[LeetCode] 639. Decode Ways 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.
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:
  1. The length of the input string will fit in range [1, 105].
  2. The input string will only contain the character '*' and digits '0' - '9'.

Thought process:
Dynamic programming.
  1. Sub-problem: find the total number of ways to decode a sub-string of s.
  2. Function:
    1. If s[i] == 0:
      1. If 0 < s[i - 1] <= 2, f[i] = f[i - 2].
      2. If s[i - 1] == *, * can be 1 or 2. f[i] = f[i - 2] * 2.
      3. Else, there is no way to decode it. Return 0.
    2. If s[i] > 0:
      1. If s[i - 1] != 0 and s[i - 1, i] <= 26, f[i] = f[i - 1] + f[i - 2].
      2. If s[i - 1] == *:
        1. If s[i] <= 6, * can be 1 or 2. f[i] = f[i - 1] + f[i - 2] * 2.
        2. Else, * can only be 1. f[i] = f[i - 1] + f[i - 2].
      3. Else, f[i] = f[i - 1].
    3. If s[i] == *:
      1. If s[i - 1] == 0, f[i] = f[i - 1] * 9.
      2. If s[i - 1] == 1, f[i] = f[i - 1] * 9 + f[i - 2] * 9.
      3. If s[i - 1] == 2, f[i] = f[i - 1] * 9 + f[i - 2] * 6.
      4. If s[i - 1] == *, f[i] = f[i - 1] * 9 + f[i - 2] * 15 
      5. Else, f[i] = f[i - 1] * 9.
  3. Initialization: f[0] = 1.
  4. Answer: f[s.length()].

Solution:

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class Solution {
    public int numDecodings(String s) {
        if (s.length() == 0) {
            return 0;
        }
        
        long[] f = new long[s.length() + 1];
        f[0] = 1;
        int M = 1000000007;
        
        for (int i = 1; i < f.length; i++) {
            char c = s.charAt(i - 1);
            char previous = '#';
            if (i > 1) {
                previous = s.charAt(i - 2);
            }
            
            if (c == '0') {
                if (previous > '0' && previous <= '2') {
                    f[i] = f[i - 2];
                } else if (previous == '*') {
                    f[i] = (2 * f[i - 2]) % M;
                } else {
                    return 0;
                }
            } else if (c > '0') {
                if (previous == '*') {
                    if (c <= '6') {
                        f[i] = (f[i - 1] + 2 * f[i - 2]) % M;
                    } else {
                        f[i] = (f[i - 1] + f[i - 2]) % M;
                    }
                } else if (i > 1 && previous != '0' && Integer.parseInt(s.substring(i - 2, i)) <= 26) {
                    f[i] = (f[i - 1] + f[i - 2]) % M;
                } else {
                    f[i] = f[i - 1];
                }
            } else {
                f[i] = 9 * f[i - 1];
                if (previous == '1') {
                    f[i] = (f[i] + 9 * f[i - 2]) % M;
                } else if (previous == '2') {
                    f[i] = (f[i] + 6 * f[i - 2]) % M;
                } else if (previous == '*') {
                    f[i] = (f[i] + 15 * f[i - 2]) % M;
                }
            }
        }
        return (int) f[s.length()];
    }
}

Location: Philadelphia, PA, USA

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