Question

10. Regular Expression Matching

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Description

Given an input string s and a pattern p, implement regular expression matching with support for '.' and '*' where:

• '.' Matches any single character.​​​​
• '*' Matches zero or more of the preceding element.

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 = "a*"
Output: true
Explanation: '*' means zero or more of the preceding element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".

Example 3:

Input: s = "ab", p = ".*"
Output: true
Explanation: ".*" means "zero or more (*) of any character (.)".

 

Constraints:

• 1 <= s.length <= 20
• 1 <= p.length <= 20
• s contains only lowercase English letters.
• p contains only lowercase English letters, '.', and '*'.
• It is guaranteed for each appearance of the character '*', there will be a previous valid character to match.

Solutions

Solution 1: Memoization Search

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

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

• If $j$ has reached the end of $p$, then if $i$ has also reached the end of $s$, the match is successful, otherwise, the match fails.
• If the next character of $j$ is '*', we can choose to match $0$ $s[i]$ characters, which is $dfs(i, j + 2)$. If $i \lt m$ and $s[i]$ matches $p[j]$, we can choose to match $1$ $s[i]$ character, which is $dfs(i + 1, j)$.
• If the next character of $j$ is not '*', then if $i \lt m$ and $s[i]$ matches $p[j]$, it is $dfs(i + 1, j + 1)$. Otherwise, the match fails.

During the process, we can use memoization search to avoid repeated calculations.

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:
    @cache
    def dfs(i, j):
      if j >= n:
        return i == m
      if j + 1 < n and p[j + 1] == '*':
        return dfs(i, j + 2) or (
          i < m and (s[i] == p[j] or p[j] == '.') and dfs(i + 1, j)
        )
      return i < m and (s[i] == p[j] or p[j] == '.') and dfs(i + 1, j + 1)

    m, n = len(s), len(p)
    return dfs(0, 0)

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

  public boolean isMatch(String s, String p) {
    m = s.length();
    n = p.length();
    f = new Boolean[m + 1][n + 1];
    this.s = s;
    this.p = p;
    return dfs(0, 0);
  }

  private boolean dfs(int i, int j) {
    if (j >= n) {
      return i == m;
    }
    if (f[i][j] != null) {
      return f[i][j];
    }
    boolean res = false;
    if (j + 1 < n && p.charAt(j + 1) == '*') {
      res = dfs(i, j + 2)
        || (i < m && (s.charAt(i) == p.charAt(j) || p.charAt(j) == '.') && dfs(i + 1, j));
    } else {
      res = i < m && (s.charAt(i) == p.charAt(j) || p.charAt(j) == '.') && dfs(i + 1, j + 1);
    }
    return f[i][j] = res;
  }
}

class Solution {
public:
  bool isMatch(string s, string p) {
    int m = s.size(), n = p.size();
    int f[m + 1][n + 1];
    memset(f, 0, sizeof f);
    function<bool(int, int)> dfs = [&](int i, int j) -> bool {
      if (j >= n) {
        return i == m;
      }
      if (f[i][j]) {
        return f[i][j] == 1;
      }
      int res = -1;
      if (j + 1 < n && p[j + 1] == '*') {
        if (dfs(i, j + 2) or (i < m and (s[i] == p[j] or p[j] == '.') and dfs(i + 1, j))) {
          res = 1;
        }
      } else if (i < m and (s[i] == p[j] or p[j] == '.') and dfs(i + 1, j + 1)) {
        res = 1;
      }
      f[i][j] = res;
      return res == 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 j >= n {
      return i == m
    }
    if f[i][j] != 0 {
      return f[i][j] == 1
    }
    res := -1
    if j+1 < n && p[j+1] == '*' {
      if dfs(i, j+2) || (i < m && (s[i] == p[j] || p[j] == '.') && dfs(i+1, j)) {
        res = 1
      }
    } else if i < m && (s[i] == p[j] || p[j] == '.') && dfs(i+1, j+1) {
      res = 1
    }
    f[i][j] = res
    return res == 1
  }
  return dfs(0, 0)
}

impl Solution {
  #[allow(dead_code)]
  pub fn is_match(s: String, p: String) -> bool {
    let n = s.len();
    let m = p.len();
    let s = s.chars().collect::<Vec<char>>();
    let p = p.chars().collect::<Vec<char>>();

    let mut dp = vec![vec![false; m + 1]; n + 1];

