Given an integer ‘x’, write a C function that returns true if binary representation of x is palindrome else return false.
For example a numbers with binary representation as 10..01 is palindrome and number with binary representation as 10..00 is not palindrome.
The idea is similar to checking a string is palindrome or not. We start from leftmost and rightmost bits and compare bits one by one. If we find a mismatch, then return false.
Method#1: We follow the following logic to check binary of number is Palindrome or not:
- Find number of bits in x using sizeof() operator.
- Initialize left and right positions as 1 and n respectively.
- Do following while left ‘l’ is smaller than right ‘r’.
- If bit at position ‘l’ is not same as bit at position ‘r’, then return false.
- Increment ‘l’ and decrement ‘r’, i.e., do l++ and r–-.
- If we reach here, it means we didn’t find a mismatching bit.
- To find the bit at a given position, we can use an idea similar to this post. The expression “x & (1 << (k-1))” gives us non-zero value if bit at k’th position from right is set and gives a zero value if if k’th bit is not set.
Following is the implementation of the above algorithm
C++
#include<iostream>
using namespace std;
bool isKthBitSet(unsigned int x, unsigned int k)
{
return (x & (1 << (k - 1))) ? true : false;
}
bool isPalindrome(unsigned int x)
{
int l = 1;
int r = sizeof(unsigned int) * 8;
while (l < r)
{
if (isKthBitSet(x, l) != isKthBitSet(x, r))
return false;
l++; r--;
}
return true;
}
int main()
{
unsigned int x = 1 << 15 + 1 << 16;
cout << isPalindrome(x) << endl;
x = 1 << 31 + 1;
cout << isPalindrome(x) << endl;
return 0;
}
|
Java
class GFG
{
static int isKthBitSet(long x, long k)
{
int rslt = ((x & (1 << (k - 1))) != 0) ? 1 : 0;
return rslt;
}
static int isPalindrome( long x)
{
long l = 1;
long r = (Integer.SIZE/8 )* 8;
while (l < r)
{
if (isKthBitSet(x, l) != isKthBitSet(x, r))
{
return 0;
}
l++; r--;
}
return 1;
}
public static void main (String[] args)
{
long x = 1 << 15 + 1 << 16 ;
System.out.println(isPalindrome(x));
x = (1 << 31) + 1 ;
System.out.println(isPalindrome(x));
}
}
|
Python3
import sys
def isKthBitSet(x, k):
if ((x & (1 << (k - 1))) !=0):
return True
else:
return False
def isPalindrome(x):
l = 1
r = 2 * 8
while (l < r):
if (isKthBitSet(x, l) != isKthBitSet(x, r)):
return False
l += 1
r -= 1
return True
if __name__ =='__main__':
x = 1 << 15 + 1 << 16
print(int(isPalindrome(x)))
x = 1 << 31 + 1
print(int(isPalindrome(x)))
|
C#
using System;
class GFG
{
static int isKthBitSet(long x, long k)
{
int rslt = ((x & (1 << (int)(k - 1))) != 0) ? 1 : 0;
return rslt;
}
static int isPalindrome( long x)
{
long l = 1;
long r = 4 * 8;
while (l < r)
{
if (isKthBitSet(x, l) != isKthBitSet(x, r))
{
return 0;
}
l++; r--;
}
return 1;
}
public static void Main ()
{
long x = 1 << 15 + 1 << 16 ;
Console.WriteLine(isPalindrome(x));
x = (1 << 31) + 1 ;
Console.WriteLine(isPalindrome(x));
}
}
|
Javascript
<script>
function isKthBitSet(x, k)
{
let rslt = ((x & (1 << (k - 1))) != 0) ? 1 : 0;
return rslt;
}
function isPalindrome(x)
{
let l = 1;
let r = 4 * 8;
while (l < r)
{
if (isKthBitSet(x, l) !=
isKthBitSet(x, r))
{
return 0;
}
l++; r--;
}
return 1;
}
let x = 1 << 15 + 1 << 16;
document.write(isPalindrome(x) + "</br>");
x = (1 << 31) + 1;
document.write(isPalindrome(x));
</script>
|
PHP
<?php
function isKthBitSet($x, $k)
{
return ($x & (1 << ($k - 1))) ? true : false;
}
function isPalindrome($x)
{
$l = 1;
$r = sizeof(4) * 8;
while ($l < $r)
{
if (isKthBitSet($x, $l) != isKthBitSet($x, $r))
return false;
$l++;
$r--;
}
return true;
}
$x = 1 << 15 + 1 << 16;
echo isPalindrome($x), "\n";
$x = 1 << 31 + 1;
echo isPalindrome($x), "\n";
?>
|
Time Complexity: O(x)
Auxiliary Space: O(1)
Method#2: Using reverse() function:
- When user inputs an integer, it is passed to method which will evaluate the result.
