Postfix to Infix Conversion
Prefix, Infix, Postfix Conversion Problems DSA practice problem on Onlearn.
Difficulty: medium.
Topics: How do you convert a postfix (Reverse Polish Notation) expression to an infix expression?, Stack, String Manipulation, Algorithm, expression parsing, expression notation, stack, Infix, Prefix, Postfix Notations.
Problem Statement Given a valid postfix expression, convert it to its equivalent infix expression. A postfix expression consists of operands (lowercase English letters) and binary operators (+, , , /). The resulting infix expression must be fully parenthesized to explicitly define the order of operations. Input The first and only line of input contains a string S representing a valid postfix expression. Output Output a single string, the infix expression equivalent to S, fully parenthesized. Constraints 1 <= |S| <= 100 S will only contain lowercase English letters (operands) and the operators +, , , /. S will represent a valid postfix expression. Example Input: Output: Explanation: 1. a is an operand, push 'a' to stack. Stack: [a] 2. b is an operand, push 'b' to stack. Stack: [a, b] 3. + is an operator. Pop b (operand2), pop a (operand1). Form (a+b). Push (a+b) to stack. Stack: [(a+b)] 4. c is an operand, push 'c' to stack. Stack: [(a+b), c] 5. is an operator. Pop c (operand2), pop (a+b) (operand1). Form ((a+b) c). Push ((a+b) c) to stack. Stack: [((a+b) c)] End of expression. The result is ((a+b) c).