Wildcard Matching LeetCode Solution | Easy Approach

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Wildcard Matching | 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 '*'.

Wildcard Matching Solutions

Time:O(∣s∣∣p∣)
Space:O(∣s∣∣p∣)

C++

class Solution {
 public:
  bool isMatch(string s, string p) {
    const int m = s.length();
    const int n = p.length();
    // dp[i][j] := true if s[0..i) matches p[0..j)
    vector<vector<bool>> dp(m + 1, vector<bool>(n + 1));
    dp[0][0] = true;

    auto isMatch = [&](int i, int j) -> bool {
      return j >= 0 && p[j] == '?' || s[i] == p[j];
    };

    for (int j = 0; j < p.length(); ++j)
      if (p[j] == '*')
        dp[0][j + 1] = dp[0][j];

    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j)
        if (p[j] == '*') {
          const bool matchEmpty = dp[i + 1][j];
          const bool matchSome = dp[i][j + 1];
          dp[i + 1][j + 1] = matchEmpty || matchSome;
        } else if (isMatch(i, j)) {
          dp[i + 1][j + 1] = dp[i][j];
        }

    return dp[m][n];
  }
};
 

Java

 class Solution {
  public boolean isMatch(String s, String p) {
    final int m = s.length();
    final int n = p.length();
    // dp[i][j] := true if s[0..i) matches p[0..j)
    boolean[][] dp = new boolean[m + 1][n + 1];
    dp[0][0] = true;

    for (int j = 0; j < p.length(); ++j)
      if (p.charAt(j) == '*')
        dp[0][j + 1] = dp[0][j];

    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j)
        if (p.charAt(j) == '*') {
          final boolean matchEmpty = dp[i + 1][j];
          final boolean matchSome = dp[i][j + 1];
          dp[i + 1][j + 1] = matchEmpty || matchSome;
        } else if (isMatch(s, i, p, j)) {
          dp[i + 1][j + 1] = dp[i][j];
        }

    return dp[m][n];
  }

  private boolean isMatch(final String s, int i, final String p, int j) {
    return j >= 0 && p.charAt(j) == '?' || s.charAt(i) == p.charAt(j);
  }
}

Python

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

    def isMatch(i: int, j: int) -> bool:
      return i >= 0 and p[j] == '?' or s[i] == p[j]

    for j, c in enumerate(p):
      if c == '*':
        dp[0][j + 1] = dp[0][j]

    for i in range(m):
      for j in range(n):
        if p[j] == '*':
          matchEmpty = dp[i + 1][j]
          matchSome = dp[i][j + 1]
          dp[i + 1][j + 1] = matchEmpty or matchSome
        elif isMatch(i, j):
          dp[i + 1][j + 1] = dp[i][j]

    return dp[m][n]
  

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