I knew it was stack-related pretty fast but fumbled the parentheses logic for a minute.
Use a stack to evaluate the RPN expression, but instead of computing numeric values, build expression trees or strings. For each token, if it's an operand, push it; if it's an operator, pop two operands, combine them into a parenthesized expression, and push the result. At the end, the stack contains the fully parenthesized infix expression.
Pro tip: Clarify the expected handling of multi-digit numbers and negative numbers, and mention that you assume valid RPN input. Also, discuss whether to add parentheses around every operation or only when necessary for precedence, but for fully parenthesized, always add them.
Restate the problem: convert a comma-separated RPN string to a fully parenthesized infix string. Ask about input format (e.g., multi-digit numbers, negative numbers, spaces) and confirm that the output should have parentheses around every binary operation.
Use a stack to store intermediate expressions. Each element can be a string representing a sub-expression. This naturally handles the postfix evaluation order.
Split the input string by commas. For each token: if it's a number, push it onto the stack; if it's an operator, pop the top two elements (right operand first, then left), form a string like '(left op right)', and push it back.
After processing all tokens, the stack should contain exactly one element: the final infix expression. If the stack has more or fewer, the input was invalid. Return that string.
Time complexity is O(n) where n is the number of tokens, as each token is processed once. Space complexity is O(n) for the stack and output strings. Walk through an example to verify correctness.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.