To check a number is palindrome or not without using any extra space
Given a number ‘n’ and our goal is to find out it is palindrome or not without using
any extra space. We can’t make a new copy of the number .
Examples:
Input : 2332 Output : Yes it is Palindrome. Explanation: original number = 2332 reversed number = 2332 Both are same hence the number is palindrome. Input :1111 Output :Yes it is Palindrome. Input : 1234 Output : No not Palindrome.
A recursive solution is discussed in below post.
Check if a number is Palindrome
In this post a different solution is discussed.
1) We can compare the first digit and the last digit, then we repeat the process.
2) For the first digit, we need the order of the number. Say, 12321. Dividing this by 10000 would get us the leading 1. The trailing 1 can be retrieved by taking the mod with 10.
3 ) Now, to reduce this to 232.
(12321 % 10000)/10 = (2321)/10 = 232
4 ) And now, the 10000 would need to be reduced by a factor of 100.
Here is the implementation of the above algorithm :
C++
// C++ program to find number is palindrome // or not without using any extra space #include <bits/stdc++.h> using namespace std; bool isPalindrome( int ); bool isPalindrome( int n) { // Find the appropriate divisor // to extract the leading digit int divisor = 1; while (n / divisor >= 10) divisor *= 10; while (n != 0) { int leading = n / divisor; int trailing = n % 10; // If first and last digit // not same return false if (leading != trailing) return false ; // Removing the leading and trailing // digit from number n = (n % divisor) / 10; // Reducing divisor by a factor // of 2 as 2 digits are dropped divisor = divisor / 100; } return true ; } // Driver code int main() { isPalindrome(1001) ? cout << "Yes, it is Palindrome" : cout << "No, not Palindrome" ; return 0; } |
Java
// Java program to find number is palindrome // or not without using any extra space public class GFG { static boolean isPalindrome( int n) { // Find the appropriate divisor // to extract the leading digit int divisor = 1 ; while (n / divisor >= 10 ) divisor *= 10 ; while (n != 0 ) { int leading = n / divisor; int trailing = n % 10 ; // If first and last digit // not same return false if (leading != trailing) return false ; // Removing the leading and trailing // digit from number n = (n % divisor) / 10 ; // Reducing divisor by a factor // of 2 as 2 digits are dropped divisor = divisor / 100 ; } return true ; } // Driver code public static void main(String args[]) { if (isPalindrome( 1001 )) System.out.println( "Yes, it is Palindrome" ); else System.out.println( "No, not Palindrome" ); } } // This code is contributed by Sumit Ghosh |
Python3
# Python program to find number # is palindrome or not without # using any extra space # Function to check if given number # is palindrome or not without # using the extra space def isPalindrome(n): # Find the appropriate divisor # to extract the leading digit divisor = 1 while (n / divisor > = 10 ): divisor * = 10 while (n ! = 0 ): leading = n / / divisor trailing = n % 10 # If first and last digit # not same return false if (leading ! = trailing): return False # Removing the leading and # trailing digit from number n = (n % divisor) / / 10 # Reducing divisor by a factor # of 2 as 2 digits are dropped divisor = divisor / 100 return True # Driver code if (isPalindrome( 1001 )): print ( 'Yes, it is palindrome' ) else : print ( 'No, not palindrome' ) # This code is contributed by Danish Raza |
C#
// C# program to find number // is palindrome or not without // using any extra space using System; class GFG { static bool isPalindrome( int n) { // Find the appropriate // divisor to extract // the leading digit int divisor = 1; while (n / divisor >= 10) divisor *= 10; while (n != 0) { int leading = n / divisor; int trailing = n % 10; // If first and last digit // not same return false if (leading != trailing) return false ; // Removing the leading and // trailing digit from number n = (n % divisor) / 10; // Reducing divisor by // a factor of 2 as 2 // digits are dropped divisor = divisor / 100; } return true ; } // Driver code static public void Main () { if (isPalindrome(1001)) Console.WriteLine( "Yes, it " + "is Palindrome" ); else Console.WriteLine( "No, not " + "Palindrome" ); } } // This code is contributed by m_kit |
PHP
<?php // PHP program to find // number is palindrome // or not without using // any extra space function isPalindrome( $n ) { // Find the appropriate divisor // to extract the leading digit $divisor = 1; while ( $n / $divisor >= 10) $divisor *= 10; while ( $n != 0) { $leading = floor ( $n / $divisor ); $trailing = $n % 10; // If first and last digit // not same return false if ( $leading != $trailing ) return false; // Removing the leading and // trailing digit from number $n = ( $n % $divisor ) / 10; // Reducing divisor by a // factor of 2 as 2 digits // are dropped $divisor = $divisor / 100; } return true; } // Driver code if (isPalindrome(1001) == true) echo "Yes, it is Palindrome" ; else echo "No, not Palindrome" ; // This code is contributed by ajit ?> |
Javascript
<script> // javascript program to find number is palindrome // or not without using any extra space function isPalindrome(n) { // Find the appropriate divisor // to extract the leading digit var divisor = 1; while (parseInt(n / divisor) >= 10) divisor *= 10; while (n != 0) { var leading = parseInt(n / divisor); var trailing = n % 10; // If first and last digit // not same return false if (leading != trailing) return false ; // Removing the leading and trailing // digit from number n = parseInt((n % divisor) / 10); // Reducing divisor by a factor // of 2 as 2 digits are dropped divisor = divisor / 100; } return true ; } // Driver code if (isPalindrome(1001)) document.write( "Yes, it is Palindrome" ); else document.write( "No, not Palindrome" ); // This code is contributed by todaysgaurav </script> |
Yes, it is Palindrome
Time Complexity: O(d), where d is the number of digits in a given number
Auxiliary space : O(1)
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