Given two non-negative numbers m and n. Find the position of rightmost same bit in the binary representation of the numbers.
Examples:
Input : m = 10, n = 9
Output : 3
(10)10 = (1010)2
(9)10 = (1001)2
It can be seen that the 3rd bit
from the right is same.
Input : m = 16, n = 7
Output : 4
(16)10 = (10000)2
(7)10 = (111)2, can also be written as
= (00111)2
It can be seen that the 4th bit
from the right is same.
Approach: Get the bitwise xor of m and n. Let it be xor_value = m ^ n. Now, get the position of rightmost unset bit in xor_value.
Explanation: The bitwise xor operation produces a number which has unset bits only at the positions where the bits of m and n are same. Thus, the position of rightmost unset bit in xor_value gives the position of rightmost same bit.
C++
// C++ implementation to find the position// of rightmost same bit#include <bits/stdc++.h>using namespace std;// Function to find the position of// rightmost set bit in 'n'int getRightMostSetBit(unsigned int n){ return log2(n & -n) + 1;}// Function to find the position of// rightmost same bit in the// binary representations of 'm' and 'n'int posOfRightMostSameBit(unsigned int m, unsigned int n){ // position of rightmost same bit return getRightMostSetBit(~(m ^ n));}// Driver program to test aboveint main(){ int m = 16, n = 7; cout << "Position = " << posOfRightMostSameBit(m, n); return 0;} |
Java
// Java implementation to find the position// of rightmost same bitclass GFG { // Function to find the position of // rightmost set bit in 'n' static int getRightMostSetBit(int n) { return (int)((Math.log(n & -n))/(Math.log(2))) + 1; } // Function to find the position of // rightmost same bit in the // binary representations of 'm' and 'n' static int posOfRightMostSameBit(int m,int n) { // position of rightmost same bit return getRightMostSetBit(~(m ^ n)); } //Driver code public static void main (String[] args) { int m = 16, n = 7; System.out.print("Position = " + posOfRightMostSameBit(m, n)); }}// This code is contributed by Anant Agarwal. |
Python3
# Python3 implementation to find the # position of rightmost same bitimport math# Function to find the position # of rightmost set bit in 'n'def getRightMostSetBit(n): return int(math.log2(n & -n)) + 1# Function to find the position of# rightmost same bit in the binary# representations of 'm' and 'n'def posOfRightMostSameBit(m, n): # position of rightmost same bit return getRightMostSetBit(~(m ^ n))# Driver Codem, n = 16, 7print("Position = ", posOfRightMostSameBit(m, n))# This code is contributed by Anant Agarwal. |
C#
// C# implementation to find the position// of rightmost same bitusing System;class GFG{ // Function to find the position of // rightmost set bit in 'n' static int getRightMostSetBit(int n) { return (int)((Math.Log(n & -n)) / (Math.Log(2))) + 1; } // Function to find the position of // rightmost same bit in the // binary representations of 'm' and 'n' static int posOfRightMostSameBit(int m,int n) { // position of rightmost same bit return getRightMostSetBit(~(m ^ n)); } //Driver code public static void Main () { int m = 16, n = 7; Console.Write("Position = " + posOfRightMostSameBit(m, n)); }}//This code is contributed by Anant Agarwal. |
PHP
<?php// PHP implementation to// find the position// of rightmost same bit// Function to find the position of// rightmost set bit in 'n'function getRightMostSetBit($n){ return log($n & -$n) + 1;}// Function to find the position of// rightmost same bit in the// binary representations of 'm' and 'n'function posOfRightMostSameBit($m, $n){ // position of rightmost same bit return getRightMostSetBit(~($m ^ $n));} // Driver Code $m = 16; $n = 7; echo "Position = " , ceil(posOfRightMostSameBit($m, $n)); // This code is contributed by anuj_67.?> |
Javascript
<script>// JavaScript implementation to find the position// of rightmost same bit// Function to find the position of// rightmost set bit in 'n'function getRightMostSetBit(n){ return Math.log2(n & -n) + 1;}// Function to find the position of// rightmost same bit in the// binary representations of 'm' and 'n'function posOfRightMostSameBit(m, n){ // position of rightmost same bit return getRightMostSetBit(~(m ^ n));}// Driver program to test above let m = 16, n = 7; document.write("Position = " + posOfRightMostSameBit(m, n));// This code is contributed by Manoj.</script> |
Output:
Position = 4
Time Complexity: O(1)
Auxiliary Space: O(1)
Alternate Approach: Until both the value becomes zero, check last bits of both numbers and right shift. At any moment, both bits are same, return counter.
