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Data Structures & Algorithms June 27, 2020 3 min read

Leetcode Regular Expression Matching Java Solution

// SOLVING THIS WITH AN AI ASSISTANT (2026)

If you are working through this problem with an AI coding assistant — Claude, ChatGPT, Cursor chat, Gemini, GitHub Copilot, Aider, or any agent — the goal isn’t to ask for the answer. It is to use the tool to understand the pattern. The prompt sequence I’d run:

  1. Spec it back to me first. “In your own words, what is this problem actually testing? What’s the smallest example that fails the naive approach?”
  2. Brute-force first, optimize after. “Write the simplest correct solution, even if it’s O(n²). Don’t optimize. Just make it correct, with comments explaining each step.”
  3. Ask for the upgrade. “Now show me the optimal solution. What insight makes it possible? What pattern is this an instance of?”
  4. Stress-test it. “Generate 10 edge cases — empty input, single element, duplicates, max size, sorted, reverse-sorted. Run my solution against each.”

The pattern matters more than the answer. If the agent just hands you optimized code, you’ve trained yourself to lose interviews.

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:

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

Solution:

Explanation:

1, init dp[0][0] && dp[0][j], to avoid corner cases;
2, transform function:
a, when p == ‘‘:
compare char[i-1][j-1];
if same, “char” can count as 0, 1, multiple;
Important: if multiple, dp[i+ 1][j+ 1] = dp[i][j+1];
think backwards, not forwards.

Or simplify:
dp[i][j+1] represents 1/multiple, think previous step, represents 0 or multiple
dp[i+ 1][j+ 1] = dp[i + 1][j – 1] || dp[i][j] || dp[i][j+1]; // count 0, 1, multiple
here, counting 1 is redundant
So we can just use
dp[i+ 1][j+ 1] = dp[i + 1][j – 1] || dp[i][j+1] 

class HackerHeap {
    public boolean isMatch(String s, String p) {
        int m = s.length(), n = p.length();
        if (n == 0) return m == 0;
        boolean[][] dp = new boolean[m + 1][n + 1];
        dp[0][0] = true;
        for (int i = 1; i < n; i+=2) {
            if (p.charAt(i) != '*') break;
            else dp[0][i + 1] = true;
        }
        
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                char a = s.charAt(i), b = p.charAt(j);
                if (b == '.' || a == b) dp[i + 1][j + 1] = dp[i][j];
                else if (b == '*') dp[i + 1][j + 1] = dp[i + 1][j - 1]  // zero
                    || ((s.charAt(i) == p.charAt(j - 1) || p.charAt(j - 1) == '.') &&               dp[i][j + 1]); // 1 or more;  // consider ".*"
            }
        }
        
        return dp[m][n];
    }
}

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