# Check whether the two numbers differ at one bit position only

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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Following are the steps:

1. Calculate num = a ^ b.
2. 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 ` `using` `namespace` `std; ` ` `  `// function to check if x is power of 2 ` `bool` `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 only ` `bool` `differAtOneBitPos(unsigned ``int` `a, ` `                       ``unsigned ``int` `b) ` `{ ` `    ``return` `isPowerOfTwo(a ^ b); ` `} ` ` `  `// Driver program to test above ` `int` `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 only ` `import` `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 2 ` `def` `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 only ` `def` `differAtOneBitPos( a , b ): ` `    ``return` `isPowerOfTwo(a ^ b) ` `     `  `# Driver code to test above ` `a ``=` `13` `b ``=` `9` `if` `(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 only ` `using` `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

 ` `

Output:

```Yes
```

Time Complexity: O(1).

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Improved By : Sam007, rachana soma, ihritik

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