LeetCode Problem

Problem Description

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

Examples

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

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

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

Constraints

1 <= s.length <= 20
1 <= p.length <= 30
s contains only lowercase English letters.
p contains only lowercase English letters, '.', and '*'.

Follow-up

Approach to Solve

Use dynamic programming to match the input string with the pattern.

Code Implementation

class Solution:
    def isMatch(self, s: str, p: str) -> bool:
        m, n = len(s), len(p)
        dp = [[False] * (n + 1) for _ in range(m + 1)]
        dp[0][0] = True
        for j in range(1, n + 1):
            if p[j-1] == '*':
                dp[0][j] = dp[0][j-2]
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if p[j-1] == '.' or p[j-1] == s[i-1]:
                    dp[i][j] = dp[i-1][j-1]
                elif p[j-1] == '*':
                    dp[i][j] = dp[i][j-2] or (dp[i-1][j] and (p[j-2] == '.' or p[j-2] == s[i-1]))
        return dp[m][n]

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

class Solution {
public:
    bool isMatch(string s, string p) {
        int m = s.length(), n = p.length();
        vector<vector<bool>> dp(m + 1, vector<bool>(n + 1, false));
        dp[0][0] = true;
        for (int j = 1; j <= n; j++) {
            if (p[j - 1] == '*') {
                dp[0][j] = dp[0][j - 2];
            }
        }
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (p[j - 1] == '.' || p[j - 1] == s[i - 1]) {
                    dp[i][j] = dp[i - 1][j - 1];
                } else if (p[j - 1] == '*') {
                    dp[i][j] = dp[i][j - 2] || (dp[i - 1][j] && (p[j - 2] == '.' || p[j - 2] == s[i - 1]));
                }
            }
        }
        return dp[m][n];
    }
};

Explanation

The Regular Expression Matching problem is solved using dynamic programming. Here's a detailed explanation of the approach:

We create a 2D DP table where dp[i][j] represents whether the first i characters of s match the first j characters of p.
Base case: An empty pattern matches an empty string, so dp[0][0] = true.
Handle patterns starting with '': For the first row (empty s), we set dp[0][j] = dp[0][j-2] if p[j-1] is ''. This is because '*' can match zero occurrences of the preceding element.
Fill the DP table:
- If p[j-1] is '.' or matches s[i-1], then dp[i][j] = dp[i-1][j-1].
- If p[j-1] is '*', we have two cases: a) Zero occurrence: dp[i][j] = dp[i][j-2] b) One or more occurrences: dp[i][j] = dp[i-1][j] if p[j-2] is '.' or matches s[i-1]
The final answer is in dp[m][n], where m and n are lengths of s and p respectively.

Time Complexity: O(mn) as we fill each cell in the mn DP table once. Space Complexity: O(m*n) for the DP table.

This approach efficiently handles all cases including '.' (matches any single character) and '*' (matches zero or more of the preceding element), making it a comprehensive solution for regular expression matching.

Complexity

Time Complexity: O(m*n)
Space Complexity: O(m*n)