))), 10) + 2; } let m = 52, n = 4; document.write( "Position = " + posOfRightMostDiffBit(m, n) ); </script> |
Position = 5
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