Balanced Parentheses (Stack)
Given a string of parentheses, determine if the parentheses are balanced and properly nested. Use a Stack data structure.
Why Interviewers Ask This
Apple interviewers ask this to evaluate a candidate's fundamental grasp of stack mechanics and their ability to handle state management during traversal. They specifically test logical precision in tracking open versus closed delimiters, ensuring you can solve problems requiring Last-In-First-Out (LIFO) logic without relying on built-in regex libraries.
How to Answer This Question
1. Clarify the input: Confirm if the string contains only parentheses or other characters like brackets and braces, as Apple often tests edge cases with mixed types. 2. Define the strategy: Explicitly state that you will use a stack to track opening symbols and match them against closing ones. 3. Walk through the algorithm: Describe iterating through the string character by character. Explain pushing opening brackets onto the stack and popping when a matching closing bracket is found. 4. Handle edge cases: Mention checking for an empty stack when a closing bracket appears or a non-empty stack at the end of the string. 5. Analyze complexity: Conclude by stating the time complexity is O(n) since you traverse once, and space complexity is O(n) for the worst-case scenario where all characters are open brackets. This structured approach demonstrates systematic problem-solving skills valued at Apple.
Key Points to Cover
- Explicitly mention the LIFO property of stacks as the core mechanism
- Demonstrate handling of mismatched pairs and empty stack scenarios
- Clearly articulate O(n) time and O(n) space complexity analysis
- Show awareness of edge cases like strings starting with closing brackets
- Connect the solution to real-world parsing tasks common in system design
Sample Answer
To solve the balanced parentheses problem efficiently, I would implement a solution using a Stack data structure. First, I'd initialize an empty stack to keep track of the opening delimiters encountered. As I iterate through the input string character by character, I need to determine the nature of each symbol. If I encounter an opening parenthesis, such as '(', '[', or '{', I push it directly onto the stack. This records the expectation that a corresponding closing delimiter must appear later. When I encounter a closing delimiter, the logic becomes more critical. I must first check if the stack is empty; if it is, the string is immediately unbalanced because there is no matching opener. If the stack is not empty, I pop the top element and verify if it matches the current closing delimiter. For instance, if I see ')', the popped item must be '('. If they don't match, the nesting is invalid. After processing the entire string, the final validation step is crucial: the stack must be completely empty. If any elements remain, it means some opening delimiters were never closed. Regarding efficiency, this approach runs in O(n) time complexity because we visit each character exactly once. The space complexity is O(n) in the worst case, occurring when the string consists entirely of opening brackets. This method ensures strict adherence to proper nesting rules, which is essential for parsing code or validating expressions in systems like those Apple develops.
Common Mistakes to Avoid
- Using a counter variable instead of a stack, which fails for mixed bracket types like '([)]'
- Forgetting to check if the stack is empty before attempting to pop a closing bracket
- Neglecting to verify that the stack is empty after the loop finishes, leaving unclosed brackets
- Assuming the input only contains round parentheses without clarifying if other types exist
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