# Check if a number has two adjacent set bits

• Difficulty Level : Easy
• Last Updated : 15 Jul, 2022

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`

Time Complexity : O(1)

Auxiliary Space  : O(1)

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