Question

Write a recursive a Java code that checks if a number is Palindrome. A
palindrome number is a number that reads the same from beginning to end and
from end to beginning, in other words, a palindrome number remains the same
when its digits are reversed. For example, 13431 is a palindrome number. 2332
is another one. (Note: Your algorithm should define and work with an integer
number)
The functionality of your code should be commented to explain what you do in
each part of the code.
You need to run your code for the following test cases and show the result:
1) 0          (output: yes)
2) 1234554321     (output: yes)
3) 123454321    (output: yes)
4) 1221       (output: yes)
5)   1234       (output: no)
6) 7676       (output: no)
7) -121      (output: yes)
8) -456      (output: no)
What is the time complexity of your algorithm? Cleary justify your
answer.

Answer 1

import java.util.*;
public class Main
{
   public static void main(String[] args) {
       Scanner s=new Scanner(System.in);
       int num=s.nextInt();//reading number
       int temp,res=0;
       if(num<0)//if it is negative make positive
       num=-num;
       temp=num;//assigning to temparory variable
       while(temp!=0)
       {
       res=(res*10)+temp%10;//making it reverse
       temp=temp/10;
       }
       if(res==num)//checking original and result number
       {
       System.out.println("output:yes");
       }
       else
       System.out.println("output:No");
   }
}

Time Complexity

The time complexity for this one is O(log n) because we are doing num%10 && num/10. so, these type of Operations gives log n complexity and remaining statements are considered constant 1 and finally it gives O(log n) complexity.