    // Initialize the dp vector
    dp[0][0] = true;

    for i in 1..=m {
      if p[i - 1] == '*' {
        dp[0][i] = dp[0][i - 2];
      }
    }

    // Begin the actual dp process
    for i in 1..=n {
      for j in 1..=m {
        if s[i - 1] == p[j - 1] || p[j - 1] == '.' {
          dp[i][j] = dp[i - 1][j - 1];
        }
        if p[j - 1] == '*' {
          if j >= 2 && (s[i - 1] == p[j - 2] || p[j - 2] == '.') {
            dp[i][j] = dp[i - 1][j] || dp[i][j - 2];
          } else if j >= 2 && s[i - 1] != p[j - 2] {
            dp[i][j] = dp[i][j - 2];
          }
        }
      }
    }

    dp[n][m]
  }
}

/**
 * @param {string} s
 * @param {string} p
 * @return {boolean}
 */
var isMatch = function (s, p) {
  const m = s.length;
  const n = p.length;
  const f = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0));
  const dfs = (i, j) => {
    if (j >= n) {
      return i === m;
    }
    if (f[i][j]) {
      return f[i][j] === 1;
    }
    let res = -1;
    if (j + 1 < n && p[j + 1] === '*') {
      if (dfs(i, j + 2) || (i < m && (s[i] === p[j] || p[j] === '.') && dfs(i + 1, j))) {
        res = 1;
      }
    } else if (i < m && (s[i] === p[j] || p[j] === '.') && dfs(i + 1, j + 1)) {
      res = 1;
    }
    f[i][j] = res;
    return res === 1;
  };
  return dfs(0, 0);
};

public class Solution {
  private string s;
  private string p;
  private int m;
  private int n;
  private int[,] f;

  public bool IsMatch(string s, string p) {
    m = s.Length;
    n = p.Length;
    f = new int[m + 1, n + 1];
    this.s = s;
    this.p = p;
    return dfs(0, 0);
  }

  private bool dfs(int i, int j) {
    if (j >= n) {
      return i == m;
    }
    if (f[i, j] != 0) {
      return f[i, j] == 1;
    }
    int res = -1;
    if (j + 1 < n && p[j + 1] == '*') {
      if (dfs(i, j + 2) || (i < m && (s[i] == p[j] || p[j] == '.') && dfs(i + 1, j))) {
        res = 1;
      }
    } else if (i < m && (s[i] == p[j] || p[j] == '.') && dfs(i + 1, j + 1)) {
      res = 1;
    }
    f[i, j] = res;
    return res == 1;
  }
}

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 $p$. The answer is $f[m][n]$. Initialize $f[0][0] = true$, indicating that the empty string and the empty regular expression match.

Similar to Solution 1, we can discuss different cases.

• If $p[j - 1]$ is '*', we can choose to match $0$ $s[i - 1]$ characters, which is $f[i][j] = f[i][j - 2]$. If $s[i - 1]$ matches $p[j - 2]$, we can choose to match $1$ $s[i - 1]$ character, which is $f[i][j] = f[i][j] \lor f[i - 1][j]$.
• If $p[j - 1]$ is not '*', then if $s[i - 1]$ matches $p[j - 1]$, it is $f[i][j] = f[i - 1][j - 1]$. Otherwise, the match fails.

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)
    f = [[False] * (n + 1) for _ in range(m + 1)]
    f[0][0] = True
    for i in range(m + 1):
      for j in range(1, n + 1):
        if p[j - 1] == "*":
          f[i][j] = f[i][j - 2]
          if i > 0 and (p[j - 2] == "." or s[i - 1] == p[j - 2]):
            f[i][j] |= f[i - 1][j]
        elif i > 0 and (p[j - 1] == "." or s[i - 1] == p[j - 1]):
          f[i][j] = f[i - 1][j - 1]
    return f[m][n]

class Solution {
  public boolean isMatch(String s, String p) {
    int m = s.length(), n = p.length();
    boolean[][] f = new boolean[m + 1][n + 1];
    f[0][0] = true;
    for (int i = 0; i <= m; ++i) {
      for (int j = 1; j <= n; ++j) {
        if (p.charAt(j - 1) == '*') {
          f[i][j] = f[i][j - 2];
          if (i > 0 && (p.charAt(j - 2) == '.' || p.charAt(j - 2) == s.charAt(i - 1))) {
            f[i][j] |= f[i - 1][j];
          }
        } else if (i > 0
          && (p.charAt(j - 1) == '.' || p.charAt(j - 1) == s.charAt(i - 1))) {
          f[i][j] = f[i - 1][j - 1];
        }
      }
    }
    return f[m][n];
  }
}

