← Bytedance Interview Insights
Classic stack problem, got through it fine.
Use a stack to track opening brackets, pushing each opening bracket and popping when a matching closing bracket is encountered. At the end, the stack should be empty for the string to be valid. Handle edge cases like empty string, single bracket, and mismatched types.
Pro tip: Mention that you can early return if the string length is odd, and discuss the trade-off between using a stack (O(n) time, O(n) space) and a counter-based approach for single bracket type. Also, clarify if the input can contain other characters and how to handle them.
Ask if the string can contain non-bracket characters, if only parentheses or multiple types, and confirm expected behavior for empty string. This shows attention to detail.
Explain that a stack is ideal because brackets must be closed in LIFO order. For single type, a counter works, but for multiple types, a stack is necessary.
Iterate through each character: if opening bracket, push onto stack; if closing bracket, check if stack is empty or top doesn't match, return false; otherwise pop. After loop, return stack is empty.
State time complexity O(n) and space O(n) worst case. Provide test cases: valid, invalid, empty, single type, mixed types, and nested.
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