Solution Explanation:

This problem asks to find the number of ways to decode a string consisting of digits and '' characters, where '' can represent any digit from '1' to '9'. The solution uses dynamic programming to efficiently solve this.

Approach:

The core idea is to build a dynamic programming solution where dp[i] represents the number of ways to decode the substring s[0...i-1]. We iterate through the string, considering both one-digit and two-digit decoding possibilities at each step. The presence of '' adds complexity, as we need to consider multiple possibilities for each ''.

State Transition:

For each position i, we calculate dp[i] based on the previous states:

• One-digit decoding: If s[i-1] is a digit other than '0', it contributes dp[i-1] ways (because we can append this digit independently). If s[i-1] is '', it contributes 9 * dp[i-1] ways (since '' can be any digit from 1 to 9).

• Two-digit decoding: If s[i-2] and s[i-1] form a valid two-digit code (between 10 and 26, inclusive), this adds dp[i-2] ways. We need to handle the cases where s[i-2] or s[i-1] or both are ''. Each of these "" cases requires careful consideration of the possible combinations.

Base Cases:

• dp[0] = 1 (empty substring has one decoding)
• dp[1] depends on whether s[0] is a digit, '0', or '*'.

Modulo Operation:

To handle potentially large numbers of decoding possibilities, the modulo operator (% 1000000007) is used throughout the calculation.

Time Complexity:

The solution iterates through the input string once. The calculation for each position takes constant time, even with the handling of '*' characters, which is a set of fixed conditions. Therefore, the overall time complexity is O(n), where n is the length of the string.

Space Complexity:

The solution utilizes constant extra space to store the three variables, a, b, and c that represent the dp states. Hence, the space complexity is O(1).

Code Explanation (Python):

class Solution:
    def numDecodings(self, s: str) -> int:
        mod = int(1e9 + 7)
        n = len(s)
 
        # dp[i - 2], dp[i - 1], dp[i]
        a, b, c = 0, 1, 0  # Initialize dp states
        for i in range(1, n + 1):
            # 1 digit
            if s[i - 1] == "*":
                c = 9 * b % mod # * can be any digit 1-9
            elif s[i - 1] != "0":
                c = b # valid single digit
            else:
                c = 0 # 0 cannot be a single digit
 
            # 2 digits
            if i > 1:
                # Handle all combinations of '*' in two digit codes
                if s[i - 2] == "*" and s[i - 1] == "*":
                    c = (c + 15 * a) % mod
                elif s[i - 2] == "*":
                    if s[i - 1] > "6":
                        c = (c + a) % mod
                    else:
                        c = (c + 2 * a) % mod
                elif s[i - 1] == "*":
                    if s[i - 2] == "1":
                        c = (c + 9 * a) % mod
                    elif s[i - 2] == "2":
                        c = (c + 6 * a) % mod
                elif (
                    s[i - 2] != "0"
                    and (ord(s[i - 2]) - ord("0")) * 10 + ord(s[i - 1]) - ord("0") <= 26
                ):
                    c = (c + a) % mod
 
            a, b = b, c  # Update dp states
 
        return c

The Java and Go code follow the same logic with minor syntax changes to accommodate the respective language features. The core dynamic programming strategy and the handling of '*' remain consistent.