DSA interview question

Postfix to Infix Conversion

Easy · Data structures · Topics: stack, string

The problem

A debugger for a stack-based virtual machine shows the bytecode's expressions in postfix, like ab*c+, but developers want to read them as ordinary fully parenthesised infix, like ((a*b)+c). Postfix is already in evaluation order, so the conversion mirrors evaluation. Read left to right, push operands, and when an operator appears, pop the right operand (top) and the left operand, then push "(" + left + op + right + ")".

Example: ab*c+ → ((a*b)+c).

Function signature

string postfixToInfix(const string& post);  // single-character operands; binary operators

Test cases

Input	Output	Explanation
`ab*c+`	`((a*b)+c)`
`ABC/-AK/L-*`	`((A-(B/C))*((A/K)-L))`
`ab+`	`(a+b)`
`abc++`	`(a+(b+c))`	The grouping comes from the postfix order
`a`	`a`

(Confirm: the postfix expression is valid; every binary operation is wrapped in parentheses.)

How interviewers phrase it

"Given a postfix expression, convert it to infix" (GeeksforGeeks "Postfix to Infix Conversion").

Clarifying questions to ask

Fully parenthesised output? (Yes.)
Only binary operators and single-character operands?
Is the input guaranteed valid?

Follow-ups to expect

"Which operand is popped first?"
"Why parenthesise everything?"
"Can you output minimal parentheses?"
"How is this related to evaluation?"
"Complexity?"

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