Leetcode 639. Decode Ways II

Seawolf

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.

Also, since the answer may be very large, you should return the output mod 109 + 7.

Example 1:
Input: "*"
Output: 9
Explanation: The encoded message can be decoded to the string: "A", "B", "C", "D", "E", "F", "G", "H", "I".
Example 2:
Input: "1*"
Output: 9 + 9 = 18
Note:
The length of the input string will fit in range [1, 105].
The input string will only contain the character '*' and digits '0' - '9'.

1. class Solution:
   
2.     def numDecodings(self, s: str) -> int:
   
3.         MOD = 1000000007
   
4.         n = len(s)
   
5.         dp = [0 for _ in range(n + 1)]
   
6.         dp[0] = 1
   
7.         
   
8.         def getCnt(c):
   
9.             if c == '0': return 0
   
10.             if c == '*': return 9
    
11.             return 1
    
12.         
    
13.         def getCnt2(c2, c1):
    
14.             if c2 == '0' or (c2 >= '3' and c2 <= '9'): return 0
    
15.             if c2 == '1':
    
16.                 if c1 == '*': return 9
    
17.                 else: return 1
    
18.             if c2 == '2':
    
19.                 if c1 >= '0' and c1 <= '6': return 1
    
20.                 if c1 >= '7' and c1 <= '9': return 0
    
21.                 return 6
    
22.             
    
23.             if c1 >= '0' and c1 <= '6': return 2
    
24.             if c1 >= '7' and c1 <= '9': return 1
    
25.             return 15
    
26.         
    
27.         for i in range(1, n + 1):
    
28.             dp[i] = 0
    
29.             c = getCnt(s[i - 1])
    
30.             dp[i] += c * dp[i - 1]
    
31.             
    
32.             if i > 1:
    
33.                 c = getCnt2(s[i - 2], s[i - 1])
    
34.                 dp[i] += c * dp[i - 2]
    
35.             
    
36.             dp[i] = dp[i] % MOD
    
37.         
    
38.         return dp[n]

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