Evaluate Postfix Expression Using Stack in DSA Python - Time & Space Complexity

Time Complexity: Evaluate Postfix Expression Using Stack

O(n)

Understanding Time Complexity

We want to understand how the time taken to evaluate a postfix expression changes as the expression gets longer.

Specifically, how does the number of steps grow when we use a stack to solve this?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

def evaluate_postfix(expression):
    stack = []
    for char in expression:
        if char.isdigit():
            stack.append(int(char))
        else:
            right = stack.pop()
            left = stack.pop()
            if char == '+':
                stack.append(left + right)
            elif char == '*':
                stack.append(left * right)
    return stack.pop()
    

This code evaluates a postfix expression with single-digit numbers and '+' or '*' operators using a stack.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Looping through each character in the expression once.
• How many times: Exactly once for each character in the input expression.

How Execution Grows With Input

Each character in the expression is processed once, so the steps grow directly with the length of the expression.

Input Size (n)	Approx. Operations
10	About 10 steps
100	About 100 steps
1000	About 1000 steps

Pattern observation: The number of operations increases in a straight line as the input size grows.

Final Time Complexity

Time Complexity: O(n)

This means the time to evaluate grows directly in proportion to the length of the postfix expression.

Common Mistake

[X] Wrong: "Evaluating postfix expressions takes more than linear time because of the stack operations."

[OK] Correct: Each character is handled once, and stack operations (push/pop) are very fast and happen only once per character, so overall time stays linear.

Interview Connect

Understanding this linear time behavior helps you explain how stacks efficiently solve expression evaluation problems, a common topic in interviews.

Self-Check

"What if the expression contained multi-digit numbers separated by spaces? How would the time complexity change?"

Practice

1. What does evaluating a postfix expression using a stack mainly involve?

easy

A. Using recursion to evaluate the expression

B. Converting the postfix expression to infix before evaluation

C. Pushing operands onto the stack and popping two operands for each operator to apply the operation

D. Ignoring operators and only summing operands

Solution

1. Step 1: Understand postfix evaluation process

   In postfix evaluation, operands are pushed onto a stack. When an operator is encountered, two operands are popped, the operation is applied, and the result is pushed back.

2. Step 2: Identify correct method

   Pushing operands onto the stack and popping two operands for each operator to apply the operation correctly describes this stack-based process. Other options describe unrelated or incorrect methods.

3. Final Answer:

   Pushing operands onto the stack and popping two operands for each operator to apply the operation -> Option C

4. Quick Check:

   Postfix evaluation uses stack push/pop [OK]

Hint: Remember: push operands, pop two for operator, push result [OK]

Common Mistakes:

• Thinking postfix needs conversion to infix first
• Trying to evaluate from left to right without stack
• Ignoring the order of operands when popping

2. Which of the following Python code snippets correctly pushes operands and evaluates a postfix expression using a stack?

easy

A. stack.append(int(char)) when char is digit; pop two, apply operator, push result

B. stack.push(char) for operands; pop one operand for operator

C. Use stack.pop() to push operands; stack.append() to pop

D. Directly evaluate expression without stack

Solution

1. Step 1: Check correct stack operations

   In Python, list's append() adds elements (push), pop() removes last element (pop). For postfix, push operands as integers, pop two for operator.

2. Step 2: Match with options

   stack.append(int(char)) when char is digit; pop two, apply operator, push result correctly uses stack.append(int(char)) for digits and pops two operands for operators. Others misuse push/pop or skip stack.

3. Final Answer:

   stack.append(int(char)) when char is digit; pop two, apply operator, push result -> Option A

4. Quick Check:

   Use append to push, pop to remove [OK]

Hint: Use append() to push and pop() to remove from stack [OK]

Common Mistakes:

• Using stack.push() which is not a Python list method
• Popping only one operand for operator
• Not converting char to int before pushing

3. What is the output of the following Python code that evaluates the postfix expression '231*+9-'?

def evaluate_postfix(expr):
    stack = []
    for char in expr:
        if char.isdigit():
            stack.append(int(char))
        else:
            b = stack.pop()
            a = stack.pop()
            if char == '+':
                stack.append(a + b)
            elif char == '-':
                stack.append(a - b)
            elif char == '*':
                stack.append(a * b)
            elif char == '/':
                stack.append(a // b)
    return stack.pop()

print(evaluate_postfix('231*+9-'))

medium

A. -4

B. -2

C. -3

D. 4

Solution

1. Step 1: Evaluate expression step-by-step

   Expression: '2 3 1 * + 9 -' - Push 2, 3, 1 - '*' operator: pop 1 and 3, multiply 3*1=3, push 3 - '+' operator: pop 3 and 2, add 2+3=5, push 5 - Push 9 - '-' operator: pop 9 and 5, subtract 5-9=-4, push -4

2. Step 2: Final result

   Pop final value from stack: -4

3. Final Answer:

   -4 -> Option A

4. Quick Check:

   Postfix '231*+9-' = -4 [OK]

Hint: Evaluate stepwise: push digits, apply operators on last two [OK]

Common Mistakes:

• Mixing operand order when popping (a, b vs b, a)
• Using float division instead of integer division
• Misreading the expression order

4. The following code is intended to evaluate a postfix expression but throws an error. What is the error and how to fix it?

def eval_postfix(expr):
    stack = []
    for char in expr:
        if char.isdigit():
            stack.append(char)
        else:
            a = stack.pop()
            b = stack.pop()
            if char == '+':
                stack.append(a + b)
            elif char == '-':
                stack.append(a - b)
    return stack.pop()

print(eval_postfix('23+'))

medium

A. SyntaxError due to missing colon; fix by adding colon after if

B. TypeError due to adding strings; fix by converting char to int before pushing

C. IndexError due to popping from empty stack; fix by checking stack size

D. No error; output is '5'

Solution

1. Step 1: Identify error cause

   The code pushes characters as strings onto stack. When popping, it tries to add strings, causing TypeError.

2. Step 2: Fix by converting to int

   Convert char to int before pushing: stack.append(int(char)) to allow arithmetic operations.

3. Final Answer:

   TypeError due to adding strings; fix by converting char to int before pushing -> Option B

4. Quick Check:

   Push integers, not strings, to avoid TypeError [OK]

Hint: Always convert digit chars to int before pushing [OK]

Common Mistakes:

• Pushing strings instead of integers
• Swapping operand order when popping
• Ignoring operator cases other than '+' and '-'

5. Given the postfix expression '56+82/-', what is the result after evaluating it using a stack? Show the step-by-step evaluation.

hard

A. 5

B. 1

C. 9

D. 7

Solution

1. Step 1: Evaluate postfix expression stepwise

   Expression: '5 6 + 8 2 / -' - Push 5, 6 - '+' operator: pop 6 and 5, add 5+6=11, push 11 - Push 8, 2 - '/' operator: pop 2 and 8, integer divide 8//2=4, push 4 - '-' operator: pop 4 and 11, subtract 11-4=7, push 7

2. Step 2: Final result

   Pop final value from stack: 7

3. Final Answer:

   7 -> Option D

4. Quick Check:

   Postfix '56+82/-' = 7 [OK]

Hint: Process operators on last two stack values in order [OK]

Common Mistakes:

• Dividing operands in wrong order (2/8 instead of 8/2)
• Using float division instead of integer division
• Mixing operand order when subtracting