1+ '''
2+ 20. Valid Parentheses
3+
4+ Given a string s containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.
5+ An input string is valid if:
6+ 1. Open brackets must be closed by the same type of brackets.
7+ 2. Open brackets must be closed in the correct order.
8+ 3. Every close bracket has a corresponding open bracket of the same type.
9+ Note that an empty string is also considered valid.
10+
11+ Example 1:
12+ Input: s = "()"
13+ Output: true
14+ Example 2:
15+ Input: s = "()[]{}"
16+ Output: true
17+ Example 3:
18+ Input: s = "(]"
19+ Output: false
20+ '''
21+ # Brute Force Approach
22+ # Time Complexity: O(n^2)
23+ # Space Complexity: O(n)
24+ 25+ '''
26+ In the brute force approach, we can use a while loop to check if there are any pairs of brackets that can be removed from the string.
27+ We can keep removing pairs of brackets until no more pairs can be removed.
28+ If the string becomes empty, it means all brackets were matched and we return True. If there are still brackets left in the string, we return False.
29+ '''
30+ 31+ class Solution :
32+ def isValid (self , s ):
33+ while "()" in s or "[]" in s or "{}" in s :
34+ s = s .replace ("()" , "" )
35+ s = s .replace ("[]" , "" )
36+ s = s .replace ("{}" , "" )
37+ 38+ return s == ""
39+ 40+ # Stack Approach
41+ # Time Complexity: O(n), where n is the length of the string.
42+ # Space Complexity: O(n), in the worst case, we need to store all opening brackets in the stack.
43+ 44+ '''
45+ Below is the code to solve the problem:
46+ 1. Create a stack to keep track of the opening brackets.
47+ 2. Create a mapping of opening brackets to closing brackets.
48+ 3. Iterate through each character in the string:
49+ - If the character is an opening bracket, push it onto the stack.
50+ - If the character is a closing bracket, check if the stack is empty or if the top of the stack does not match the corresponding opening bracket.
51+ If either condition is true, return False.
52+ - If the character is a closing bracket and the stack is not empty, pop the top of the stack.
53+ 4. After iterating through the string, check if the stack is empty. If it is, return True; otherwise, return False.
54+ '''
55+ class Solution :
56+ def isValid (self , s ):
57+ stack = []
58+ mapp = {"(" :")" ,"[" :"]" ,"{" :"}" }
59+ for ch in s :
60+ if ch in mapp :
61+ stack .append (ch )
62+ elif ch in mapp .values ():
63+ if not stack or mapp [stack .pop ()]!= ch :
64+ return False
65+ return True if not stack else False
66+ 67+ 68+ class Solution :
69+ def isValid (self , s ):
70+ stack = []
71+ mapp = { ")" : "(" , "]" : "[" , "}" : "{" }
72+ 73+ for c in s :
74+ if c in mapp :
75+ if stack and stack [- 1 ] == mapp [c ]:
76+ stack .pop ()
77+ else :
78+ return False
79+ else :
80+ stack .append (c )
81+ 82+ return True if not stack else False
83+ 84+ obj = Solution ()
85+ s = "([{}])"
86+ print (obj .isValid (s )) # Output: True
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