Check if binary representation is palindrome

Given an integer x, determine whether its binary representation is a palindrome. Return true if it is a palindrome; otherwise, return false.

Input: x = 9
Output: true
Explanation: The binary representation of 9 is 1001, which is a palindrome.

Input: x = 10
Output: false
Explanation: The binary representation of 10 is 1010, which is not a palindrome.

Try it on GfG Practice

Table of Content

Compare Bits from Both Ends
Using reverse() Function
Using Bitwise Reverse

Compare Bits from Both Ends

Compare bits from the leftmost and rightmost positions. If all corresponding bits match, the binary representation is a palindrome.

Find the number of bits in x.
Set l = 0 (leftmost) and r = n - 1 (rightmost).
While l < r, compare bits at positions l and r.
If they differ, return false; otherwise move inward.
If no mismatch is found, return true.

C++

#include <iostream>
#include <cmath>
using namespace std;

bool isBinaryPalindrome(int x) {
    
    if (x == 0) return true;
    
    // Find number of bits
    int n = log2(x) + 1; 
    int l = 0, r = n - 1;

    while (l < r) {
        
        // Compare bits at positions l and r
        if (((x >> l) & 1) != ((x >> r) & 1)) {
            return false;
        }
        l++;
        r--;
    }
    return true;
}

int main() {
    cout << (isBinaryPalindrome(9) ? "true" : "false") << endl;  
    cout << (isBinaryPalindrome(10) ? "true" : "false") << endl; 
    return 0;
}

Java

class GfG {
    
    static boolean isBinaryPalindrome(int x) {
        
        if (x == 0) return true;
        
        // Find number of bits
        int n = (int)(Math.log(x) / Math.log(2)) + 1;
        int l = 0, r = n - 1;

        while (l < r) {
            
            // Compare bits at positions l and r
            if (((x >> l) & 1) != ((x >> r) & 1)) {
                return false;
            }
            l++;
            r--;
        }
        return true;
    }

    public static void main(String[] args) {
        System.out.println(isBinaryPalindrome(9) ? "true" : "false");  
        System.out.println(isBinaryPalindrome(10) ? "true" : "false"); 
    }
}

Python

import math

def isBinaryPalindrome(x):
    
    if (x == 0) return true;
    
    # Find number of bits
    n = int(math.log2(x)) + 1  
    l, r = 0, n - 1

    while l < r:
        
        # Compare bits at positions l and r
        if ((x >> l) & 1) != ((x >> r) & 1):
            return False
        l += 1
        r -= 1
    return True

if __name__ == "__main__":
    print("true" if isBinaryPalindrome(9) else "false")  
    print("true" if isBinaryPalindrome(10) else "false")

using System;

class GfG {
    public static bool IsBinaryPalindrome(int x) {
        
        if (x == 0) return true;
        
        // Find number of bits
        int n = (int)(Math.Log(x) / Math.Log(2)) + 1; 
        int l = 0, r = n - 1;

        while (l < r) {
            
            // Compare bits at positions l and r
            if (((x >> l) & 1) != ((x >> r) & 1)) {
                return false;
            }
            l++;
            r--;
        }
        return true;
    }

    public static void Main() {
        Console.WriteLine(IsBinaryPalindrome(9) ? "true" : "false");  
        Console.WriteLine(IsBinaryPalindrome(10) ? "true" : "false"); 
    }
}

JavaScript

function isBinaryPalindrome(x) {
    
    if (x == 0) return true;
    
    // Find number of bits
    let n = Math.floor(Math.log2(x)) + 1; 
    let l = 0, r = n - 1;

    while (l < r) {
        
        // Compare bits at positions l and r
        if (((x >> l) & 1) !== ((x >> r) & 1)) {
            return false;
        }
        l++;
        r--;
    }
    return true;
}

// Driver Code
console.log(isBinaryPalindrome(9) ? "true" : "false");  
console.log(isBinaryPalindrome(10) ? "true" : "false");

Output

true
false

Time Complexity: O(log x), as the loop runs once for each bit in x.
Space Complexity: O(1).

Using reverse() Function

Convert the integer to its binary string form and check if it is equal to its reverse using the reverse() function.

Convert x to a binary string (without leading zeros).
Store the binary string.
Reverse it using reverse().
Compare original and reversed strings.
If equal, return true; otherwise, return false.

