LeetCode 639: Decode Ways II Solution

Master LeetCode problem 639 (Decode Ways II), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

639. Decode Ways II

Hard

String Dynamic Programming

Problem Explanation

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).

Solution Code

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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