Vpropel VIT | POD | 18/05/2020 | Cryptography-II | 08

cryptography_II.cpp

//Dp
#include <iostream>
#include<string>
using namespace std;
unsigned long int MOD=1000000007;
unsigned long int counter(string digits, unsigned long int n)
{
    unsigned long int count[n+1]={0};
    count[0] = 1;
    if(digits[0]=='*')
        count[1]=9;
    else
        count[1] = 1;
    for (int i = 2; i <= n; i++)
    {

        if(digits[i-2]=='*' && digits[i-1]!='*')
        {
            count[i]=count[i-1];
            char varnum='1';
            while(varnum<='9')
            {
                if((varnum<='2' && digits[i-1]<='6')||varnum=='1')
                    count[i]=(count[i]+count[i-2])%MOD;
                varnum++;
            }
        }
        else if(digits[i-1]=='*' && digits[i-2]!='*')
        {
            count[i]=(9*count[i-1])%MOD;
            char varnum='1';
            while(varnum<='9')
            {
                if((varnum<='6' && digits[i-2]<='2')||digits[i-2]=='1')
                    count[i]=(count[i]+count[i-2])%MOD;
                varnum++;
            }
        }
        else if(digits[i-1]=='*' && digits[i-2]=='*')
        {
            count[i]=(9*count[i-1])%MOD;
            char v1='1';
            char v2;
            for(v1;v1<='9';v1++)
                for(v2='1';v2<='9';v2++)
                {
                    if((v1<='2'&& v2<='6')||v1=='1')
                        count[i]=(count[i]+count[i-2])%MOD;;
                }
        }
        else
        {
            count[i] = count[i-1];
            if((digits[i-2] <='2' && digits[i-1] <='6')||digits[i-2]=='1')
                count[i]=(count[i]+count[i-2])%MOD;;
        }
    }
    /*
    // To see how Table is formed and DP applied :)
    for(int i=0;i<=n;i++)
        cout<<digits[i]<<"\t";
    cout<<endl;
    for(int i=0;i<=n;i++)
        cout<<count[i]<<'\t';
    */
    return (unsigned long int)count[n];
}

int main()
{
    string digits;
    cin>>digits;
    unsigned long int n = digits.length();

    cout<< counter(digits,n);
    return 0;
}
//                                          ### Thank YOu :) 
//                                            @cs_jawanda
//                                     .Chiranjeet Singh Jawanda.
//                                            VIT VELLORE

cryptography_II.java

//Dp
import java.util.Scanner;

public class Main
{
    public static void counter(char []digits,int n)
    {   

        final long MOD=1000000007;
        long  count[]=new long[n+1];
        count[0] = 1;
        if(digits[0]=='*')
            count[1]=9;
        else
            count[1] = 1;
        for (int i = 2; i <= n; i++)
            {

                if(digits[i-2]=='*' && digits[i-1]!='*')
                {
                    count[i]=count[i-1];
                    count[i]=(count[i]+count[i-2]*(digits[i-1]<='6'?2:1))%MOD;    
                }
                 else if(digits[i-1]=='*' && digits[i-2]!='*')
                {
                    count[i]=(9*count[i-1])%MOD;
                    char varnum='1';
                    while(varnum<='9')
                    {
                        if((varnum<='6' && digits[i-2]<='2')||digits[i-2]=='1')
                            count[i]=(count[i]+count[i-2])%MOD;
                            varnum++;
                    }
                }
                else if(digits[i-1]=='*' && digits[i-2]=='*')
                {
                    count[i]=(9*count[i-1])%MOD;
                    count[i]=(count[i]+count[i-2]*15)%MOD;
                }
                else
                {
                    count[i] = count[i-1];
                    if((digits[i-2] <='2' && digits[i-1] <='6')||digits[i-2]=='1')
                        count[i]=(count[i]+count[i-2])%MOD;;
                }
            }
            System.out.print(count[n]);
    }
    public static void main(String []args)
    {
        Scanner s=new Scanner(System.in);
        String str=s.next();
        char digits[]=str.toCharArray();
        counter(digits,digits.length);
    }
}
//                                          ### Thank YOu :) 
//                                            @cs_jawanda
//                                     .Chiranjeet Singh Jawanda.
//                                            VIT VELLORE

cryptography_II2.java

//Recursion using Memoisation
import java.util.Scanner;

 class Solve
{
	long M=1000000007;
	
	public int numDecodings(String s)
	{
		Integer[]memo=new Integer[s.length()];
		return ways(s,s.length()-1,memo);
	}
	
	public int ways(String s,int i,Integer[] memo)
	{
		if(i<0) return 1;
		if(memo[i]!=null) return memo[i];
		if(s.charAt(i)=='*')
		{
			long res=9*ways(s,i-1,memo);
			if(i>0 && s.charAt(i-1)=='1')
				res=(res+9*ways(s,i-2,memo))%M;
			else if(i>0 && s.charAt(i-1)=='2')
				res=(res+6*ways(s,i-2,memo))%M;
			else if(i>0 && s.charAt(i-1)=='*')
				res=(res+15*ways(s,i-2,memo))%M;
			memo[i]=(int)res;
			return memo[i];
		}
		long res=s.charAt(i)!='0'?ways(s,i-1,memo):0;
		if(i>0 && (s.charAt(i-1)=='1' ||(s.charAt(i-1)=='2' && s.charAt(i)<='6')))
			res=(res + ways(s,i-2,memo))%M;
		else if(i>0 && s.charAt(i-1)=='*')
			res=(res+(s.charAt(i)<='6'?2:1)*ways(s,i-2,memo))%M;
		memo[i]=(int)res;
		return memo[i];
			
	}
}

public class Practice
{
	public static void main(String []args) throws Exception
	{
		Scanner s=new Scanner(System.in);
		String str=s.next(); s.close();
		Solve obj=new Solve();
		System.out.println(obj.numDecodings(str));
	}
}

Question(cryptography_II).txt

Cryptography-II
A message containing letters from A-Z is being encoded to numbers using the following mapping way:'A' -> 1,'B' -> 2,'Z' -> 26

Beyond that, now the encoded string can also contain the character '*', which can be treated as one of the numbers from 1 to 9.

Given the encoded message containing digits and the character '*', return the total number of ways to decode it.

For example, 1* has got 18 decoding possibilities as shown below

(i) 1* - 1 may be decoded as A and * can take values from 1 to 9

(ii) 1* - Keep 1 as 1 and * can take values from 1 to 9 then characters corressponding to 11 to 19 will be deocoded

Input Format

First line contains the encoded string

Output Format

Print the decoded string

Also, since the answer may be very large, you should return the output mod 10^9 + 7.

https://leetcode.com/articles/decode-ways-ii/