Prefix to Postfix Conversion

Overview

In computer science, prefix notation (Polish Notation) and postfix notation (Reverse Polish Notation) are ways of writing mathematical expressions without needing parentheses to define operator precedence.

• Prefix: Operator appears before operands (+ A B).
• Postfix: Operator appears after operands (A B +).

The core intuition for converting prefix to postfix is to process the expression from right to left. By starting at the end, we encounter operands first. When we eventually encounter an operator, we know the two most recently processed sub-expressions are its operands. We then bundle them into a new postfix string and continue.

Brute Force

While a recursive approach is often considered the "natural" way to handle expression trees, a literal brute-force string manipulation approach involves repeatedly scanning the string for the pattern [operator][operand][operand] and replacing it with [operand][operand][operator].

In this approach, we scan the string from right to left. We use a list to store operands and sub-expressions. When we see an operand, we treat it as a standalone "postfix" string. When we see an operator, we pop the last two postfix strings we've built, concatenate them, add the operator at the end, and push the result back.

def prefixToPostfix(prefix: str) -> str:
    # Use a stack to keep track of operands/sub-expressions
    stack = []
    
    # Iterate through the prefix string from right to left
    # We use reversed() because in prefix, the operator 
    # precedes the operands it acts upon.
    for char in reversed(prefix):
        # Check if the character is an operator
        if char in "+-*/":
            # Pop two operands from the stack
            # Since we are moving right-to-left, the first pop
            # is actually the 'left' operand and the second is 'right'.
            op1 = stack.pop()
            op2 = stack.pop()
            
            # Combine them in postfix order: operand1 + operand2 + operator
            new_expr = op1 + op2 + char
            
            # Push the new sub-expression back to the stack
            stack.append(new_expr)
        else:
            # If it's an operand (letter or digit), push it to the stack
            stack.append(char)
            
    # The final element in the stack is the complete postfix expression
    return stack[0]

• Time complexity: O(N^2) — In each step of the loop, we perform string concatenation. In Python, strings are immutable, so concatenating strings of length $k$ takes $O(k)$ time.
• Space complexity: O(N^2) — The stack stores intermediate strings which, in the worst case (a heavily nested expression), can result in a cumulative space usage proportional to the square of the input length.

Optimal / Best Approach

The logic remains similar to the stack-based approach, but we can conceptualize this as a recursive problem to better handle the nested nature of expressions. However, for a conversion task with constraints of length 100, the iterative right-to-left stack method is highly efficient and preferred for its simplicity and lack of recursion overhead.

To optimize "string building" in some languages, one might use a list of characters, but since the result is a str, the stack-based approach is generally the standard.

def prefixToPostfix(prefix: str) -> str:
    # A stack is ideal for Reverse Polish conversions
    stack = []
    
    # Process from right to left to ensure operands are 
    # available for their preceding operators
    for i in range(len(prefix) - 1, -1, -1):
        symbol = prefix[i]
        
        if symbol in "+-*/":
            # Take the two most recent sub-expressions
            operand1 = stack.pop()
            operand2 = stack.pop()
            
            # Form: (Operand1)(Operand2)(Operator)
            # This follows the Postfix rule
            combined = f"{operand1}{operand2}{symbol}"
            stack.append(combined)
        else:
            # It's an operand (A-Z or 0-9)
            stack.append(symbol)
            
    return stack.pop()

• Time complexity: O(N^2) — Similar to the previous approach, string concatenation in the loop remains the dominant factor. While the logic is $O(N)$, the string immutable properties make the actual runtime $O(N^2)$.
• Space complexity: O(N^2) — Storing intermediate postfix strings in the stack as we build the final result.

Why this works

When we read a prefix expression like *+AB-CD from right to left:

1. We see D, then C. Stack: ['D', 'C']
2. We see -. We pop C and D, create CD-, push it. Stack: ['D', 'C', 'CD-'] ... actually, to maintain correct order for non-commutative operators (like subtraction/division), we ensure the character encountered first during the right-to-left scan is the left operand of the operation when viewed from a standard perspective.
3. Because we iterate backwards, the first operand we pop was the one appearing earliest in the original string relative to its partner.

This ensures that for an input like /AB (which is $A/B$), the stack pops A then B, resulting in AB/.

Fast