Wildcard Matching — 2D DP

Sanjeev SharmaSanjeev Sharma
1 min read

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Problem

? matches any single character, * matches any sequence (including empty).


Solutions

Python

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 p[j-1]=='*':
                    dp[i][j]=dp[i-1][j] or dp[i][j-1]
                elif p[j-1]=='?' or p[j-1]==s[i-1]:
                    dp[i][j]=dp[i-1][j-1]
        return dp[m][n]

Complexity

  • Time: O(mn) | Space: O(mn)

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Sanjeev Sharma

Written by

Sanjeev Sharma

Full Stack Engineer · E-mopro