639. Decode Ways II

Explanation

To solve this problem, we can use dynamic programming. We will iterate through the input string and calculate the number of ways to decode it based on the previous characters.

We will define two main dynamic programming arrays:

dp1[i] represents the number of ways to decode the substring s.substring(0, i) where s[i] is not a '*'.
dp2[i] represents the number of ways to decode the substring s.substring(0, i) where s[i] is a '*'.

For each character in the input string:

If s[i] is a digit from '1' to '9', we can update dp1[i] based on the previous characters.
If s[i] is '*', we can update dp2[i] based on the previous characters:
- If s[i-1] is '1', '2', or '', we can have 9 possibilities for '' (mapping to '1' to '9').
- If s[i-1] is '0', we can have 6 possibilities for '*' (mapping to '1' to '6').

The final answer will be the sum of dp1[n] and dp2[n] (where n is the length of the input string).

View Leetcode Problem

class Solution {
    public int numDecodings(String s) {
        int n = s.length();
        long MOD = 1000000007;
        long[] dp1 = new long[n + 1];
        long[] dp2 = new long[n + 1];
        dp1[0] = 1;
        dp2[0] = 0;

        for (int i = 1; i <= n; i++) {
            char c = s.charAt(i - 1);
            if (c == '0') {
                dp1[i] = 0;
                dp2[i] = (dp2[i - 1] * 2) % MOD;
            } else if (c == '*') {
                dp1[i] = (dp1[i - 1] * 9 + dp2[i - 1] * 9) % MOD;
                dp2[i] = (dp1[i - 1] + dp2[i - 1] * 6) % MOD;
            } else {
                dp1[i] = c == '0' ? 0 : dp1[i - 1];
                dp2[i] = (dp1[i - 1] * (c > '6' ? 0 : 1) + dp2[i - 1] * (c > '6' ? 0 : 1)) % MOD;
            }
        }

        return (int) ((dp1[n] + dp2[n]) % MOD);
    }
}

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