Regular Expression Matching — DSA · Engineering Playbook

Problem#

Pattern uses . (any single char) and * (zero or more of the preceding element). Return true iff the entire s matches p.

Examples#

s = "aa", p = "a*" → true
s = "ab", p = ".*" → true
s = "mississippi", p = "mis*is*p*." → false

Approach#

dp[i][j] = does s[:i] match p[:j]. Base: dp[0][0] = true. For p[j-1] == '*': dp[i][j] = dp[i][j-2] (zero copies) OR (dp[i-1][j] AND s[i-1] matches p[j-2]). Else literal/dot: dp[i][j] = dp[i-1][j-1] && (p[j-1] == '.' || p[j-1] == s[i-1]).

Solution#

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

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

class Solution:
    def isMatch(self, s: str, p: str) -> bool:
        n, m = len(s), len(p)
        dp = [[False] * (m + 1) for _ in range(n + 1)]
        dp[0][0] = True
        for j in range(2, m + 1):
            if p[j - 1] == '*':
                dp[0][j] = dp[0][j - 2]
        for i in range(1, n + 1):
            for j in range(1, m + 1):
                if p[j - 1] == '*':
                    dp[i][j] = dp[i][j - 2] or (
                        dp[i - 1][j] and (p[j - 2] == '.' or p[j - 2] == s[i - 1])
                    )
                else:
                    dp[i][j] = dp[i - 1][j - 1] and (
                        p[j - 1] == '.' or p[j - 1] == s[i - 1]
                    )
        return dp[n][m]

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

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

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

Editorial#

The * case has two branches: “zero copies of the preceding element” (drops two chars from p) or “one more match of that element” (drops one char from s). The first row pre-handles empty s matched against a*b*c* style patterns.

Complexity#

Time: O(n * m).
Space: O(n * m).

Concept revision#

Wildcard Matching

Problem#

Examples#

Approach#

Solution#

Editorial#

Complexity#

Concept revision#

Related#