10. Regular Expression Matching

• Difficulty: Hard

• Topics: String, Dynamic Programming, Backtracking

• Similar Questions:

  • Wildcard Matching

Problem:

Given an input string (s) and a pattern (p), implement regular expression matching with support for '.' and '*'.

'.' Matches any single character.
'*' Matches zero or more of the preceding element.

The matching should cover the entire input string (not partial).

Note:

• s could be empty and contains only lowercase letters a-z.
• p could be empty and contains only lowercase letters a-z, and characters like . or *.

Example 1:

Input:
s = "aa"
p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".

Example 2:

Input:
s = "aa"
p = "a*"
Output: true
Explanation: '*' means zero or more of the preceding element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".

Example 3:

Input:
s = "ab"
p = ".*"
Output: true
Explanation: ".*" means "zero or more (*) of any character (.)".

Example 4:

Input:
s = "aab"
p = "c*a*b"
Output: true
Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore, it matches "aab".

Example 5:

Input:
s = "mississippi"
p = "mis*is*p*."
Output: false

Solutions:

class Solution {
public:
    bool isMatch(string s, string p) {
        int m = s.length();
        int n = p.length();

        vector<vector<bool>> dp(m + 1, vector<bool>(n + 1, false));
        dp[0][0] = true;

        // when s is empty
        for (int j = 1; j <= n; ++j) {
            char c = p[j-1];
            if (isalpha(c)) {
                dp[0][j] = false;
            } else if (c == '.'){
                dp[0][j] = false;
            } else {
                dp[0][j] = dp[0][j-2];
            }
        }

        // when p is empty
        for (int i = 1; i <= m; ++i) {
            dp[i][0] = false;
        }

        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                if (p[j-1] == '*') {
                    if (s[i-1] == p[j-2] || p[j-2] == '.') {
                        dp[i][j] = dp[i-1][j] || dp[i][j-2];
                    } else {
                        dp[i][j] = dp[i][j-2];
                    }
                    continue;
                }

                if (p[j-1] == '.' || s[i-1] == p[j-1]) {
                    dp[i][j] = dp[i-1][j-1];
                    continue;
                }

                dp[i][j] = false;
            }
        }

        return dp[m][n];
    }
};