Infix to Postfix Conversion Using Stack in DSA Javascript - Build from Scratch

Infix to Postfix Conversion Using Stack

📖 Scenario: You are building a simple calculator that converts mathematical expressions written in the usual way (called infix notation) into a form easier for computers to calculate (called postfix notation).For example, the expression (A + B) * C in infix becomes AB+C* in postfix.

🎯 Goal: Build a program that converts an infix expression string into a postfix expression string using a stack.

📋 What You'll Learn

Create a string variable called infix with the exact value 'A+B*(C-D)'.

Create a stack using an array called stack to help with the conversion.

Write a function called precedence that returns the precedence of operators '+', '-', '*', and '/'.

Use a for loop to process each character in infix and convert it to postfix using the stack.

Print the final postfix expression string.

💡 Why This Matters

🌍 Real World

Converting infix to postfix is a key step in calculators and compilers to evaluate expressions efficiently.

💼 Career

Understanding stacks and expression parsing is important for software developers working on interpreters, compilers, and calculators.

1

Create the Infix Expression

Create a string variable called infix and set it to the exact value 'A+B*(C-D)'.

DSA Javascript

# Create a string variable called infix with value 'A+B*(C-D)'
# Your code here

Hint

Use const infix = 'A+B*(C-D)'; to create the infix expression string.

2

Create the Stack and Precedence Function

Create an empty array called stack to use as a stack. Then write a function called precedence that takes an operator character and returns 2 for '*' and '/', 1 for '+' and '-', and 0 for any other character.

DSA Javascript

const infix = 'A+B*(C-D)';
# Create an empty array called stack
# Write function precedence(op) to return operator precedence
# Your code here

Hint

Use an array stack = [] and a function with if statements to return precedence values.

3

Convert Infix to Postfix Using Stack

Create an empty string variable called postfix. Use a for loop with variable char to go through each character in infix. If char is a letter (A-Z), add it to postfix. If char is '(', push it to stack. If char is ')', pop from stack to postfix until '(' is found and remove '('. If char is an operator, pop from stack to postfix while the top of stack has higher or equal precedence, then push char to stack. After the loop, pop all remaining operators from stack to postfix.

DSA Javascript

const infix = 'A+B*(C-D)';
const stack = [];

function precedence(op) {
    if (op === '+' || op === '-') return 1;
    if (op === '*' || op === '/') return 2;
    return 0;
}

// Create postfix string and convert infix to postfix
// Your code here

Hint

Use a loop to check each character and apply stack rules for operators and parentheses.

4

Print the Postfix Expression

Write a console.log statement to print the postfix string.

DSA Javascript

const infix = 'A+B*(C-D)';
const stack = [];

function precedence(op) {
    if (op === '+' || op === '-') return 1;
    if (op === '*' || op === '/') return 2;
    return 0;
}

let postfix = '';
for (let i = 0; i = 'A' && char = precedence(char)) {
            postfix += stack.pop();
        }
        stack.push(char);
    }
}
while (stack.length) {
    postfix += stack.pop();
}

// Print the postfix expression
# Your code here

Hint

Use console.log(postfix); to show the final postfix expression.

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