Infix to Postfix Conversion Using Stack in DSA Python - Time & Space Complexity

Time Complexity: Infix to Postfix Conversion Using Stack

O(n)

Understanding Time Complexity

We want to understand how the time taken to convert an infix expression to postfix grows as the expression gets longer.

Specifically, how does the number of steps change when the input expression size increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

def infix_to_postfix(expression):
    stack = []
    result = ''
    for char in expression:
        if char.isalnum():
            result += char
        elif char == '(': 
            stack.append(char)
        elif char == ')':
            while stack and stack[-1] != '(': 
                result += stack.pop()
            stack.pop()
        else:
            while stack and precedence(stack[-1]) >= precedence(char):
                result += stack.pop()
            stack.append(char)
    while stack:
        result += stack.pop()
    return result

This code converts an infix expression (like "a+b*c") to postfix notation (like "abc*+") using a stack to manage operators.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: One main loop that goes through each character of the input expression once.
• How many times: Exactly once for each character in the expression (n times).
• Inside this loop, there are some smaller loops that pop from the stack, but each operator is pushed and popped only once in total.

How Execution Grows With Input

As the expression length grows, the number of steps grows roughly in direct proportion.

Input Size (n)	Approx. Operations
10	About 10 main steps plus some stack operations
100	About 100 main steps plus stack operations
1000	About 1000 main steps plus stack operations

Pattern observation: The total work grows linearly with the input size.

Final Time Complexity

Time Complexity: O(n)

This means the time to convert grows in a straight line with the length of the expression.

Common Mistake

[X] Wrong: "Because there are nested loops, the time must be quadratic O(n²)."

[OK] Correct: The inner loops only pop operators that were pushed once, so total operations still add up to a single pass over the input and stack, keeping it linear.

Interview Connect

Understanding this linear time complexity helps you explain why stack-based expression parsing is efficient and practical in real coding tasks.

Self-Check

"What if the input expression contains multi-digit numbers or variables with multiple letters? How would the time complexity change?"

Practice

1. What is the main purpose of using a stack in converting an infix expression to postfix?

easy

A. To reverse the order of operands

B. To temporarily hold operators and manage their precedence

C. To store operands permanently

D. To count the number of operators

Solution

1. Step 1: Understand the role of stack in infix to postfix

   The stack is used to hold operators temporarily while processing the expression to respect operator precedence and associativity.

2. Step 2: Clarify what the stack does not do

   The stack does not store operands permanently or reverse operands; it only manages operators.

3. Final Answer:

   To temporarily hold operators and manage their precedence -> Option B

4. Quick Check:

   Stack manages operators [OK]

Hint: Stack holds operators, not operands, during conversion [OK]

Common Mistakes:

• Thinking stack stores operands
• Confusing stack role with reversing expression
• Ignoring operator precedence handling

2. Which of the following is the correct way to push an operator onto a stack in Python?

easy

A. stack.append(operator)

B. stack.push(operator)

C. stack.insert(operator)

D. stack.add(operator)

Solution

1. Step 1: Recall Python list methods for stack operations

   In Python, a list is used as a stack and the method to add an element on top is append().

2. Step 2: Identify incorrect methods

   push() and add() are not valid list methods; insert() inserts at a specific index, not top of stack.

3. Final Answer:

   stack.append(operator) -> Option A

4. Quick Check:

   Use append() to push on stack [OK]

Hint: Use append() to push items on Python list stack [OK]

Common Mistakes:

• Using push() which is not a Python list method
• Using add() which is for sets, not lists
• Using insert() incorrectly for stack push

3. What is the postfix expression for the infix expression A + B * C?

medium

A. ABC*+

B. AB+C*

C. A+BC*

D. *ABC+

Solution

1. Step 1: Apply operator precedence rules

   Multiplication (*) has higher precedence than addition (+), so B * C is evaluated first.

2. Step 2: Convert to postfix respecting precedence

   Postfix places operands first, then operator: B C * for multiplication, then A and + last, so expression is A B C * +.

3. Final Answer:

   ABC*+ -> Option A

4. Quick Check:

   Multiplication before addition = ABC*+ [OK]

Hint: Multiply before add; postfix puts operator after operands [OK]

Common Mistakes:

• Ignoring operator precedence
• Placing operators before operands
• Mixing operand order

4. Given the following code snippet for infix to postfix conversion, what is the error?

def precedence(op):
    if op == '+' or op == '-':
        return 1
    if op == '*' or op == '/':
        return 2
    return 0

stack = []
postfix = ''
expr = 'A+B*C'

for char in expr:
    if char.isalpha():
        postfix += char
    else:
        while stack and precedence(stack[-1]) >= precedence(char):
            postfix += stack.pop()
        stack.append(char)

while stack:
    postfix += stack.pop()

print(postfix)

medium

A. The precedence function returns 0 for unknown operators, causing errors

B. The postfix string is not updated with operands

C. The stack is not initialized properly

D. The code does not handle parentheses in the expression

Solution

1. Step 1: Analyze code handling of operators and operands

   The code correctly appends operands and manages operator precedence using the stack.

2. Step 2: Check for missing features

   The code does not check or handle parentheses, which are common in infix expressions and affect precedence.

3. Final Answer:

   The code does not handle parentheses in the expression -> Option D

4. Quick Check:

   Missing parentheses handling [OK]

Hint: Remember to handle parentheses explicitly in infix to postfix [OK]

Common Mistakes:

• Assuming precedence 0 causes errors (it just means lowest)
• Ignoring parentheses handling
• Thinking stack initialization is wrong

5. Convert the infix expression (A+B)*(C-D)/E to postfix notation using stack. What is the correct postfix expression?

hard

A. AB+CD-E*/

B. AB+CD-E/*

C. AB+CD-*E/

D. AB+CD*-E/

Solution

1. Step 1: Handle parentheses and operator precedence

   Parentheses force A+B and C-D to be evaluated first, so their postfix forms are AB+ and CD- respectively.

2. Step 2: Apply multiplication and division in order

   After parentheses, multiply the two results: AB+ CD- * then divide by E: AB+CD-* E /.

3. Final Answer:

   AB+CD-*E/ -> Option C

4. Quick Check:

   Parentheses first, then * and / = AB+CD-*E/ [OK]

Hint: Convert inside parentheses first, then apply * and / left to right [OK]

Common Mistakes:

• Ignoring parentheses order
• Mixing operator order for * and /
• Placing operators before operands