- Actual logic inside the method focuses on following:
- It first convert the integer to binary form of integer in string format.
- It reverse the string using reverse method.
- It is palindrome if both the string is equal else not.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
string bin(unsigned n)
{ string ans;
while(n > 0){
ans = (to_string(n&1)) + ans;
n >>= 1;
}
return ans;
}
bool checkPalindrome( unsigned int n){
string s1 = bin(n);
string s2 = s1;
reverse(s2.begin(), s2.end());
return s1 == s2;
}
int main() {
unsigned int x = 1 << 15 + 1 << 16;
cout << checkPalindrome(x) << endl;
x = 10;
cout << checkPalindrome(x) << endl;
return 0;
}
|
Java
class GFG
{
static String bin(int n)
{
String ans = "";
while(n > 0){
ans = (Integer.toString(n&1)) + ans;
n >>= 1;
}
return ans;
}
static int checkPalindrome(int n){
String s1 = bin(n);
StringBuilder s2 = new StringBuilder(s1);
s2 = s2.reverse();
return s1.equals(s2.toString()) ? 1 : 0;
}
public static void main(String[] args) {
int x = 9;
System.out.println(checkPalindrome(x));
x = 10;
System.out.println(checkPalindrome(x));
}
}
|
Python
def bin(n):
ans="";
while n > 0:
ans = (str(n&1)) + ans;
n >>= 1;
return ans;
def checkPalindrome(x):
s1 = bin(x)
s2 = s1[::-1]
return 1 if s1 == s2 else 0
x = 9;
print(checkPalindrome(x))
x = 10
print(checkPalindrome(x))
|
C#
using System;
public class GFG
{
static string bin(int n)
{
string ans = "";
while (n > 0) {
ans = (Convert.ToString(n & 1)) + ans;
n >>= 1;
}
return ans;
}
static int checkPalindrome(int n)
{
string s1 = bin(n);
char[] charArray = s1.ToCharArray();
Array.Reverse(charArray);
string s2 = new string(charArray);
return s1.Equals(s2) ? 1 : 0;
}
public static void Main(string[] args)
{
int x = 9;
Console.WriteLine(checkPalindrome(x));
x = 10;
Console.WriteLine(checkPalindrome(x));
}
}
|
Javascript
function bin(n)
{ let ans="";
while(n > 0){
ans = ((n&1).toString()) + ans;
n >>= 1;
}
return ans;
}
function checkPalindrome(x){
let s1 = bin(x);
let s2 = s1.split("").reverse().join("");
return s1 === s2 ? 1 :0;
}
let x = 1 << 15 + 1 << 16 ;
console.log(checkPalindrome(x));
x = 10;
console.log(checkPalindrome(x));
|
Time Complexity: O(log(x))
Auxiliary Space: O(X)
Method 3: Using builtin method bitset<>
- Convert the given number into its binary form.
