Algorithm Day106 - Regular Expression Matching

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LC-10 Regular Expression Matching

📌 Problem

Implement regular expression matching with support for ‘.’ and ‘*’ where:

  • ‘.’ Matches any single character.
  • ‘*’ Matches zero or more of the preceding element.

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

💬 Examples

Example 1

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Input: s = "aa", p = "a"
Output: false

Example 2

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Input: s = "aa", p = "a*"
Output: true

Example 3

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Input: s = "ab", p = ".*"
Output: true

💡 Intuition & Brute Force

A naive recursive solution tries all possibilities for ‘*’ (match zero, one, multiple). This leads to exponential time in the worst case.

Key observation for DP: define dp[i][j] = whether s[0..i) matches p[0..j). Build up using smaller prefixes.

✅ Optimized Solution (Dynamic Programming)

Transitions:

  1. If p[j-1] is a normal char or ‘.’, then:
    dp[i][j] = dp[i-1][j-1] && (p[j-1] == s[i-1] || p[j-1] == ‘.’)

  2. If p[j-1] == ‘*’, it stands for zero or more of p[j-2]:

    • Zero occurrence: dp[i][j] = dp[i][j-2]
    • One or more: if p[j-2] matches s[i-1], dp[i][j] = dp[i-1][j]

Initialize dp[0][0] = true, and handle patterns like a*, ab that can match empty string.

🔢 Java Code (DP)

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class Solution {
    public boolean isMatch(String s, String p) {
        int m = s.length(), n = p.length();
        boolean[][] dp = new boolean[m + 1][n + 1];
        dp[0][0] = true;

        // Patterns like a*, a*b* can match empty string
        for (int j = 2; j <= n; j++) {
            if (p.charAt(j - 1) == '*') {
                dp[0][j] = dp[0][j - 2];
            }
        }

        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                char pc = p.charAt(j - 1);
                if (pc == '.' || pc == s.charAt(i - 1)) {
                    dp[i][j] = dp[i - 1][j - 1];
                } else if (pc == '*') {
                    // zero occurrence
                    dp[i][j] = dp[i][j - 2];
                    // one or more occurrence
                    char pprev = p.charAt(j - 2);
                    if (pprev == '.' || pprev == s.charAt(i - 1)) {
                        dp[i][j] = dp[i][j] || dp[i - 1][j];
                    }
                }
            }
        }
        return dp[m][n];
    }
}

⏱ Complexity Analysis

  • Time: O(m × n) where m = s.length(), n = p.length().
  • Space: O(m × n) for DP table.

✍️ Summary

  • Use DP to avoid exponential recursion.
  • Carefully handle ‘*’ by considering zero or multiple occurrences.
  • Initialization for empty string vs pattern is important.