valid-parentheses solution

Author: hyeri0903

valid-parentheses/hyeri0903.py

@@ -0,0 +1,37 @@
+class Solution:
+    def isValid(self, s: str) -> bool:
+        '''
+        문제: string s 에 대해서 valid parentheses 이면 true 아니면 false
+        conditions
+        - open brackets must be closed by the same type 
+        - // must be closed in the correct order
+        - s 최소 길이 = 1, 최대 10^4
+        solution
+        - open brackets -> st array 에 저장, close brackets -> st.pop -> check valid
+        - 마지막에 st array length 가 1 이상이면 return False
+
+        - time complexity: O(n)
+        - space complexity: O(n)
+        '''
+        
+        st = []
+    
+        for i in range(len(s)):
+            if s[i] == '(' or s[i] == '{' or s[i] == '[':
+                st.append(s[i])
+            else:
+                # check if st is empty
+                if len(st) <= 0:
+                    return False
+                cur = st.pop()
+                if s[i] == ')' and cur != '(':
+                    return False
+                if s[i] == '}' and cur != '{':
+                    return False
+                if s[i] == ']' and cur != '[':
+                    return False
+        if len(st) > 0:
+            return False
+        return True
+
+