Python dp solution

  • 0

    Here is a solution I wrote for python using dynamic programming. I hope I made it easy to understand.

    def isMatch(self, s, p):
            dp = [[False for _ in range(len(s)+1)] for _ in range(len(p)+1)]
            dp[0][0] = True
            # Base cases for comparing against empty string
            for i in range(1, len(p)+1):
                if dp[i-2][0] and p[i-1] == '*':
                    dp[i][0] = True
            for i in range(1, len(s)+1):
                if dp[0][i-2] and s[i-1] == '*':
                    dp[0][i] = True
            # fill out 2d array with recurrance relation
            for i in range(1, len(p)+1):
                for j in range(1, len(s)+1):
                    if p[i-1] == s[j-1] or p[i-1] == '.':
                        dp[i][j] = dp[i-1][j-1]
                    elif p[i-1] == '*':
                        if s[j-1] == p[i-2] or p[i-2] == '.':
                            dp[i][j] = dp[i][j-1] or dp[i-1][j] or dp[i-2][j]
                            dp[i][j] = dp[i-1][j] or dp[i-2][j]
            return dp[-1][-1]

Log in to reply

Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect.