10. Regular Expression Matching

Explanation

To solve this problem, we can use dynamic programming. We will create a 2D DP array where dp[i][j] will represent whether the pattern p[0...j-1] matches the string s[0...i-1]. We will then fill the DP array based on the following conditions:

If p[j-1] is a normal character or '.' (matches any single character), then dp[i][j] is true if dp[i-1][j-1] is true and s[i-1] matches p[j-1].
If p[j-1] is '*', then we have two cases:
- If p[j-2] matches s[i-1], then dp[i][j] is true if either dp[i][j-2] (zero occurrences of the preceding element) or dp[i-1][j] (one or more occurrences of the preceding element) is true.

Finally, the result will be stored in dp[s.length()][p.length()].

View Leetcode Problem

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;

        for (int i = 0; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (j > 1 && p.charAt(j - 1) == '*') {
                    dp[i][j] = dp[i][j - 2] || (i > 0 && (s.charAt(i - 1) == p.charAt(j - 2) || p.charAt(j - 2) == '.') && dp[i - 1][j]);
                } else {
                    dp[i][j] = i > 0 && dp[i - 1][j - 1] && (s.charAt(i - 1) == p.charAt(j - 1) || p.charAt(j - 1) == '.');
                }
            }
        }

        return dp[m][n];
    }
}

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