44. Wildcard Matching

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).

Example 1:

Input: s = "aa", p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".

Example 2:

Input: s = "aa", p = "*"
Output: true
Explanation: '*' matches any sequence.

Example 3:

Input: s = "cb", p = "?a"
Output: false
Explanation: '?' matches 'c', but the second letter is 'a', which does not match 'b'.

Constraints:

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

Solutions

Solution 1: Dynamic Programming

We define the state $dp[i][j]$ to represent whether the first $i$ characters of $s$ match the first $j$ characters of $p$.

The state transition equation is as follows:

\[dp[i][j]= \begin{cases} dp[i-1][j-1] & \text{if } s[i-1]=p[j-1] \text{ or } p[j-1]=\text{?} \\ dp[i-1][j-1] \lor dp[i-1][j] \lor dp[i][j-1] & \text{if } p[j-1]=\text{*} \\ \text{false} & \text{otherwise} \end{cases}\]

The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. Here, $m$ and $n$ are the lengths of $s$ and $p$ respectively.

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

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

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

        # if p starting with "*", then all true for dp[0][i]
        # or else, all false for dp[0][i]
        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 s[i - 1] == p[j - 1] or p[j - 1] == '?':
                    dp[i][j] = dp[i - 1][j - 1]
                # dp[i - 1][j], where j is always '*', all the way back to i-1==0
                elif p[j - 1] == '*':
                    dp[i][j] = dp[i - 1][j] or dp[i][j - 1]
        return dp[m][n]

func isMatch(s string, p string) bool {
	m, n := len(s), len(p)
	dp := make([][]bool, m+1)
	for i := range dp {
		dp[i] = make([]bool, n+1)
	}
	dp[0][0] = true
	for j := 1; j <= n; j++ {
		if p[j-1] == '*' {
			dp[0][j] = dp[0][j-1]
		}
	}
	for i := 1; i <= m; i++ {
		for j := 1; j <= n; j++ {
			if s[i-1] == p[j-1] || p[j-1] == '?' {
				dp[i][j] = dp[i-1][j-1]
			} else if p[j-1] == '*' {
				dp[i][j] = dp[i-1][j] || dp[i][j-1]
			}
		}
	}
	return dp[m][n]
}

using System.Linq;

public class Solution {
    public bool IsMatch(string s, string p) {
        if (p.Count(ch => ch != '*') > s.Length)
        {
            return false;
        }
        
        bool[,] f = new bool[s.Length + 1, p.Length + 1];
        bool[] d = new bool[s.Length + 1]; // d[i] means f[0, j] || f[1, j] || ... || f[i, j]
        for (var j = 0; j <= p.Length; ++j)
        {
            d[0] = j == 0 ? true : d[0] && p[j - 1] == '*';
            for (var i = 0; i <= s.Length; ++i)
            {
                if (j == 0)
                {
                    f[i, j] = i == 0;
                    continue;
                }
                
                if (p[j - 1] == '*')
                {
                    if (i > 0)
                    {
                        d[i] = f[i, j - 1] || d[i - 1];
                    }
                    f[i, j] = d[i];
                }
                else if (p[j - 1] == '?')
                {
                    f[i, j] = i > 0 && f[i - 1, j - 1];
                }
                else
                {
                    f[i, j] = i > 0 && f[i - 1, j - 1] && s[i - 1] == p[j - 1];
                }
            }
        }
        return f[s.Length, p.Length];
    }
}

function isMatch(s: string, p: string): boolean {
    const m = s.length;
    const n = p.length;
    const f: number[][] = Array.from({ length: m + 1 }, () =>
        Array.from({ length: n + 1 }, () => -1),
    );
    const dfs = (i: number, j: number): boolean => {
        if (i >= m) {
            return j >= n || (p[j] === '*' && dfs(i, j + 1));
        }
        if (j >= n) {
            return false;
        }
        if (f[i][j] !== -1) {
            return f[i][j] === 1;
        }
        if (p[j] === '*') {
            f[i][j] = dfs(i + 1, j) || dfs(i, j + 1) ? 1 : 0;
        } else {
            f[i][j] = (p[j] === '?' || s[i] === p[j]) && dfs(i + 1, j + 1) ? 1 : 0;
        }
        return f[i][j] === 1;
    };
    return dfs(0, 0);
}

44 - Wildcard Matching

44. Wildcard Matching

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44. Wildcard Matching

Description

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All Problems

All Solutions