Count possible decodings of a given Digit Sequence | Set 2

Given an encoded string str consisting of digits and * which can be filled by any digit 1 – 9, the task is to find the number of ways to decode that string into a sequence of alphabets A-Z.

Note: The input string contains number from 0-9 and character ‘*’ only.
Examples:

Input: str = “1*”
Output: 18
Explanation:
Since * can be replaced by any value from (1-9),
The given string can be decoded as A[A-I] + [J-R] = 9 + 9 ways

Input: str = “12*3”
Output: 28

Naive Approach: A simple solution is to solve the problem using recursion considering all the possible decodings of the string. Below is the recursion tree for the problem:



                   12*3
              /             \
           12*(3)           12(*3) 
         /     \            /      \
    12(*)(3)  1(2*)(3)  1(2)(*3)   ""
      /    \      \       /
1(2)(*)(3) ""     ""     ""   
    /
   ""

Efficient Approach: The idea is to solve the problem using Dynamic Programming using the optimal substructure for considering all the cases of the current and the previous digits of the string and their number of ways to decode the string.

Definition of the DP State: In this problem the dp[i] denotes the number of ways to decode the string upto the i^{th} index.

Intial DP States: Intially the value of the DP states are defined as below:

// Number of ways to decode 
// an empty string
dp[0] = 1

// Number of ways to decode the
// first character
if (s[0] == '*')
    dp[1] = 9  // 9 ways
else 
    dp[1] = 1

Optimal Sub-structure: There are generally two ways to decode the current character of the string:

  • Including the current character as single-digit to decode: If the current character is used as single-digit then there can be generally two cases for the character to be:
    • Case 1: If current character is equal to the "*", Then there are 9 possible ways to take any digit from [1-9] and decode it as any character from [A-Z].
      if (current == "*")
          dp[i] += 9 * dp[i-1]
      
    • Case 2: If current character is equal to any other digit from [0-9], Then the number of possible ways to decode is equal to the number of way to decode the string upto (i-1)^{th} index.
      if (current != "*")
           dp[i] += dp[i-1]
      
  • Including the current character as two-digits to decode: If the current character is to be decoded as two-digits then there are two possible cases:
    • Case 1: If the previous character is equal to the "1" or "2", then there can be two more possible subcases which will depend upon the current character:
      • Case 1.1: If the current character is equal to the "*", then total possible ways to decode are 9 if the previous character is 1, otherwise 6 if previous character is 2.
      • Case 1.2: If the current character is less than equal to 6, Then the total number of possible way to decode the string will depend only on the number of way to decode upto the previous character. That is dp[i-2]
    • Case 2: If the previous character is "*", then there can be two more possible subcases which will depend upon the current character:
      • Case 2.1: If the current character is also "*", Then the total number of cases will be 15 * dp[i-2], because the single digits of decoding ways of previous character must be already included.
      • Case 2.2: If the current character is less than 6, Then the total number of ways will be 2*dp[i-2], because then the number of ways to choose the digits of the first character is 2. That is [1, 2].
      • Case 2.3: If the current character is any digits, Then the total number of ways will be the number of ways to decode uptill the previous digit only. That is dp[i-2].

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation to count the 
// possible decodings of the given 
// digit sequence
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to count the number of 
// ways to decode the given digit 
// sequence 
int waysToDecode2(string s) {
    int n = s.size();
      
    // Array to store the dp states 
    vector<int> dp(n+1,0);
      
    // Case of empty string
    dp[0]=1; 
      
    // Condition to check if the 
    // first character of string is 0
    if(s[0]=='0')
        return 0;
          
    // Base case for single length 
    // string
    dp[1]= ((s[0]=='*')? 9 : 1);
      
    // Bottom-up dp for the string
    for(int i=2;i<=n;i++)
    {
        // Previous character
        char first= s[i-2];
          
        // Current character
        char second= s[i-1]; 
          
        // Case to include the Current
        // digit as a single digit for 
        // decoding the string
        if(second=='*')
        {
            dp[i]+= 9*dp[i-1];
        }
        else if(second>'0')
            dp[i]+=dp[i-1];
          
        // Case to include the current 
        // character as two-digit for 
        // decoding the string
        if(first=='1'|| first=='2')
        {
              
            // Condition to check if the 
            // current character is "*"
            if(second=='*')
            {
                if(first=='1')
                    dp[i]+= 9 * dp[i-2];
                else if(first=='2')
                    dp[i]+= 6 * dp[i-2]; 
            }
              
            // Condition to check if the 
            // current character is less than 
            // or equal to 26
            else if(((first-'0')* 10 + 
                    (second-'0'))<= 26)
                dp[i]+=dp[i-2];
        }
          
        // Condition to check if the 
        // Previous digit is equal to "*"
        else if(first=='*')
        {
            if(second=='*')
            {
                dp[i]+= 15 * dp[i-2];
            }
            else if(second<='6')
                dp[i]+= 2* dp[i-2];
            else
                dp [i]+= dp[i-2];
        }
    }
    return dp[n];
}
  
