#Padovan string, C/C++ Programming

#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).
Posted Date: 8/31/2012 6:37:58 AM | Location : United States





program 1 :
package test.padovanstring;

public class PadovanString {
public int stringOccurrences(int n, String str){

if(n >= 40)
return -1;
if(str.replaceAll("X|Y|Z","").length()>0)
return -1;
String res=pad(n);
return(res.length() - res.replaceAll(str,"").length())/(str.length());
}
public String pad(int n){
if(n == 0) return "X";
if(n == 1) return "Y";
if(n == 2) return "Z";
else return pad(n-2) + pad(n-3);

}
public static void main(String []args)
{
PadovanString p = new PadovanString();
System.out.println(p.stringOccurrences(21,"YZ"));
}
}



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));
}
}

Padovan String
Problem Description
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.
Instruction to work with Open PBT Client:

Specify the work directory path in the ''Work Directory Path'' field. The path should correspond to your solution project directory.
Download the support files by clicking the Get Dev Files.
You will find the following three folders:
bin
src
lib
in your work directory.
Code the solution in . java file inside the src folder
All required files will be downloaded to your work directory. Creating additional files is strongly discouraged.

Step 1:
In your solution file:

Implement your logic in public int stringOccurrences(int n, String str) in class PadovanString.
int n: n is an integer represents n(th) for which full main string has to formed, from which occurrence of string str has to be found.
String str : str is a string represents sub string whose occurrence has to be found in the main string.

You can create more methods but, do it in the same class.

Step 2:
In your solution keep in mind the following constraints:.

In this problem you have to write a program that counts the number of occurrences of the string in the n-th Padovan string P(n).
Padovan formula is P(n) = P(n-2) + P(n-3) where n>2 and n is the integer.
In the above Padovan formula + sign means string concatenation.
For the Padovan formula value of P(0),P(1) and P(2) is given. You have to use those value only.
n should not be greater than 40 otherwise return -1
string str should be consist of only X,Y and Z in upper case otherwise return -1

The Prototype of the method is:
public int stringOccurrences(int n, String str).
This method takes following arguments.

n is the value for which full main string has to formed, from which occurrence of string str has to be found.
str represents sub string whose occurrence has to be find in the main string.

This method returns a number of occurrence of string str in the main string.

The constraints are:

n should not be greater than 40 otherwise return -1
string str should be consist of only X,Y and Z in upper case otherwise return -1

Example 1
Input :
int n = 6
string str = ZY
Output :
1
Explanation : This question is explained in problem description.
Example 2
Input :
int n = 6
String str = ZYmm
Output :
-1
Explanation :
String str is containing character other than X, Z and Y.
Example 3
Input :
int n = 8
String str = XZ
Output :
Posted by diana | Posted Date: 9/4/2012 4:13:30 AM


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