Infix to Postfix Conversion Using Stack in DSA Typescript - 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: string): string {
  const stack: string[] = [];
  let result = '';
  for (const 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(stack[stack.length - 1]) >= precedence(char)) {
        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

• Primary operation: A single loop over each character in the input expression.
• How many times: Exactly once for each character (n times if expression length is n).
• Inside this loop, there are inner 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 steps plus some stack pops
100	About 100 main steps plus stack operations
1000	About 1000 main steps plus stack operations

Pattern observation: The total steps increase roughly linearly with 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 complexity must be quadratic O(n²)."

[OK] Correct: The inner loops pop operators only once per operator, so total work is proportional to n, not n squared.

Interview Connect

Understanding this linear time complexity helps you explain how stacks efficiently handle operator precedence in expressions, a common skill in coding interviews.

Self-Check

"What if the input expression included multi-character operands separated by spaces? 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 store operands permanently

C. To temporarily hold operators and manage their precedence

D. To count the number of operators in the expression

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 ensure correct precedence and associativity.

2. Step 2: Identify what is stored in the stack

   Operands are directly added to the output, while operators are pushed to the stack to manage order based on precedence.

3. Final Answer:

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

4. Quick Check:

   Stack holds operators for precedence [OK]

Hint: Stack holds operators, not operands, for precedence [OK]

Common Mistakes:

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

2. Which of the following is the correct way to check if the stack is empty in TypeScript when implementing infix to postfix conversion?

easy

A. if (stack.length === 0) {}

B. if (stack.isEmpty()) {}

C. if (stack == null) {}

D. if (stack.size() === 0) {}

Solution

1. Step 1: Understand stack implementation in TypeScript

   Stacks are often implemented as arrays in TypeScript, so checking if empty means checking if length is zero.

2. Step 2: Identify correct syntax for empty check

   Using stack.length === 0 is the standard way to check if an array-based stack is empty.

3. Final Answer:

   if (stack.length === 0) {} -> Option A

4. Quick Check:

   Array length zero means empty stack [OK]

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

Common Mistakes:

• Using non-existent isEmpty() method
• Checking stack == null instead of length
• Using size() which is not standard for arrays

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

medium

A. A+BC*

B. ABC*+

C. AB+C*

D. *ABC+

Solution

1. Step 1: Identify operator precedence in infix

   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 * then A +, resulting in ABC*+.

3. Final Answer:

   ABC*+ -> Option B

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
• Confusing order of operands

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

function isOperator(c: string): boolean {
  return c === '+' || c === '-' || c === '*' || c === '/';
}

function precedence(c: string): number {
  if (c === '+' || c === '-') return 1;
  if (c === '*' || c === '/') return 2;
  return 0;
}

function infixToPostfix(expression: string): string {
  const stack: string[] = [];
  let result = '';
  for (const char of expression) {
    if (char.match(/[a-zA-Z0-9]/)) {
      result += char;
    } else if (isOperator(char)) {
      while (stack.length && precedence(stack[stack.length - 1]) >= precedence(char)) {
        result += stack.pop();
      }
      stack.push(char);
    }
  }
  while (stack.length) {
    result += stack.pop();
  }
  return result;
}

medium

A. The code does not handle parentheses in the expression

B. The precedence function returns wrong values

C. The stack is not used to store operators

D. Operands are not added to the result

Solution

1. Step 1: Analyze code handling of parentheses

   The code does not check for '(' or ')' characters, so it cannot handle expressions with parentheses correctly.

2. Step 2: Confirm other parts are correct

   Precedence and operator checks are correct; operands are added properly; stack stores operators as expected.

3. Final Answer:

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

4. Quick Check:

   Missing parentheses handling causes error [OK]

Hint: Check if parentheses are handled explicitly [OK]

Common Mistakes:

• Assuming parentheses handled automatically
• Confusing precedence values
• Ignoring stack usage for operators

5. Convert the infix expression (A + B) * (C - D) / E to postfix notation using stack-based conversion.

hard

A. AB+CD-E*/

B. A+B*CD-E/

C. AB+CD-*E

D. AB+CD-*E/

Solution

1. Step 1: Handle parentheses and operator precedence

   Parentheses force evaluation order: (A + B) and (C - D) first, then multiply, then divide by E.

2. Step 2: Convert each part to postfix and combine

   (A + B) -> AB+; (C - D) -> CD-; multiply results: AB+CD-*; then divide by E: AB+CD-*E/.

3. Final Answer:

   AB+CD-*E/ -> Option D

4. Quick Check:

   Parentheses respected, operators after operands [OK]

Hint: Convert inside parentheses first, then apply operators [OK]

Common Mistakes:

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