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

Time Complexity: Infix to Postfix Conversion Using Stack

O(n)

Understanding Time Complexity

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

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.

function infixToPostfix(expression) {
  const stack = [];
  let result = '';
  for (let char of expression) {
    if (isOperand(char)) {
      result += char;
    } else if (char === '(') {
      stack.push(char);
    } else if (char === ')') {
      while (stack.length && stack[stack.length - 1] !== '(') {
        result += stack.pop();
      }
      stack.pop();
    } else {
      while (stack.length && precedence(char) < precedence(stack[stack.length - 1])) {
        result += stack.pop();
      }
      stack.push(char);
    }
  }
  while (stack.length) {
    result += stack.pop();
  }
  return result;
}

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

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: A single loop over each character in the input expression.
• How many times: Exactly once for each character, so n times if the expression length is n.
• Inside this loop, there are while loops that pop from the stack, but each operator is pushed and popped at most once.

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 loop steps plus some stack operations
100	About 100 main loop steps plus stack operations, roughly 10 times more
1000	About 1000 main loop steps plus stack operations, roughly 10 times more than n=100

Pattern observation: The total steps increase linearly as the input size increases.

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 while loops, the time complexity must be quadratic O(n²)."

[OK] Correct: Each character is pushed and popped at most once, so the inner loops do not multiply the total steps beyond linear.

Interview Connect

Understanding this linear time complexity shows you can analyze stack-based algorithms well, a useful skill for many coding challenges.

Self-Check

"What if the input expression included multi-character operands instead of single characters? How would the time complexity change?"

Practice

1. What is the main purpose of using a stack in Infix to Postfix Conversion?

easy

A. To temporarily hold operators and manage their precedence

B. To store the final postfix expression

C. To reverse the infix expression

D. To count the number of operands

Solution

1. Step 1: Understand the role of stack in conversion

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

2. Step 2: Differentiate stack usage from other tasks

   The postfix expression is built separately, and the stack is not used to store operands or reverse the expression.

3. Final Answer:

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

4. Quick Check:

   Stack holds operators [OK]

Hint: Stack holds operators to manage precedence during conversion [OK]

Common Mistakes:

• Thinking stack stores operands
• Confusing stack role with reversing expression
• Assuming stack counts operands

2. Which of the following is the correct way to check if the stack is empty in JavaScript during infix to postfix conversion?

easy

A. if (stack === undefined)

B. if (stack == null)

C. if (stack.length === 0)

D. if (stack.isEmpty())

Solution

1. Step 1: Recall JavaScript array properties

   Stacks are usually implemented using arrays; checking if stack.length === 0 confirms if it is empty.

2. Step 2: Evaluate other options

   stack == null or stack === undefined check if the variable exists, not if it's empty. stack.isEmpty() is not a built-in method in JavaScript arrays.

3. Final Answer:

   if (stack.length === 0) -> Option C

4. Quick Check:

   Array length zero means empty stack [OK]

Hint: Use stack.length === 0 to check empty stack in JavaScript [OK]

Common Mistakes:

• Using undefined or null checks instead of length
• Calling non-existent isEmpty() method
• Confusing variable existence with emptiness

3. What is the postfix expression output for the infix expression A + B * C using stack-based conversion?

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 step-by-step

   Postfix for B * C is BC*. Then add A and + operator: ABC*+.

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. Identify the error in this JavaScript snippet for infix to postfix conversion:

if (stack.length = 0) { output += token; }

medium

A. Output variable is not declared

B. Missing semicolon after if condition

C. Stack length property is incorrect

D. Assignment operator used instead of comparison

Solution

1. Step 1: Check the if condition syntax

   The condition uses single equals = which assigns 0 to stack.length instead of comparing it.

2. Step 2: Understand correct comparison operator

   Use === or == to compare values, not =.

3. Final Answer:

   Assignment operator used instead of comparison -> Option D

4. Quick Check:

   Use === for comparison, not = [OK]

Hint: Use === for comparison, not = assignment [OK]

Common Mistakes:

• Using = instead of === in conditions
• Ignoring syntax errors in if statements
• Assuming missing semicolon causes error here

5. Given the infix expression (A + B) * (C - D) / E, what is the correct postfix expression after conversion using stack?

hard

A. AB+CD-E*/

B. AB+CD-*E/

C. AB+CD-E/*

D. AB+CD*E/-

Solution

1. Step 1: Convert each parenthesis group separately

   (A + B) becomes AB+, and (C - D) becomes CD-.

2. Step 2: Apply multiplication and division in order

   Multiply results: AB+ CD- *, then divide by E: AB+CD-*E/.

3. Final Answer:

   AB+CD-*E/ -> Option B

4. Quick Check:

   Parentheses handled first, then * and / left to right [OK]

Hint: Handle parentheses first, then * and / left to right [OK]

Common Mistakes:

• Ignoring parentheses order
• Mixing operator precedence
• Placing division before multiplication incorrectly