Check if actual binary representation of a number is palindrome

Last Updated : 01 Aug, 2022

Given a non-negative integer n. The problem is to check if binary representation of n is palindrome or not. Note that the actual binary representation of the number is being considered for palindrome checking, no leading 0’s are being considered.

Examples :

Input : 9
Output : Yes
(9)₁₀ = (1001)₂

Input : 10
Output : No
(10)₁₀ = (1010)₂

Approach: Following are the steps:

Get the number obtained by reversing the bits in the binary representation of n. Refer this post. Let it be rev.
If n == rev, then print "Yes" else "No".

Implementation:

C++

// C++ implementation to check whether binary 
// representation of a number is palindrome or not
#include <bits/stdc++.h>
using namespace std;

// function to reverse bits of a number
unsigned int reverseBits(unsigned int n)
{
    unsigned int rev = 0;

    // traversing bits of 'n' from the right
    while (n > 0) {

        // bitwise left shift 'rev' by 1
        rev <<= 1;

        // if current bit is '1'
        if (n & 1 == 1)
            rev ^= 1;

        // bitwise right shift 'n' by 1
        n >>= 1;
    }

    // required number
    return rev;
}

// function to check whether binary representation
// of a number is palindrome or not
bool isPalindrome(unsigned int n)
{
    // get the number by reversing bits in the
    // binary representation of 'n'
    unsigned int rev = reverseBits(n);

    return (n == rev);
}

// Driver program to test above
int main()
{
    unsigned int n = 9;
    if (isPalindrome(n))
        cout << "Yes";
    else
        cout << "No";
    return 0;
}

Java

// Java code  to check whether  
// binary representation of a 
// number is palindrome or not
import java.util.*;
import java.lang.*;

public class GfG
{
    // function to reverse bits of a number
    public static long reverseBits(long n)
    {
        long rev = 0;

        // traversing bits of 'n' 
        // from the right
        while (n > 0) 
        {
            // bitwise left shift 'rev' by 1
            rev <<= 1;

            // if current bit is '1'
            if ((n & 1) == 1)
                rev ^= 1;

            // bitwise right shift 'n' by 1
            n >>= 1;
        }

        // required number
        return rev;
    }

    // function to check a number
    // is palindrome or not
    public static boolean isPalindrome(long n)
    {
        // get the number by reversing
        // bits in the  binary
        // representation of 'n'
        long rev = reverseBits(n);

        return (n == rev);
    }
    
    // Driver function 
    public static void main(String argc[])
    {
        long n = 9;
        if (isPalindrome(n))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
    
}

// This code is contributed by Sagar Shukla

Python3

# Python 3 implementation to check 
# whether binary representation of
# a number is palindrome or not

# function to reverse bits of a number
def reverseBits(n) :
    rev = 0
    
  # traversing bits of 'n' from the right
  while (n > 0) :

     # bitwise left shift 'rev' by 1
     rev = rev << 1

     # if current bit is '1'
     if (n & 1 == 1) :
     rev = rev ^ 1

     # bitwise right shift 'n' by 1
     n = n >> 1
    
     # required number
     return rev
    
# function to check whether binary
# representation of a number is
# palindrome or not
def isPalindrome(n) :

    # get the number by reversing 
    # bits in the binary
    # representation of 'n'
    rev = reverseBits(n)

    return (n == rev)


# Driver program to test above
n = 9
if (isPalindrome(n)) :
    print("Yes")
else :
    print("No")
    
# This code is contributed by Nikita Tiwari.

// C# code to check whether 
// binary representation of a 
// number is palindrome or not
using System;

public class GfG
{
    // function to reverse bits of a number
    public static long reverseBits(long n)
    {
        long rev = 0;

        // traversing bits of 'n' 
        // from the right
        while (n > 0) 
        {
            // bitwise left shift 'rev' by 1
            rev <<= 1;

            // if current bit is '1'
            if ((n & 1) == 1)
                rev ^= 1;

            // bitwise right shift 'n' by 1
            n >>= 1;
        }

        // required number
        return rev;
    }

    // function to check a number
    // is palindrome or not
    public static bool isPalindrome(long n)
    {
        // get the number by reversing
        // bits in the binary
        // representation of 'n'
        long rev = reverseBits(n);

        return (n == rev);
    }
    
    // Driver function 
    public static void Main()
    {
        long n = 9;
        if (isPalindrome(n))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
    
}

// This code is contributed by vt_m

PHP

<?php
// PHP implementation to check 
// whether binary representation 
// of a number is palindrome or not

// function to reverse bits of a number
function reverseBits($n)
{
    $rev = 0;

    // traversing bits of 'n' 
    // from the right
    while ($n > 0) 
    {

        // bitwise left shift 'rev' by 1
        $rev <<= 1;

        // if current bit is '1'
        if ($n & 1 == 1)
            $rev ^= 1;

        // bitwise right shift 'n' by 1
        $n >>= 1;
    }

    // required number
    return $rev;
}

// function to check whether 
// binary representation of a
// number is palindrome or not
function isPalindrome($n)
{
    // get the number by reversing 
    // bits in the binary representation 
    // of 'n'
    $rev = reverseBits($n);

    return ($n == $rev);
}

// Driver code
$n = 9;
if (isPalindrome($n))
    echo "Yes";
else
    echo "No";
return 0;

// This code is contributed by mits 
?>

JavaScript

<script>

// JavaScript program  to check whether  
// binary representation of a 
// number is palindrome or not

    // function to reverse bits of a number
    function reverseBits(n)
    {
        let rev = 0;
  
        // traversing bits of 'n' 
        // from the right
        while (n > 0) 
        {
            // bitwise left shift 'rev' by 1
            rev <<= 1;
  
            // if current bit is '1'
            if ((n & 1) == 1)
                rev ^= 1;
  
            // bitwise right shift 'n' by 1
            n >>= 1;
        }
  
        // required number
        return rev;
    }
  
    // function to check a number
    // is palindrome or not
    function isPalindrome(n)
    {
        // get the number by reversing
        // bits in the  binary
        // representation of 'n'
        let rev = reverseBits(n);
  
        return (n == rev);
    }

// Driver code

        let n = 9;
        if (isPalindrome(n))
            document.write("Yes");
        else
            document.write("No");

</script>

Output

Yes

Time Complexity: O(num), where num is the number of bits in the binary representation of n.
Auxiliary Space: O(1).

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