Explanation: Rightmost bit of two values m and n are equal only when both values are either odd or even.
C++
// C++ implementation to find the position // of rightmost same bit #include <iostream>using namespace std; // Function to find the position of // rightmost same bit in the binary // representations of 'm' and 'n' static int posOfRightMostSameBit(int m, int n) { // Initialize loop counter int loopCounter = 1; while (m > 0 || n > 0) { // Check whether the value 'm' is odd bool a = m % 2 == 1; // Check whether the value 'n' is odd bool b = n % 2 == 1; // Below 'if' checks for both // values to be odd or even if (!(a ^ b)) { return loopCounter; } // Right shift value of m m = m >> 1; // Right shift value of n n = n >> 1; loopCounter++; } // When no common set is found return -1; } // Driver code int main(){ int m = 16, n = 7; cout << "Position = " << posOfRightMostSameBit(m, n); } // This code is contributed by shivanisinghss2110 |
Java
// Java implementation to find the position // of rightmost same bit class GFG { // Function to find the position of // rightmost same bit in the // binary representations of 'm' and 'n' static int posOfRightMostSameBit(int m,int n) { int loopCounter = 1; // Initialize loop counter while (m > 0 || n > 0){ boolean a = m%2 == 1; //Check whether the value 'm' is odd boolean b = n%2 == 1; //Check whether the value 'n' is odd // Below 'if' checks for both values to be odd or even if (!(a ^ b)){ return loopCounter;} m = m >> 1; //Right shift value of m n = n >> 1; //Right shift value of n loopCounter++; } return -1; //When no common set is found } //Driver code public static void main (String[] args) { int m = 16, n = 7; System.out.print("Position = " + posOfRightMostSameBit(m, n)); } } |
Python3
# Python3 implementation to find the position# of rightmost same bit# Function to find the position of# rightmost same bit in the# binary representations of 'm' and 'n'def posOfRightMostSameBit(m, n): # Initialize loop counter loopCounter = 1 while (m > 0 or n > 0): # Check whether the value 'm' is odd a = m % 2 == 1 # Check whether the value 'n' is odd b = n % 2 == 1 # Below 'if' checks for both # values to be odd or even if (not (a ^ b)): return loopCounter # Right shift value of m m = m >> 1 # Right shift value of n n = n >> 1 loopCounter += 1 # When no common set is found return -1# Driver codeif __name__ == '__main__': m, n = 16, 7 print("Position = ", posOfRightMostSameBit(m, n))# This code is contributed by mohit kumar 29 |
C#
// C# implementation to find the position // of rightmost same bit using System;class GFG { // Function to find the position of // rightmost same bit in the // binary representations of 'm' and 'n' static int posOfRightMostSameBit(int m, int n) { int loopCounter = 1; // Initialize loop counter while (m > 0 || n > 0) { Boolean a = m % 2 == 1; // Check whether the value 'm' is odd Boolean b = n % 2 == 1; // Check whether the value 'n' is odd // Below 'if' checks for both values to be odd or even if (!(a ^ b)) { return loopCounter; } m = m >> 1; // Right shift value of m n = n >> 1; // Right shift value of n loopCounter++; } return -1; // When no common set is found } // Driver code public static void Main (String[] args) { int m = 16, n = 7; Console.Write("Position = " + posOfRightMostSameBit(m, n)); } } // This code is contributed by shivanisinghss2110 |
Javascript
<script>// JavaScript implementation to find the position// of rightmost same bit // Function to find the position of// rightmost same bit in the binary// representations of 'm' and 'n'function posOfRightMostSameBit(m, n){ // Initialize loop counter let loopCounter = 1; while (m > 0 || n > 0) { // Check whether the value 'm' is odd let a = m % 2 == 1; // Check whether the value 'n' is odd let b = n % 2 == 1; // Below 'if' checks for both // values to be odd or even if (!(a ^ b)) { return loopCounter; } // Right shift value of m m = m >> 1; // Right shift value of n n = n >> 1; loopCounter++; } // When no common set is found return -1;}// Driver code let m = 16, n = 7; document.write("Position = " + posOfRightMostSameBit(m, n));// This code is contributed by Manoj.</script> |
Output:
Position = 4
Time Complexity: O(1)
Auxiliary Space: O(1)
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