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# 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 ``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

 `= 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

 ``

Output

`Yes, it is Palindrome`

Time Complexity: O(d), where d is the number of digits in a given number
Auxiliary space : O(1)

This article is contributed by Abhijit Shankhdhar. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.