A Padovan string P(n) for a natural number n is defined as: P(0) = ‘X’ P(1) = ‘Y’ P(2) = ‘Z’ P(n) = P(n-2) + P(n-3), n>2 where + denotes string concatenation. For a string of the characters ‘X’ , ‘Y’ and ‘Z’ only, and given value of n, write a program that counts the number of occurrences of the string in the n-th Padovan string P(n). An example is given below. For n = 6 and the string ZY, the program should count the occurrences of ZY in P(6). P(0) = ‘X’ P(1) = ‘Y’ P(2) = ‘Z’ P(n) = P(n-2) + P(n-3), n>2 P(3) = P(1)+P(0) P(3) = YX P(4) = P(2)+P(1) P(4) = ZY P(5) = P(3)+P(2) P(5) = YXZ P(6) = P(4)+P(3) P(6) = ZYYX So, the number of occurrences of the string ZY in P(6) is 1.

program in java

// aakash , suraj , prem sasi kumar kamaraj college
program 1 :

Program set 2 :

import java.util.Scanner;

import java.util.ArrayList;

public class PadovanSeries

{

    public static void main(String[] arg)

    {

        Scanner read = new Scanner(System.in);

        System.out.println("Enter starting no. : ");

        int start = read.nextInt();

        System.out.println("Enter ending no. : ");

        int end = read.nextInt();

        int[] ans = getSeries(start, end);

        System.out.println("Padovan series : ");

        for (int a : ans)

            System.out.print(a + " ");

    }

    public static int[] getSeries(int s, int e)

    {

        ArrayList list = new ArrayList();

        int i, j = 0;

        for (i = s; i <= e; i++, j++)

            list.add(getPadovan(i));

        int[] ans = new int[j];

        for (i = 0; i < j; i++)

            ans[i] = list.get(i);

        return ans;

    }

    public static int getPadovan(int p)

    {

        if (p == 0 || p == 1 || p == 2)

            return 1;

        return (getPadovan(p - 2) + getPadovan(p - 3));

    }

}

Padova