Check if a number has two adjacent set bits

0 0 2 minutes read

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 <iostream> 
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

<?php 
// PHP program to check  
// if there are two 
// adjacent set bits. 
  
function adjacentSet($n) 
{ 
    return ($n & ($n >> 1)); 
} 
  
// Driver Code 
$n = 3; 
adjacentSet($n) ?  
   print("Yes") :  
     print("No"); 
  
// This code is contributed by Sam007. 
?> 

Javascript

<script> 
  
// Javascript program to check 
// if there are two 
// adjacent set bits. 
   
function adjacentSet(n) 
    { 
        let x = (n & (n >> 1)); 
          
        if(x > 0) 
            return true; 
        else
            return false; 
    } 
  
// driver program 
  
           let n = 3; 
          
        if(adjacentSet(n)) 
            document.write("Yes"); 
        else
            document.write("No");  
          
</script> 

Output :

Yes

Time Complexity : O(1)

Auxiliary Space : O(1)

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