# Regular Expression Matching in Python

PythonServer Side ProgrammingProgramming

#### Beyond Basic Programming - Intermediate Python

Most Popular

36 Lectures 3 hours

#### Practical Machine Learning using Python

Best Seller

91 Lectures 23.5 hours

#### Practical Data Science using Python

22 Lectures 6 hours

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 −

Live Demo

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
Updated on 26-May-2020 11:20:39