44

44. Wildcard Matching (Hard)

Given an input string (s) and a pattern (p), implement wildcard pattern matching with support for '?' and '*'.

'?' Matches any single character.
'*' Matches any sequence of characters (including the empty sequence).

The matching should cover the entire input string (not partial).

Note:

• s could be empty and contains only lowercase letters a-z.
• p could be empty and contains only lowercase letters a-z, and characters like ? or *.

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'.

Example 4:

Input:
s = "adceb"
p = "*a*b"
Output: true
Explanation: The first '*' matches the empty sequence, while the second '*' matches the substring "dce".

Example 5:

Input:
s = "acdcb"
p = "a*c?b"
Output: false

Related Topics:
String, Dynamic Programming, Backtracking, Greedy

Similar Questions:

• Regular Expression Matching (Hard)

Solution 1. DP Top-down

// OJ: https://leetcode.com/problems/wildcard-matching/
// Author: github.com/lzl124631x
// Time: O(MN)
// Space: O(MN)
class Solution {
    int M, N;
    vector<vector<int>> m;
    bool isMatch(string &s, string &p, int i, int j) {
        if (m[i][j] != -1) return m[i][j];
        int &ans = m[i][j];
        for (; j < N; ++j) {
            if (p[j] != '*' && i >= M) return ans = false;
            if (p[j] == '?') ++i;
            else if (p[j] == '*') {
                while (j + 1 < N && p[j + 1] == '*') ++j;
                for (int k = 0; i + k <= M; ++k) {
                    if (isMatch(s, p, i + k, j + 1)) return ans = true;
                }
            } else if (s[i++] != p[j]) return ans = false;
        }
        return ans = i >= M;
    }
public:
    bool isMatch(string s, string p) {
        M = s.size(), N = p.size();
        m.assign(M + 1, vector<int>(N + 1, -1));
        return isMatch(s, p, 0, 0);
    }
};

Solution 2. DP Bottom-up

Let dp[i][j] be a boolean indicating whether s[0..(i-1)] and p[0..(j-1)] matches, where 0 <= i <= M, 1 <= j <= N.

Trivial case: dp[0][0] = true.

If i > 0 && (p[j - 1] == '?' || s[i - 1] == p[j - 1]), dp[i][j] = dp[i - 1][j - 1].

If p[j - 1] == '*', dp[i][j] = dp[i][j - 1] || dp[i-1][j-1] || dp[i-2][j-1] || ... || dp[0][j-1]. Since dp[i-1][j] = dp[i-1][j-1] || dp[i-2][j-1] || ... || dp[0][j-1], so dp[i][j] = dp[i][j-1] || (i > 0 && dp[i-1][j])

Otherwise dp[i][j] = false.

In sum:

dp[i][j] = dp[i][j-1] || (i > 0 && dp[i-1][j])   // If p[j-1] == '*'
         = dp[i-1][j-1]                          // If i > 0 && (p[j - 1] == '?' || s[i - 1] == p[j - 1])
         = false                                 // otherwise
dp[0][0] = true

// OJ: https://leetcode.com/problems/wildcard-matching/
// Author: github.com/lzl124631x
// Time: O(MN)
// Space: O(MN)
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 i = 0; i <= M; ++i) {
            for (int j = 1; j <= N; ++j) {
                if (i > 0 && (p[j - 1] == '?' || s[i - 1] == p[j - 1])) dp[i][j] = dp[i - 1][j - 1];
                else if (p[j - 1] == '*') dp[i][j] = dp[i][j - 1] || (i > 0 && dp[i - 1][j]);
            }
        }
        return dp[M][N];
    }
};

Solution 3.

Once we find a '*' in p, log the indexes when we met '*' into star (the index pointing to the '*') and ss (s[ss] is the first character we match against p[star + 1]).

Continue matching the rest. If we can't match the rest, reset the indexes to j = star + 1 and i = ++ss.

What if we the second '*' overwrites the indexes we stored in star and ss? For example s=ababac, p=a*b*c.

This still works because when there are multiple matches to the pattern in between two '*', picking either one works. So here we just pick the first match, and let the second '*' overwrite the state.

// OJ: https://leetcode.com/problems/wildcard-matching/
// Author: github.com/lzl124631x
// Time: O(MN)
// Space: O(1)
// Ref: https://leetcode.com/problems/wildcard-matching/discuss/17810/Linear-runtime-and-constant-space-solution
class Solution { public:
    bool isMatch(string s, string p) {
        int M = s.size(), N = p.size(), star = -1, ss = 0, i = 0, j = 0;
        while (i < M) {
            if (p[j] == '?' || p[j] == s[i]) {
                ++i;
                ++j;
                continue;
            }
            if (p[j] == '*') {
                star = j++;
                ss = i;
                continue;
            }
            if (star != -1) {
                j = star + 1;
                i = ++ss;
                continue;
            }
            return false;
        }
        while (p[j] == '*') ++j;
        return j == N;
    }
};