class Solution {
public:
  bool isMatch(string s, string p) {
    int m = s.size(), n = p.size();
    bool f[m + 1][n + 1];
    memset(f, false, sizeof f);
    f[0][0] = true;
    for (int i = 0; i <= m; ++i) {
      for (int j = 1; j <= n; ++j) {
        if (p[j - 1] == '*') {
          f[i][j] = f[i][j - 2];
          if (i && (p[j - 2] == '.' || p[j - 2] == s[i - 1])) {
            f[i][j] |= f[i - 1][j];
          }
        } else if (i && (p[j - 1] == '.' || p[j - 1] == s[i - 1])) {
          f[i][j] = f[i - 1][j - 1];
        }
      }
    }
    return f[m][n];
  }
};

func isMatch(s string, p string) bool {
  m, n := len(s), len(p)
  f := make([][]bool, m+1)
  for i := range f {
    f[i] = make([]bool, n+1)
  }
  f[0][0] = true
  for i := 0; i <= m; i++ {
    for j := 1; j <= n; j++ {
      if p[j-1] == '*' {
        f[i][j] = f[i][j-2]
        if i > 0 && (p[j-2] == '.' || p[j-2] == s[i-1]) {
          f[i][j] = f[i][j] || f[i-1][j]
        }
      } else if i > 0 && (p[j-1] == '.' || p[j-1] == s[i-1]) {
        f[i][j] = f[i-1][j-1]
      }
    }
  }
  return f[m][n]
}

/**
 * @param {string} s
 * @param {string} p
 * @return {boolean}
 */
var isMatch = function (s, p) {
  const m = s.length;
  const n = p.length;
  const f = Array.from({ length: m + 1 }, () => Array(n + 1).fill(false));
  f[0][0] = true;
  for (let i = 0; i <= m; ++i) {
    for (let j = 1; j <= n; ++j) {
      if (p[j - 1] === '*') {
        f[i][j] = f[i][j - 2];
        if (i && (p[j - 2] === '.' || p[j - 2] === s[i - 1])) {
          f[i][j] |= f[i - 1][j];
        }
      } else if (i && (p[j - 1] === '.' || p[j - 1] === s[i - 1])) {
        f[i][j] = f[i - 1][j - 1];
      }
    }
  }
  return f[m][n];
};

public class Solution {
  public bool IsMatch(string s, string p) {
    int m = s.Length, n = p.Length;
    bool[,] f = new bool[m + 1, n + 1];
    f[0, 0] = true;
    for (int i = 0; i <= m; ++i) {
      for (int j = 1; j <= n; ++j) {
        if (p[j - 1] == '*') {
          f[i, j] = f[i, j - 2];
          if (i > 0 && (p[j - 2] == '.' || p[j - 2] == s[i - 1])) {
            f[i, j] |= f[i - 1, j];
          }
        } else if (i > 0 && (p[j - 1] == '.' || p[j - 1] == s[i - 1])) {
          f[i, j] = f[i - 1, j - 1];
        }
      }
    }
    return f[m, n];
  }
}

class Solution {
  /**
   * @param string $s
   * @param string $p
   * @return boolean
   */

  function isMatch($s, $p) {
    $m = strlen($s);
    $n = strlen($p);

    $dp = array_fill(0, $m + 1, array_fill(0, $n + 1, false));
    $dp[0][0] = true;

    for ($j = 1; $j <= $n; $j++) {
      if ($p[$j - 1] == '*') {
        $dp[0][$j] = $dp[0][$j - 2];
      }
    }

    for ($i = 1; $i <= $m; $i++) {
      for ($j = 1; $j <= $n; $j++) {
        if ($p[$j - 1] == '.' || $p[$j - 1] == $s[$i - 1]) {
          $dp[$i][$j] = $dp[$i - 1][$j - 1];
        } elseif ($p[$j - 1] == '*') {
          $dp[$i][$j] = $dp[$i][$j - 2];
          if ($p[$j - 2] == '.' || $p[$j - 2] == $s[$i - 1]) {
            $dp[$i][$j] = $dp[$i][$j] || $dp[$i - 1][$j];
          }
        } else {
          $dp[$i][$j] = false;
        }
      }
    }

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