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Wildcard Matching in Python
Suppose we have an input string s and another input string p. Here is the main string and p is the pattern. We have to define one method, that can match pattern in the string. So we have to implement this for a regular expression, that supports wildcard characters like ‘?’ And ‘*’.
Dot ‘?’ Matches any single character
Star ‘*’ Matches zero or more characters.
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 1 to ps −
if p[i] = star, then
dp[0, i] := dp[0, i - 1]
for i in range 1 to ss
for j in range 1 to ps
if s[i] is p[j], or p[j] is ‘?’, then
dp[i, j] := dp[i – 1, j – 1]
otherwise when p[j] is star, then
dp[i, j] := max of dp[i – 1, j] and dp[i, j – 1]
return dp[ss, ps]
Example (Python)
Let us see the following implementation to get better understanding −
class Solution(object):
def isMatch(self, s, p):
sl = len(s)
pl = len(p)
dp = [[False for i in range(pl+1)] for j in range(sl+1)]
s = " "+s
p = " "+p
dp[0][0]=True
for i in range(1,pl+1):
if p[i] == '*':
dp[0][i] = dp[0][i-1]
for i in range(1,sl+1):
for j in range(1,pl+1):
if s[i] == p[j] or p[j] == '?':
dp[i][j] = dp[i-1][j-1]
elif p[j]=='*':
dp[i][j] = max(dp[i-1][j],dp[i][j-1])
return dp[sl][pl]
ob = Solution()
print(ob.isMatch("aa", "a?"))
print(ob.isMatch("aaaaaa", "a*"))Input
"aa", "a." "aaaaaa", "a*"
Output
True True
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