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 codeint 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 spacepublic 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 codeif(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 spaceusing 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 spacefunction 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 codeif(isPalindrome(1001) == true)
echo "Yes, it is Palindrome" ;
elseecho "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> |
Output
Yes, it is Palindrome
Time Complexity: O(d), where d is the number of digits in a given number
Auxiliary space : O(1)