Check if actual binary representation of a number is palindrome
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”.
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 numberunsigned 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 notbool 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 aboveint 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 notimport 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 numberdef 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 notdef isPalindrome(n) : # get the number by reversing # bits in the binary # representation of 'n' rev = reverseBits(n) return (n == rev)# Driver program to test aboven = 9if (isPalindrome(n)) : print("Yes")else : print("No") # This code is contributed by Nikita Tiwari. |
C#
// C# code to check whether// binary representation of a// number is palindrome or notusing 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 numberfunction 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 $rev;}// function to check whether// binary representation of a// number is palindrome or notfunction isPalindrome($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.
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