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Regular Expression Matching in Python
Suppose we have an input string s and another input string p. Here s is the main string and p is the pattern. We have to define one method, that can match patterns in the string. So we have to implement this for a regular expression, that supports ‘.’ And ‘*’.
Dot ‘.’ Matches any single character
Star ‘*’ Matches zero or more of the preceding element.
So for example, if the input is like s = “aa” and p = “a.”, then it will be true, for the same input string, if the patter is “.*”, then it will be true.
To solve this, we will follow these steps −
ss := size of s and ps := size of p
make dp a matrix of size ss x ps, and fill this using false value
Update p and s by adding one blank space before these
For i in range 2 to ps −
dp[0, i] := dp[0, i - 2] when p[i] is star, otherwise False
for i in range 1 to ss
for j in range 1 to ps
if s[i] is p[j], or p[j] is dot, then
dp[i, j] := dp[i – 1, j – 1]
otherwise when p[j] is star, then
dp[i, j] := dp[i, j - 2]
if s[i] is p[j – 1] or p[j – 1] is dot, then
dp[i, j] := max of dp[i, j] and dp[i – 1, j]
return dp[ss, ps]
Example
Let us see the following implementation to get better understanding −
class Solution(object):
def isMatch(self, s, p):
ss = len(s)
ps = len(p)
dp = [[False for i in range(ps+1)] for j in range(ss+1)]
p = " "+p
s = " " + s
dp[0][0]=True
for i in range(2,ps+1):
dp[0][i] = dp[0][i-2] if p[i]=='*'else False
for i in range(1,ss+1):
for j in range(1,ps+1):
if s[i] ==p[j] or p[j]=='.':
dp[i][j]= dp[i-1][j-1]
elif p[j] == '*':
dp[i][j] = dp[i][j-2]
if s[i] == p[j-1] or p[j-1]=='.':
dp[i][j] = max(dp[i][j],dp[i-1][j])
return dp[ss][ps]
ob = Solution()
print(ob.isMatch("aa", "a."))
print(ob.isMatch("aaaaaa", "a*"))Input
"aa", "a." "aaaaaa", "a*"
Output
True True
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