C++

#include <iostream>
#include <bitset>
#include <algorithm>
using namespace std;

bool isBinaryPalindrome(int x) {
    
    // Convert to binary
    string binary = bitset<32>(x).to_string(); 
    
    // Remove leading zeros
    binary = binary.substr(binary.find('1')); 
    
    string revBinary = binary;
    
     // Reverse the string
    reverse(revBinary.begin(), revBinary.end());
    
    return binary == revBinary;
}

int main() {
    cout << (isBinaryPalindrome(9) ? "true" : "false") << endl;
    cout << (isBinaryPalindrome(10) ? "true" : "false") << endl;
    return 0;
}

Java

import java.util.*;

class GfG {
    static boolean isBinaryPalindrome(int x) {
        
        // Convert to binary
        String binary = Integer.toBinaryString(x);
        
        // Reverse the string
        String revBinary = new StringBuilder(binary).reverse().toString(); 
        
        return binary.equals(revBinary); 
    }

    public static void main(String[] args) {
        System.out.println(isBinaryPalindrome(9) ? "true" : "false");
        System.out.println(isBinaryPalindrome(10) ? "true" : "false");
    }
}

Python

def isBinaryPalindrome(x):
    
    # Convert to binary and remove leading zeros
    binary = bin(x)[2:]  
    
    # Reverse and compare
    return binary == binary[::-1]  

if __name__ == "__main__":
    print("true" if isBinaryPalindrome(9) else "false")
    print("true" if isBinaryPalindrome(10) else "false")

using System;

class GfG {
    public static bool IsBinaryPalindrome(int x) {
        
        // Convert to binary
        string binary = Convert.ToString(x, 2); 
        char[] revBinary = binary.ToCharArray();
        
        // Reverse the string
        Array.Reverse(revBinary); 
        
        return binary == new string(revBinary);
    }

    public static void Main() {
        Console.WriteLine(IsBinaryPalindrome(9) ? "true" : "false");
        Console.WriteLine(IsBinaryPalindrome(10) ? "true" : "false");
    }
}

JavaScript

function isBinaryPalindrome(x) {
    
    // Convert to binary
    let binary = x.toString(2); 
    
    // Reverse the string
    let revBinary = binary.split("").reverse().join(""); 
    
    return binary === revBinary;
}

// Driver Code
console.log(isBinaryPalindrome(9) ? "true" : "false");
console.log(isBinaryPalindrome(10) ? "true" : "false");

Output

true
false

Time Complexity: O(log x), as operations are performed on all bits of x.
Space Complexity: O(log x), due to storing the binary string and its reverse.

Using Bitwise Reverse

Reverse the binary representation using bitwise operations and compare it with the original number.

Create a copy of the input and initialize rev = 0.
Extract bits one by one and build the reversed number using OR.
Compare the reversed number with the original.
If equal, return true; otherwise, return false.

C++

#include <iostream>
using namespace std;

bool isBinaryPalindrome(int x) {
    int rev = 0, temp = x;

    while (temp > 0) {
        
        // Append LSB to rev
        rev = (rev << 1) | (temp & 1); 
        
        // Shift temp right
        temp >>= 1; 
    }

    return x == rev;
}

int main() {
    cout << (isBinaryPalindrome(9) ? "true" : "false") << endl;
    cout << (isBinaryPalindrome(10) ? "true" : "false") << endl;
    return 0;
}

Java

class GfG {
    static boolean isBinaryPalindrome(int x) {
        int rev = 0, temp = x;

        while (temp > 0) {
            
            // Append LSB to rev
            rev = (rev << 1) | (temp & 1);
            
            // Shift temp right
            temp >>= 1; 
        }

        return x == rev;
    }

    public static void main(String[] args) {
        System.out.println(isBinaryPalindrome(9) ? "true" : "false");
        System.out.println(isBinaryPalindrome(10) ? "true" : "false");
    }
}

Python

def isBinaryPalindrome(x):
    rev, temp = 0, x

    while temp > 0:
        
        # Append LSB to rev
        rev = (rev << 1) | (temp & 1)  
        
        # Shift temp right
        temp >>= 1  

    return x == rev

if __name__ == "__main__":
    print("true" if isBinaryPalindrome(9) else "false")
    print("true" if isBinaryPalindrome(10) else "false")

using System;

class GfG {
    public static bool IsBinaryPalindrome(int x) {
        int rev = 0, temp = x;

        while (temp > 0) {
            
            // Append LSB to rev
            rev = (rev << 1) | (temp & 1);
            
            // Shift temp right
            temp >>= 1; 
        }

        return x == rev;
    }

    public static void Main() {
        Console.WriteLine(IsBinaryPalindrome(9) ? "true" : "false");
        Console.WriteLine(IsBinaryPalindrome(10) ? "true" : "false");
    }
}

JavaScript

function isBinaryPalindrome(x) {
    let rev = 0, temp = x;

    while (temp > 0) {
        
        // Append LSB to rev
        rev = (rev << 1) | (temp & 1); 
        
        // Shift temp right
        temp >>= 1; 
    }

    return x === rev;
}

// Driver Code
console.log(isBinaryPalindrome(9) ? "true" : "false");
console.log(isBinaryPalindrome(10) ? "true" : "false");

Output

true
false

Time Complexity: O(log x)
Space Complexity: O(1)

Check if binary representation is palindrome

Compare Bits from Both Ends

Using reverse() Function

Using Bitwise Reverse

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