// Driver Code
int main() {
    string str = "12*3";
      
    // Function Call
    cout << waysToDecode2(str) << endl;
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation to count the 
// possible decodings of the given 
// digit sequence
class GFG{
  
// Function to count the number of 
// ways to decode the given digit 
// sequence 
static int waysToDecode2(char []s) 
{
      
    int n = s.length;
      
    // Array to store the dp states 
    int []dp = new int[n + 1];
      
    // Case of empty String
    dp[0] = 1
      
    // Condition to check if the 
    // first character of String is 0
    if(s[0] == '0')
        return 0;
          
    // Base case for single length 
    // String
    dp[1] = ((s[0] == '*') ? 9 : 1);
      
    // Bottom-up dp for the String
    for(int i = 2; i <= n; i++)
    {
         
       // Previous character
       char first = s[i - 2];
         
       // Current character
       char second = s[i - 1]; 
         
       // Case to include the Current
       // digit as a single digit for 
       // decoding the String
       if(second == '*')
       {
           dp[i] += 9 * dp[i - 1];
       }
       else if(second > '0')
           dp[i] += dp[i - 1];
         
       // Case to include the current 
       // character as two-digit for 
       // decoding the String
       if(first == '1' || first == '2')
       {
             
           // Condition to check if the 
           // current character is "*"
           if(second == '*')
           {
               if(first == '1')
                  dp[i] += 9 * dp[i - 2];
               else if(first == '2')
                  dp[i] += 6 * dp[i - 2]; 
           }
             
           // Condition to check if the 
           // current character is less than 
           // or equal to 26
           else if(((first - '0') * 10
                   (second - '0')) <= 26)
           {
               dp[i] += dp[i - 2];
           }
       }
         
       // Condition to check if the 
       // previous digit is equal to "*"
       else if(first == '*')
       {
           if(second == '*')
           {
               dp[i] += 15 * dp[i - 2];
           }
           else if(second <= '6')
           {
               dp[i] += 2 * dp[i - 2];
           }
           else
           {
               dp[i] += dp[i - 2];
           }
       }
    }
    return dp[n];
}
  
// Driver Code
public static void main(String[] args)
{
    String str = "12*3";
      
    // Function Call
    System.out.print(waysToDecode2(
                     str.toCharArray()) + "\n");
}
}
  
// This code is contributed by amal kumar choubey

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation to count the 
// possible decodings of the given 
// digit sequence
using System;
  
class GFG{
  
// Function to count the number of 
// ways to decode the given digit 
// sequence 
static int waysToDecode2(char []s) 
{
    int n = s.Length;
      
    // Array to store the dp states 
    int []dp = new int[n + 1];
      
    // Case of empty String
    dp[0] = 1; 
      
    // Condition to check if the 
    // first character of String is 0
    if(s[0] == '0')
        return 0;
          
    // Base case for single length 
    // String
    dp[1] = ((s[0] == '*') ? 9 : 1);
      
    // Bottom-up dp for the String
    for(int i = 2; i <= n; i++)
    {
         
       // Previous character
       char first = s[i - 2];
         
       // Current character
       char second = s[i - 1]; 
         
       // Case to include the current
       // digit as a single digit for 
       // decoding the String
       if(second == '*')
       {
           dp[i] += 9 * dp[i - 1];
       }
       else if(second > '0')
       {
           dp[i] += dp[i - 1];
       }
         
       // Case to include the current 
       // character as two-digit for 
       // decoding the String
       if(first == '1' || first == '2')
       {
             
           // Condition to check if the 
           // current character is "*"
           if(second == '*')
           {
               if(first == '1')
               {
                   dp[i] += 9 * dp[i - 2];
               }
               else if(first == '2')
               {
                   dp[i] += 6 * dp[i - 2];
               
           }
             
           // Condition to check if the 
           // current character is less than 
           // or equal to 26
           else if(((first - '0') * 10 + 
                   (second - '0')) <= 26)
           {
               dp[i] += dp[i - 2];
           }
       }
         
       // Condition to check if the 
       // previous digit is equal to "*"
       else if(first == '*')
       {
           if(second == '*')
           {
               dp[i] += 15 * dp[i - 2];
           }
           else if(second <= '6')
           {
               dp[i] += 2 * dp[i - 2];
           }
           else
           {
               dp[i] += dp[i - 2];
           }
       }
    }
    return dp[n];
}
  
// Driver Code
public static void Main(String[] args)
{
    String str = "12*3";
      
    // Function Call
    Console.Write(waysToDecode2(
                  str.ToCharArray()) + "\n");
}
}
  
// This code is contributed by amal kumar choubey

chevron_right


Output:

28

Performance Analysis:

  • Time Complexity: O(N)
  • Auxiliary Space: O(N)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.