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We are going to implement a function that checks if a string is a valid open bracket sequence.
The idea is to use a stack to keep track of the opening brackets.
We traverse the string, and for every character:
If it's an opening bracket (i.e., one of '(', '{', '['), we push it onto the stack.
If it's a closing bracket, we check if the stack is empty (which means no matching opening bracket) -> invalid.
Otherwise, we pop the top of the stack and check if the popped opening bracket matches the closing bracket.
If not, return invalid.
After processing the entire string, if the stack is not empty, there are unmatched opening brackets -> invalid.
Otherwise, valid.
Steps:
Initialize an empty stack.
Traverse each character in the string:
a. If the character is an opening bracket, push it to the stack.
b. Else (it's a closing bracket):
- If stack is empty, return False.
- Else, pop the top element and check if it matches the corresponding opening bracket for the current closing bracket.
If not, return False.
After traversal, if the stack is empty, return True; else, False.
We'll use a dictionary to map closing brackets to their corresponding opening brackets.
To solve this problem, we need to determine if a given string of brackets is valid. A string is considered valid if every opening bracket has a corresponding closing bracket of the same type and they are properly nested.
Approach
The approach involves using a stack data structure to keep track of opening brackets as we iterate through the string. For each character in the string:
Opening Brackets: When encountering an opening bracket ((, {, [), push it onto the stack.
Closing Brackets: When encountering a closing bracket (), }, ]), check if the stack is empty. If it is, the string is invalid. Otherwise, pop the top element from the stack and verify if it matches the corresponding opening bracket for the current closing bracket. If not, the string is invalid.
Final Check: After processing all characters, if the stack is not empty, it means there are unmatched opening brackets, making the string invalid.
Solution Code
def isValid(s: str) -> bool:
stack = []
mapping = {')': '(', '}': '{', ']': '['}
for char in s:
if char in mapping.values():
stack.append(char)
elif char in mapping:
if not stack:
return False
top_element = stack.pop()
if mapping[char] != top_element:
return False
return not stack
Explanation
Initialization: We initialize an empty stack to keep track of opening brackets and a dictionary mapping that maps each closing bracket to its corresponding opening bracket.
Iteration through the String: For each character in the string:
If the character is an opening bracket (found in mapping.values()), it is pushed onto the stack.
If the character is a closing bracket (found in mapping keys):
If the stack is empty, return False because there's no matching opening bracket.
Pop the top element from the stack and check if it matches the corresponding opening bracket using the mapping dictionary. If not, return False.
Final Validation: After processing all characters, if the stack is empty, all opening brackets have been properly closed, so return True. Otherwise, return False.
This approach efficiently checks the validity of the bracket string by leveraging stack operations to ensure proper nesting and matching of brackets. The time complexity is O(n) where n is the length of the string, as each character is processed exactly once. The space complexity is O(n) in the worst case when all characters are opening brackets.
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发表于 2025-11-6 07:43:13