Check Palindrome (for numbers) - Algorithm, flowchart, Pseudocode, Implementation

Given a number, write a algorithm, flowchart, pseudocode to check if it is palindrome or not.

A string is said to be palindrome if reverse of the string is same as string. For example, 1221 is palindrome, but 1223 is not palindrome.

We will follow a method where we get each numbers using mathematical operations.

Logic

1. Get the number to check and store it in n.

2. Create empty variables r = 0, s = 0 andt = n. Here, s will become the reversed number, t will be a reference to original number, r will be a temporary variable.

3. Start a while loop checking whether the n != 0(if n is 0 exit the while loop).

4. Inside the while loop do the following operations.

   r = n % 10
   s = s * 10 + r
   n = n / 10

   First get the last digit of n by dividing the given number(n) by 10 and finding the remainder(the % operator) and store it in r. Then, add r to s after multiplying it by previous stored value of s by 10. And finally divide n by 10 and store it again in n. This way the number is reversed and stored in s.

5. After the loop ends, check whether s == t(whether the original number is same as reversed number).

6. If s == t, display that n palindrome, else display n is not a palindrome.

Flowchart

WIP, have a look here.

graph TD
    A((Start)) --> B[/Read Number to check as n/];
    B --> C[Assign s=0, t = n];
    C --> D{ While n!=0 }; 
    D -- True --> E["r = n % 10 <br>s = s * 10 + r<br>n = n / 10"] --> D
    D -- False --> F{Is n == t?}
    F -- Yes --> G[/Print n is palindrome/] --> I[Stop];
    F -- No --> H[/Print n is not a palindrome/] --> I((Stop));

Pseudocode

BEGIN
READ number to check as n
SET s = 0
SET t = n
WHILE n != 0 DO
	r = n % 10
	s = s * 10 + r
	n = n / 10
ENDWHILE
IF n == t
	DISPLAY n is palindrome
ELSE
	DISPLAY n is not a palindrome
ENDIF
END

Implementation

1. In C:

   #include <stdio.h>
   
   /* Function to check if n is Palindrome*/
   int is_palindrome(int n)
   {
       int s = 0;
       int t = n;
       int r = 0;
       while (n != 0 )
       {
           r = n % 10;
           s = s * 10 + r;
           n = n / 10;
       }
       // Check if rev_n and n are same or not.
       if (s == t)
           return 1;
       else
           return 0;
   }
   
   int main()
   {
       int tocheck = 1221;
       int op = is_palindrome(to_check);
       if (op == 1)
           printf("%d is a palindrome", tocheck);
       else
           printf("%d is not a palindrome", tocheck);
       return 0;
   }

2. In Python:

   def is_palindrome(n: int) -> bool:
       s = 0
       t = n
       r = 0
       while n != 0:
           r = n % 10
           s = s * 10 + r
           n = n / 10
       if t == s:
           return True
      	else:
           return False
   number = 1001
   palim = is_palindrom(number)
   if palim:
       print(f"{number} is a palindrom")
   else:
       print(f"{number} is not a palindrom")