Question

44. Wildcard Matching

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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:
  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 s[i - 1] == p[j - 1] or p[j - 1] == '?':
          dp[i][j] = dp[i - 1][j - 1]
        elif p[j - 1] == '*':
          dp[i][j] = dp[i - 1][j] or dp[i][j - 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 - 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];
  }
};

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];
  }
}