Given two non-negative integers a and b. The problem is to check whether the two numbers differ at one bit position only or not.
Examples:
Input : a = 13, b = 9 Output : Yes (13)10 = (1101)2 (9)10 = (1001)2 Both the numbers differ at one bit position only, i.e, differ at the 3rd bit from the right. Input : a = 15, b = 8 Output : No
Approach: Following are the steps:
- Calculate num = a ^ b.
- Check whether num is a power of 2 or not. Refer this post.
C++
// C++ implementation to check whether the two // numbers differ at one bit position only#include <bits/stdc++.h>using namespace std;// function to check if x is power of 2bool isPowerOfTwo(unsigned int x){ // First x in the below expression is // for the case when x is 0 return x && (!(x & (x - 1)));}// function to check whether the two numbers// differ at one bit position onlybool differAtOneBitPos(unsigned int a, unsigned int b){ return isPowerOfTwo(a ^ b);}// Driver program to test aboveint main(){ unsigned int a = 13, b = 9; if (differAtOneBitPos(a, b)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java implementation to check whether the two // numbers differ at one bit position onlyimport java.io.*;import java.util.*;class GFG { // function to check if x is power of 2 static boolean isPowerOfTwo(int x) { // First x in the below expression is // for the case when x is 0 return x!= 0 && ((x & (x - 1)) == 0); } // function to check whether the two numbers // differ at one bit position only static boolean differAtOneBitPos(int a, int b) { return isPowerOfTwo(a ^ b); } // Driver code public static void main(String args[]) { int a = 13, b = 9; if (differAtOneBitPos(a, b) == true) System.out.println("Yes"); else System.out.println("No"); }} // This code is contributed by rachana soma |
Python3
# Python3 implementation to check whether the two # numbers differ at one bit position only# function to check if x is power of 2def isPowerOfTwo( x ): # First x in the below expression is # for the case when x is 0 return x and (not(x & (x - 1)))# function to check whether the two numbers# differ at one bit position onlydef differAtOneBitPos( a , b ): return isPowerOfTwo(a ^ b) # Driver code to test abovea = 13b = 9if (differAtOneBitPos(a, b)): print("Yes")else: print( "No")# This code is contributed by "Sharad_Bhardwaj". |
C#
// C# implementation to check whether the two // numbers differ at one bit position onlyusing System;class GFG { // function to check if x is power of 2 static bool isPowerOfTwo(int x) { // First x in the below expression is // for the case when x is 0 return x != 0 && ((x & (x - 1)) == 0); } // function to check whether the two numbers // differ at one bit position only static bool differAtOneBitPos(int a, int b) { return isPowerOfTwo(a ^ b); } // Driver code public static void Main() { int a = 13, b = 9; if (differAtOneBitPos(a, b) == true) Console.WriteLine("Yes"); else Console.WriteLine("No"); }} // This code is contributed by ihritik |
PHP
<?php// PHP implementation to check // whether the two numbers differ// at one bit position only // function to check if x is power of 2 function isPowerOfTwo($x) { $y = 0; // First x in the below expression is // for the case when x is 0 if($x && (!($x & ($x - 1)))) $y = 1; return $y; } // function to check whether // the two numbers differ at // one bit position only function differAtOneBitPos($a,$b) { return isPowerOfTwo($a ^ $b); } // Driver Code $a = 13; $b = 9; if (differAtOneBitPos($a, $b)) echo "Yes"; else echo "No"; // This code is contributed by Sam007?> |
Javascript
<script>// JavaScript program to check whether the two // numbers differ at one bit position only // function to check if x is power of 2 function isPowerOfTwo(x) { // First x in the below expression is // for the case when x is 0 return x!= 0 && ((x & (x - 1)) == 0); } // function to check whether the two numbers // differ at one bit position only function differAtOneBitPos(a, b) { return isPowerOfTwo(a ^ b); } // Driver code let a = 13, b = 9; if (differAtOneBitPos(a, b) == true) document.write("Yes"); else document.write("No"); // This code is contributed by code_hunt.</script> |
Output:
Yes
Time Complexity: O(1).
Auxiliary Space: O(1).
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