Infix to Postfix Conversion Using Stack in DSA Javascript - Execution Trace

Concept Flow - Infix to Postfix Conversion Using Stack

Start with empty stack and output

↓

Read next symbol from infix

↓

Is symbol operand?

Yes→Add to output

No↓

Is symbol '(' ?

Yes→Push to stack

No↓

Is symbol ')' ?

Yes→Pop from stack to output until '('

No↓

Is symbol operator?

↓

While stack top has higher or equal precedence

↓

Pop stack top to output

↓

Push current operator to stack

↓

Repeat until infix ends

↓

Pop remaining stack to output

↓

End with postfix expression

This flow shows how each symbol in the infix expression is processed using a stack to build the postfix expression step-by-step.

Execution Sample

DSA Javascript

infix = 'A+B*C'
stack = []
output = ''
for symbol in infix:
  if symbol.isalnum():
    output += symbol
  elif symbol == '(': 
    stack.append(symbol)
  elif symbol == ')':
    while stack and stack[-1] != '(': 
      output += stack.pop()
    stack.pop()  # pop '('
  else:
    while stack and precedence(stack[-1]) >= precedence(symbol):
      output += stack.pop()
    stack.append(symbol)
while stack:
  output += stack.pop()
print(output)

Converts infix expression 'A+B*C' to postfix using stack operations.

Execution Table

Step	Operation	Current Symbol	Stack Before	Stack After	Output Before	Output After	Visual State
1	Read symbol	A	[]	[]	""	"A"	Output: A, Stack: []
2	Read symbol	+	[]	+	"A"	"A"	Output: A, Stack: [+]
3	Read symbol	B	+	+	"A"	"AB"	Output: AB, Stack: [+]
4	Read symbol	*	+	+, *	"AB"	"AB"	Output: AB, Stack: [+, *]
5	Read symbol	C	+, *	+, *	"AB"	"ABC"	Output: ABC, Stack: [+, *]
6	End of input		+, *		"ABC"	"ABC*+"	Pop * then + to output; Output: ABC*+, Stack: []

💡 All symbols processed; stack emptied to output; postfix expression complete.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6 (Final)
stack	[]	[]	+	+	+, *	+, *	[]
output	""	"A"	"A"	"AB"	"AB"	"ABC"	"ABC*+"
current_symbol	N/A	A	+	B	*	C	N/A

Key Moments - 3 Insights

Why do we push '*' on stack even though '+' is already there?

Because '*' has higher precedence than '+', so we keep '*' on stack above '+' (see Step 4 in execution_table). We only pop operators with higher or equal precedence.

Why do operands go directly to output?

Operands (like A, B, C) are part of the final postfix expression in order, so we add them immediately to output (see Steps 1, 3, 5).

What happens when input ends but stack is not empty?

We pop all remaining operators from stack to output to complete postfix (see Step 6). This ensures no operators are left behind.

Visual Quiz - 3 Questions

Look at the execution_table at Step 4. What is the stack after processing symbol '*'?

A [+, *]

B [*]

C [+]

D []

💡 Hint

Check the 'Stack After' column in Step 4 of execution_table.

At which step does the output first include the symbol 'B'?

A Step 1

B Step 3

C Step 5

D Step 6

💡 Hint

Look at the 'Output After' column for when 'B' appears in execution_table.

If the input was 'A*B+C' instead, what would happen to the stack after reading '+'?

A [+]

B [*, +]

C [*]

D []

💡 Hint

Recall that '+' has lower precedence than '*', so '*' is popped before pushing '+'.

Concept Snapshot

Infix to Postfix Conversion Using Stack:
- Read symbols left to right
- Operands go directly to output
- '(' pushed to stack
- ')' pops until '('
- Operators pop higher/equal precedence before push
- At end, pop all stack to output
- Result is postfix expression

Full Transcript

This visual trace shows how to convert an infix expression like 'A+B*C' to postfix using a stack. We read each symbol, pushing operators to stack and adding operands to output. Operators with higher precedence stay on stack until lower precedence operator arrives. Parentheses control grouping by pushing '(' and popping until ')'. At the end, all remaining operators are popped to output. The final output is the postfix expression. The execution table tracks stack and output changes step-by-step, helping beginners see how the stack manages operator precedence and order.

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