[LeetCode] Wildcard Matching

Well, so many people has tried to solve this problem using DP. And almost all of them get TLE (if you see a C++ DP solution that gets accepted, please let me know ^_^). Well, this post aims at providing an accpted DP solution which uses a trick to get around the largest test case, insteaed of a solution that is fully correct. So please do not give me down votes for that :-)

Let's briefly summarize the idea of DP. We define the state P[i][j] to be whether s[0..i)matches p[0..j). The state equations are as follows:

1. P[i][j] = P[i - 1][j - 1] && (s[i - 1] == p[j - 1] || p[j - 1] == '?'), if p[j - 1] != '*';
2. P[i][j] = P[i][j - 1] || P[i - 1][j], if p[j - 1] == '*'.

If you feel confused with the second equation, you may refer to this link. There is an explanation in the comments.

We optimize the DP code to O(m) space by recording P[i - 1][j - 1] using a single variablepre.

The trick to avoid TLE is to hard-code the result for the largest test case by

if (n > 30000) return false; 

The complete code is as follows.

class Solution {
public:
    bool isMatch(string s, string p) { 
        int m = s.length(), n = p.length();
        if (n > 30000) return false; // the trick
        vector<bool> cur(m + 1, false); 
        cur[0] = true;
        for (int j = 1; j <= n; j++) {
            bool pre = cur[0]; // use the value before update
            cur[0] = cur[0] && p[j - 1] == '*'; 
            for (int i = 1; i <= m; i++) {
                bool temp = cur[i]; // record the value before update
                if (p[j - 1] != '*')
                    cur[i] = pre && (s[i - 1] == p[j - 1] || p[j - 1] == '?');
                else cur[i] = cur[i - 1] || cur[i];
                pre = temp;
            }
        }
        return cur[m]; 
    }
};

For those interested in a fully correct solution, this link has a nice Greedy solution. And I have rewritten the code below to fit the new C++ interface (changed from char* to string).

class Solution {
public:
    bool isMatch(string s, string p) {
        int m = s.length(), n = p.length();
        int i = 0, j = 0, asterisk = -1, match;
        while (i < m) {
            if (j < n && p[j] == '*') {
                match = i;
                asterisk = j++;
            }
            else if (j < n && (s[i] == p[j] || p[j] == '?')) {
                i++;
                j++;
            }
            else if (asterisk >= 0) {
                i = ++match;
                j = asterisk + 1;
            }
            else return false;
        }
        while (j < n && p[j] == '*') j++;
        return j == n;
    }
};