- Check if it’s palindrome or not.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int isPalindrome(int N)
{
string s = bitset<32>(N).to_string();
s = s.substr(s.find('1'));
int i = 0, j = s.size() - 1;
while (i < j) {
if (s[i] != s[j])
return false;
i++;
j--;
}
return true;
}
int main()
{
int x = 16;
cout << isPalindrome(x) << endl;
x = 17;
cout << isPalindrome(x) << endl;
return 0;
}
|
Java
import java.io.*;
class Main {
public static boolean isPalindrome(int N)
{
String s = Integer.toBinaryString(N);
int i = 0, j = s.length() - 1;
while (i < j) {
if (s.charAt(i) != s.charAt(j)) {
return false;
}
i++;
j--;
}
return true;
}
public static void main(String[] args) {
int x = 16;
System.out.println(isPalindrome(x));
x = 17;
System.out.println(isPalindrome(x));
}
}
|
Python3
import math
def isPalindrome(N):
s = bin(N)[2:]
s = s[s.index('1'):]
i = 0;
j = len(s) - 1;
while (i < j):
if (s[i] != s[j]):
return 0;
i+=1;
j-=1;
return 1;
x = 16;
print(isPalindrome(x));
x = 17;
print(isPalindrome(x));
|
C#
using System;
using System.Linq;
using System.Collections.Generic;
class GFG
{
static int isPalindrome(int N)
{
string s = Convert.ToString(N,2);
int i = 0, j = s.Length - 1;
while (i < j) {
if (s[i] != s[j])
return 0;
i++;
j--;
}
return 1;
}
static public void Main()
{
int x = 16;
Console.WriteLine(isPalindrome(x));
x = 17;
Console.WriteLine(isPalindrome(x));
}
}
|
Javascript
function isPalindrome(N)
{
const s = N.toString(2);
let i = 0, j = s.length - 1;
while (i < j) {
if (s[i] != s[j])
return 0;
i++;
j--;
}
return 1;
}
let x = 16;
console.log(isPalindrome(x));
x = 17;
console.log(isPalindrome(x));
|
Time Complexity: O(k), where k is the number of bits in the given number X
Auxiliary Space: O(k)
This article is contributed by Saurabh Gupta. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Approach#4: using | operator
This approach uses bitwise operations to reverse the binary representation of a number and then checks if it is equal to the original binary representation, indicating whether or not it is a palindrome.
Algorithm
1. Create a copy of the input number and initialize a variable to hold the reversed binary representation.
2. Loop through the bits of the input number and append each bit to the reversed binary representation.
3. Convert the reversed binary representation to an integer and compare it with the original number.
4. If the two numbers are equal, return True, else return False.
C++
#include <iostream>
using namespace std;
bool is_binary_palindrome(int num) {
int rev_binary = 0;
int copy_num = num;
while (copy_num) {
rev_binary = (rev_binary << 1) | (copy_num & 1);
copy_num >>= 1;
}
return rev_binary == num;
}
int main() {
int num = 9;
cout << is_binary_palindrome(num) << endl;
num = 10;
cout << is_binary_palindrome(num) << endl;
return 0;
}
|
Java
public class Main {
public static boolean isBinaryPalindrome(int num) {
int revBinary = 0;
int copyNum = num;
while (copyNum != 0) {
revBinary = (revBinary << 1) | (copyNum & 1);
copyNum >>= 1;
}
return revBinary == num;
}
public static void main(String[] args) {
int num = 9;
System.out.println(isBinaryPalindrome(num));
num = 10;
System.out.println(isBinaryPalindrome(num));
}
}
|
Python3
def is_binary_palindrome(num):
rev_binary = 0
copy_num = num
while copy_num:
rev_binary = (rev_binary << 1) | (copy_num & 1)
copy_num >>= 1
return rev_binary == num
num=9
print(is_binary_palindrome(num))
num=10
print(is_binary_palindrome(num))
|
C#
using System;
class GFG {
public static bool IsBinaryPalindrome(int num) {
int revBinary = 0;
int copyNum = num;
while (copyNum != 0) {
revBinary = (revBinary << 1) | (copyNum & 1);
copyNum >>= 1;
}
return revBinary == num;
}
public static void Main(string[] args) {
int num = 9;
Console.WriteLine(IsBinaryPalindrome(num));
num = 10;
Console.WriteLine(IsBinaryPalindrome(num));
}
}
|
Javascript
class GFG {
static IsBinaryPalindrome(num) {
let revBinary = 0;
let copyNum = num;
while (copyNum !== 0) {
revBinary = (revBinary << 1) | (copyNum & 1);
copyNum >>= 1;
}
return revBinary === num;
}
static main() {
let num = 9;
console.log(GFG.IsBinaryPalindrome(num));
num = 10;
console.log(GFG.IsBinaryPalindrome(num));
}
}
GFG.main();
|
Time Complexity: O(log n), where n is the value of the input number.
Auxiliary Space: O(1), for storing the loop variable and reversed binary representation.