How to find Xth palindrome?

Question

Dear Friends:

• As much as strings, some numbers are also palindrome. For instance: 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, ... , 101, 111, ... ,753537, ... and so on.

• Here is the thing, We need to figure a way to find first 10.000 palindromic numbers in order to respond user's entry. Starts from 1 to 10000th palindromic number. For example if user enters 12 it means what is the 12th palindromic number between 1 and 10.000 ?

• The input consists of a series of lines with each line containing one integer value i (1 <= i <= 10000). This integer value i indicates the index of the palindrome number that is to be written to the output, where index 1 stands for the first palindrome number (1), index 2 stands for the second palindrome number (2) and so on.

EX:

Input 1 --> Output should be: 1

Input 12 --> Output should be: 33

Input 24 --> Output should be: 151

    import java.util.Scanner;

    public class xth_palindrome
    {
        // Some Code may be here

        public static void main(String[] args)
        {
            @SuppressWarnings("resource")
            Scanner read = new Scanner(System.in);
            System.out.println("Enter values as much as you want. To stop Enter \"0\" ");

            int Xth;

            do
            {
                Xth = read.nextInt();

                // Some coding here 

                System.out.println(Xth + " palindromic num is " + "????");

            } while(Xth != 0);
        }
    }

• By the way: time limit is 1 second. Considering these factors What is the right Algorithm to solve this problem ? If you could help me and show the solution code wise in Java I would very appreciate it. Thanks for checking!

Answer 1

We can iterate through the palindromes quite quickly. Note that

1. If there is an odd palindrome ABCBA, the next larger palindrome will be ABDBA where D=C+1

2. If there is an even palindrome ABCCBA, the next larger palindrome will be ABDDBA where D=C+1

The reasoning is simple. Any other number will also increment a larger MSB and hence the next higher palindrome will have changes in the centre.

Now if C = 9, we will need to increment B and reset C to 0 making the cases become AE0EA and AE00EA where E=B+1. This method is easily extensible and you can iterate through palindromes. Since we need to find at most 10,000 of them, a second should be more than enough for an iterative approach.

Answer 2

Maybe not "the very best way", but works fine.

And it does the job in less than 1 sec (depending of your hardware).

I've tested here.

import java.util.Scanner;

public class HelloWorld{

     public static void main(String []args){

            Scanner read = new Scanner(System.in);
            System.out.println("Enter values as much as you want (lower than 100000).\n To stop Enter \"0\" ");

            int Xth;

            do
            {
                Xth = read.nextInt();


                // Some coding here 
                if (Xth > 100000)
                {
                    System.out.println("Type a number lower than 100000");
                    continue;
                }
                int count = 0;
                for (int i = 1; i <= 1000000000; i++)
                {
                    if (!isPalindrome(i))
                        continue;

                    count++;
                    if (count == Xth)
                    {
                        System.out.println(Xth + "th palindromic number is " + i);
                        break;
                    }
                }
                if (count != Xth)
                    System.out.println("I can't compute that!");


            } while(Xth != 0);
     }

     private static StringBuilder sb = new StringBuilder();

     private static boolean isPalindrome(int i)
     {
        sb.setLength(0);
        sb.append(i);
        return  sb.toString().equals(sb.reverse().toString());
     }