From b031f82d97d495ef7b256295b9cbb238ba4a1819 Mon Sep 17 00:00:00 2001
From: Deven <63876261+devenperez@users.noreply.github.com>
Date: Wed, 11 Jan 2023 20:59:06 -0500
Subject: [PATCH] Time: 1054 ms (28.24%), Space: 13.9 MB (55.17%) - LeetHub

---
 .../10-regular-expression-matching.py         | 34 +++++++++++++++++++
 1 file changed, 34 insertions(+)
 create mode 100644 10-regular-expression-matching/10-regular-expression-matching.py

diff --git a/10-regular-expression-matching/10-regular-expression-matching.py b/10-regular-expression-matching/10-regular-expression-matching.py
new file mode 100644
index 0000000..71421e2
--- /dev/null
+++ b/10-regular-expression-matching/10-regular-expression-matching.py
@@ -0,0 +1,34 @@
+class Solution:
+    def isMatch(self, s: str, p: str) -> bool:
+        # Recursive solution
+        
+        # Base Case
+        if s == "" or p == "":
+            if s == "" and p == "":
+                return True
+            # Check for "*" = 0 elements
+            elif s == "" and len(p) >= 2 and p[1] == "*":
+                return self.isMatch(s,p[2:])   
+            else:
+                return False
+        
+        # Dealing with "*"
+        if len(p) >= 2 and p[1] == "*":
+            
+            # Next character compatable with wildcard
+            if s[0] == p[0] or p[0] == ".":
+                return (self.isMatch(s,p[2:]) or    # End wildcard
+                        self.isMatch(s[1:],p))      # Use character as wildcard
+            else:
+                return self.isMatch(s,p[2:])
+            
+            
+        # Dealing with "."
+        if p[0] == ".":
+            return self.isMatch(s[1:],p[1:])    
+        
+        
+        # Check first letters, recurse on the rest
+        if s[0] == p[0]:
+            return self.isMatch(s[1:],p[1:])
+            
\ No newline at end of file