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: Memoization Search

We design a function $dfs(i, j)$, which represents whether the string $s$ starting from the $i$-th character matches the string $p$ starting from the $j$-th character. The answer is $dfs(0, 0)$.

The execution process of the function $dfs(i, j)$ is as follows:

If $i \geq \textit{len}(s)$, then $dfs(i, j)$ is true only when $j \geq \textit{len}(p)$ or $p[j] = '*'$ and $dfs(i, j + 1)$ is true.
If $j \geq \textit{len}(p)$, then $dfs(i, j)$ is false.
If $p[j] = '*'$, then $dfs(i, j)$ is true if and only if $dfs(i + 1, j)$ or $dfs(i + 1, j + 1)$ or $dfs(i, j + 1)$ is true.
Otherwise, $dfs(i, j)$ is true if and only if $p[j] = '?'$ or $s[i] = p[j]$ and $dfs(i + 1, j + 1)$ is true.

To avoid repeated calculations, we use the method of memoization search and store the result of $dfs(i, j)$ in a hash table.

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

Python3JavaC++GoTypeScriptC#

class Solution:
    def isMatch(self, s: str, p: str) -> bool:
        @cache
        def dfs(i: int, j: int) -> bool:
            if i >= len(s):
                return j >= len(p) or (p[j] == "*" and dfs(i, j + 1))
            if j >= len(p):
                return False
            if p[j] == "*":
                return dfs(i + 1, j) or dfs(i + 1, j + 1) or dfs(i, j + 1)
            return (p[j] == "?" or s[i] == p[j]) and dfs(i + 1, j + 1)

        return dfs(0, 0)

class Solution {
    private Boolean[][] f;
    private char[] s;
    private char[] p;
    private int m;
    private int n;

    public boolean isMatch(String s, String p) {
        this.s = s.toCharArray();
        this.p = p.toCharArray();
        m = s.length();
        n = p.length();
        f = new Boolean[m][n];
        return dfs(0, 0);
    }

    private boolean dfs(int i, int j) {
        if (i >= m) {
            return j >= n || (p[j] == '*' && dfs(i, j + 1));
        }
        if (j >= n) {
            return false;
        }
        if (f[i][j] != null) {
            return f[i][j];
        }
        if (p[j] == '*') {
            f[i][j] = dfs(i + 1, j) || dfs(i + 1, j + 1) || dfs(i, j + 1);
        } else {
            f[i][j] = (p[j] == '?' || s[i] == p[j]) && dfs(i + 1, j + 1);
        }
        return f[i][j];
    }
}

class Solution {
public:
    bool isMatch(string s, string p) {
        int m = s.size(), n = p.size();
        int f[m + 1][n + 1];
        memset(f, -1, sizeof(f));
        function<bool(int, int)> dfs = [&](int i, int j) {
            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);
    }
};

func isMatch(s string, p string) bool {
    m, n := len(s), len(p)
    f := make([][]int, m+1)
    for i := range f {
        f[i] = make([]int, n+1)
    }
    var dfs func(i, j int) bool
    dfs = func(i, j int) bool {
        if i >= m {
            return j >= n || p[j] == '*' && dfs(i, j+1)
        }
        if j >= n {
            return false
        }
        if f[i][j] != 0 {
            return f[i][j] == 1
        }
        f[i][j] = 2
        ok := false
        if p[j] == '*' {
            ok = dfs(i+1, j) || dfs(i+1, j+1) || dfs(i, j+1)
        } else {
            ok = (p[j] == '?' || s[i] == p[j]) && dfs(i+1, j+1)
        }
        if ok {
            f[i][j] = 1
        }
        return ok
    }
    return dfs(0, 0)
}

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);
}

public class Solution {
    private bool?[,] f;
    private char[] s;
    private char[] p;
    private int m;
    private int n;

    public bool IsMatch(string s, string p) {
        this.s = s.ToCharArray();
        this.p = p.ToCharArray();
        m = s.Length;
        n = p.Length;
        f = new bool?[m, n];
        return Dfs(0, 0);
    }

    private bool Dfs(int i, int j) {
        if (i >= m) {
            return j >= n || (p[j] == '*' && Dfs(i, j + 1));
        }
        if (j >= n) {
            return false;
        }
        if (f[i, j] != null) {
            return f[i, j].Value;
        }
        if (p[j] == '*') {
            f[i, j] = Dfs(i + 1, j) || Dfs(i + 1, j + 1) || Dfs(i, j + 1);
        } else {
            f[i, j] = (p[j] == '?' || s[i] == p[j]) && Dfs(i + 1, j + 1);
        }
        return f[i, j].Value;
    }
}

Solution 2: Dynamic Programming

We can convert the memoization search in Solution 1 into dynamic programming.

Define $f[i][j]$ to represent whether the first $i$ characters of string $s$ match the first $j$ characters of string $p$. Initially, $f[0][0] = \textit{true}$, indicating that two empty strings are matching. For $j \in [1, n]$, if $p[j-1] = '*'$, then $f[0][j] = f[0][j-1]$.

Next, we consider the case of $i \in [1, m]$ and $j \in [1, n]$:

If $p[j-1] = '*'$, then $f[i][j] = f[i-1][j] \lor f[i][j-1] \lor f[i-1][j-1]$.
Otherwise, $f[i][j] = (p[j-1] = '?' \lor s[i-1] = p[j-1]) \land f[i-1][j-1]$.

The final answer is $f[m][n]$.