“Policz możliwe dekodowania” Kod odpowiedzi

Policz możliwe dekodowania

// A Dynamic Programming based C#
// implementation to count decodings
using System;
 
class GFG
{
     
// A Dynamic Programming based
// function to count decodings
static int countDecodingDP(char[] digits, int n)
{
    // A table to store results of subproblems
    int[] count = new int[n + 1];
    count[0] = 1;
    count[1] = 1;
 
    for (int i = 2; i <= n; i++)
    {
        count[i] = 0;
 
        // If the last digit is not 0,
        // then last digit must add to
        // the number of words
        if (digits[i - 1] > '0')
            count[i] = count[i - 1];
 
        // If second last digit is smaller
        // than 2 and last digit is smaller
        // than 7, then last two digits
        // form a valid character
        if (digits[i - 2] == '1' ||(digits[i - 2] == '2' && digits[i - 1] < '7'))
            count[i] += count[i - 2];
    }
    return count[n];
}
 
// Driver Code
    public static void Main()
    {
        char[] digits = {'1','2','3','4'};
        int n = digits.Length;
        Console.WriteLine("Count is " +
                countDecodingDP(digits, n));
    }
}
 
// This code is contributed
// by Akanksha Rai
// Modified by Atanu Sen