LeetCode Problem

Problem Description

Given an input string (s) and a pattern (p), implement wildcard pattern matching with support for '?' and '' where: '?' Matches any single character. '' Matches any sequence of characters (including the empty sequence). The matching should cover the entire input string (not partial).

Examples

Input: aa, a
Output: false

Input: aa, *
Output: true

Input: cb, ?a
Output: false

Constraints

0 <= s.length, p.length <= 2000
s contains only lowercase English letters.
p contains only lowercase English letters, '?' or '*'.

Follow-up

Approach to Solve

Use dynamic programming to solve the wildcard matching problem.

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-1]
        
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if p[j-1] == '*':
                    dp[i][j] = dp[i][j-1] or dp[i-1][j]
                elif p[j-1] == '?' or s[i-1] == p[j-1]:
                    dp[i][j] = dp[i-1][j-1]
        
        return dp[m][n]

class Solution {
    public boolean isMatch(String s, String p) {
        int sLen = s.length(), pLen = p.length();
        boolean[][] dp = new boolean[sLen + 1][pLen + 1];
        dp[0][0] = true;
        for (int j = 1; j <= pLen; j++) {
            if (p.charAt(j - 1) == '*') {
                dp[0][j] = dp[0][j - 1];
            }
        }
        for (int i = 1; i <= sLen; i++) {
            for (int j = 1; j <= pLen; j++) {
                if (p.charAt(j - 1) == '*') {
                    dp[i][j] = dp[i][j - 1] || dp[i - 1][j];
                } else if (p.charAt(j - 1) == '?' || s.charAt(i - 1) == p.charAt(j - 1)) {
                    dp[i][j] = dp[i - 1][j - 1];
                }
            }
        }
        return dp[sLen][pLen];
    }

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

Explanation

This solution uses dynamic programming to solve the wildcard matching problem. Here's a detailed explanation of the algorithm:

Initialize a 2D array dp with dimensions (sLen + 1) x (pLen + 1) to store the matching results, where dp[i][j] represents whether the first i characters of s match the first j characters of p.
Set dp[0][0] to true, as an empty pattern matches an empty string.
Iterate through the pattern p to update the dp array:
- If p[j-1] is '*', set dp[0][j] to dp[0][j-1], indicating that the empty sequence can match the pattern up to the current position.
Iterate through the string s and the pattern p to fill the dp array:
- If p[j-1] is '*', update dp[i][j] to dp[i][j-1] || dp[i-1][j], indicating that the current sequence can match the pattern up to the current position.
- If p[j-1] is '?' or s[i-1] matches p[j-1], update dp[i][j] to dp[i-1][j-1], indicating that the current characters match.
Return dp[sLen][pLen], which indicates whether the entire string s matches the pattern p.

The time complexity is O(mn), where m and n are the lengths of the string s and the pattern p, respectively. The space complexity is also O(mn) due to the storage of the dp array.

Complexity

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

LeetSolved

Wildcard Matching