feat: regular-expression-matching-dp challenge

Author: IbatoLionDev

3-HARD/Python/8-regular-expression-matching-dp/proposed-solution/__pycache__/regex_matcher.cpython-312.pyc

@@ -0,0 +1,17 @@
+# Challenge: Design a regular expression matching engine that supports special characters like '.' and '*',
+# using dynamic programming techniques to efficiently handle branching and backtracking.
+# Use object-oriented programming and follow the DRY principle.
+
+from regex_matcher import RegexMatcher
+
+def main():
+    text = "aab"
+    pattern = "c*a*b"
+    matcher = RegexMatcher(text, pattern)
+    result = matcher.is_match()
+    print(f"Text: {text}")
+    print(f"Pattern: {pattern}")
+    print(f"Match result: {result}")
+
+if __name__ == "__main__":
+    main()

3-HARD/Python/8-regular-expression-matching-dp/proposed-solution/main.py

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+class RegexMatcher:
+    def __init__(self, text: str, pattern: str):
+        self.text = text
+        self.pattern = pattern
+        self.dp = [[False] * (len(pattern) + 1) for _ in range(len(text) + 1)]
+
+    def is_match(self) -> bool:
+        self.dp[0][0] = True
+
+        for j in range(2, len(self.pattern) + 1):
+            if self.pattern[j - 1] == '*':
+                self.dp[0][j] = self.dp[0][j - 2]
+
+        def matches(i, j):
+            if self.pattern[j - 1] == '.' or self.pattern[j - 1] == self.text[i - 1]:
+                return self.dp[i - 1][j - 1]
+            if self.pattern[j - 1] == '*':
+                zero = self.dp[i][j - 2]
+                one_or_more = (
+                    (self.pattern[j - 2] == '.' or self.pattern[j - 2] == self.text[i - 1])
+                    and self.dp[i - 1][j]
+                )
+                return zero or one_or_more
+            return False
+
+        for i in range(1, len(self.text) + 1):
+            for j in range(1, len(self.pattern) + 1):
+                self.dp[i][j] = matches(i, j)
+
+        return self.dp[len(self.text)][len(self.pattern)]