Valid Parentheses Solution

Author: jinbeom

valid-parentheses/kayden.py

@@ -0,0 +1,19 @@
+# 시간복잡도: O(N)
+# 공간복잡도: O(N)
+class Solution:
+    def isValid(self, s: str) -> bool:
+        if len(s) % 2 != 0:
+            return False
+
+        stack = []
+        matching_brackets = {'(': ')', '{': '}', '[': ']'}
+
+        for char in s:
+            if char in matching_brackets:
+                stack.append(char)
+            elif stack and matching_brackets[stack[-1]] == char:
+                stack.pop()
+            else:
+                return False
+
+        return not stack