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Check if a number has two adjacent set bits
• Difficulty Level : Easy
• Last Updated : 03 May, 2021

Given a number you have to check whether there is pair of adjacent set bit or not.
Examples :

```Input : N = 67
Output : Yes
There is a pair of adjacent set bit
The binary representation is 100011

Input : N = 5
Output : No```

A simple solution is to traverse all bits. For every set bit, check if next bit is also set.
An efficient solution is to shift number by 1 and then do bitwise AND. If bitwise AND is non-zero then there are two adjacent set bits. Else not.

## C++

 `// CPP program to check``// if there are two``// adjacent set bits.``#include ``using` `namespace` `std;` `bool` `adjacentSet(``int` `n)``{``    ``return` `(n & (n >> 1));``}` `// Driver Code``int` `main()``{``    ``int` `n = 3;``    ``adjacentSet(n) ?``     ``cout << ``"Yes"` `:``       ``cout << ``"No"``;``    ``return` `0;``}`

## Java

 `// Java program to check``// if there are two``// adjacent set bits.``class` `GFG``{``    ` `    ``static` `boolean` `adjacentSet(``int` `n)``    ``{``        ``int` `x = (n & (n >> ``1``));``        ` `        ``if``(x > ``0``)``            ``return` `true``;``        ``else``            ``return` `false``;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `main(String args[])``    ``{` `        ``int` `n = ``3``;``        ` `        ``if``(adjacentSet(n))``            ``System.out.println(``"Yes"``);``        ``else``            ``System.out.println(``"No"``);` `    ``}``}` `// This code is contributed by Sam007.`

## Python3

 `# Python 3 program to check if there``# are two adjacent set bits.` `def` `adjacentSet(n):``    ``return` `(n & (n >> ``1``))` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``n ``=` `3``    ``if` `(adjacentSet(n)):``        ``print``(``"Yes"``)``    ``else``:``        ``print``(``"No"``)``        ` `# This code is contributed by``# Shashank_Sharma`

## C#

 `// C# program to check``// if there are two``// adjacent set bits.``using` `System;` `class` `GFG``{``    ``static` `bool` `adjacentSet(``int` `n)``    ``{``        ``int` `x = (n & (n >> 1));``        ` `        ``if``(x > 0)``            ``return` `true``;``        ``else``            ``return` `false``;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main ()``    ``{``        ``int` `n = 3;``        ` `        ``if``(adjacentSet(n))``            ``Console.WriteLine(``"Yes"``);``        ``else``            ``Console.WriteLine(``"No"``);``    ``}``        ` `}` `// This code is contributed by Sam007.`

## php

 `> 1));``}` `// Driver Code``\$n` `= 3;``adjacentSet(``\$n``) ?``   ``print``(``"Yes"``) :``     ``print``(``"No"``);` `// This code is contributed by Sam007.``?>`

## Javascript

 ``

Output :

`Yes`

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