Infix to Postfix Conversion Using Stack in DSA C - 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 stack to output until '('

No↓

Is symbol operator?

Yes→Pop stack while top has higher/equal precedence

↓

Push current operator

↓

Repeat until infix ends

↓

Pop all remaining stack to output

↓

Done: Postfix expression ready

This flow shows how each symbol in the infix expression is processed using a stack to produce postfix notation.

Execution Sample

DSA C

infix = "A+B*C"
stack = []
output = ""
for symbol in infix:
    if symbol.isalnum():
        output += symbol
    else:
        # handle operator with stack
        pass
# pop remaining stack to output

Converts infix expression 'A+B*C' to postfix by scanning each symbol and using a stack for operators.

Execution Table

Step	Operation	Symbol	Stack Before	Stack After	Output After	Visual State
1	Read symbol	A	[]	[]	A	Output: A | Stack: []
2	Read symbol	+	[]	+	A	Output: A | Stack: [+]
3	Read symbol	B	+	+	AB	Output: AB | Stack: [+]
4	Read symbol	*	+	+, *	AB	Output: AB | Stack: [+ *]
5	Read symbol	C	+, *	+, *	ABC	Output: ABC | Stack: [+ *]
6	End of input	N/A	+, *	+	ABC*	Output: ABC* | Stack: [+] (Pop '*')
7	Pop remaining	N/A	+		ABC*+	Output: ABC*+ | Stack: [] (Pop '+')
8	Done	N/A			ABC*+	Final Postfix: ABC*+

💡 All symbols processed and 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	After Step 7	Final
stack	[]	[]	[+]	[+]	[+, *]	[+, *]	[+]	[]	[]
output	""	A	A	AB	AB	ABC	ABC*	ABC*+	ABC*+
symbol	N/A	A	+	B	*	C	N/A	N/A	N/A

Key Moments - 3 Insights

Why do we pop operators from the stack when the incoming operator has lower or equal precedence?

Because operators with higher or equal precedence must appear before the incoming operator in postfix. See step 6 where '*' is popped before '+' is pushed.

What happens when we encounter a closing parenthesis ')'?

We pop operators from the stack to output until we find the matching '('. This clears the operators inside the parentheses. (Not shown in this example but part of the flow.)

Why do operands go directly to output without using the stack?

Operands appear in the same order in postfix as in infix, so we add them directly to output. See steps 1, 3, and 5 where operands A, B, C are added immediately.

Visual Quiz - 3 Questions

Look at the execution table, what is the stack after reading symbol '*' (step 4)?

A ['+']

B ['*']

C ['+', '*']

D []

💡 Hint

Check the 'Stack After' column in row for step 4.

At which step does the output first contain 'ABC'?

A Step 5

B Step 6

C Step 3

D Step 7

💡 Hint

Look at the 'Output After' column and find when 'ABC' appears.

If the input was 'A*B+C', when would '+' be pushed onto the stack?

A Immediately after reading '+'

B After popping '*' from stack

C Before reading '*'

D At the end of input

💡 Hint

Recall operator precedence and when stack operators are popped before pushing new ones.

Concept Snapshot

Infix to Postfix Conversion Using Stack:
- Scan infix left to right
- Operands go directly to output
- '(' pushed to stack
- ')' pops until '('
- Operators pop stack while top has higher/equal precedence
- Push current operator
- At end, pop all stack to output

Full Transcript

This concept converts an infix expression like 'A+B*C' into postfix notation 'ABC*+'. It uses a stack to hold operators and outputs operands immediately. Operators are pushed or popped based on precedence rules. Parentheses control grouping by pushing '(' and popping until '(' on ')'. The process ends by emptying the stack to output, producing the 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 precedence

B. To store operands permanently

C. To reverse the order of operands

D. To count the number of operators

Solution

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

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

2. Step 2: Differentiate stack usage from other data structures

   Operands are directly added to the postfix output, while operators are pushed to the stack to ensure correct order in the postfix expression.

3. Final Answer:

   To temporarily hold operators and manage precedence -> Option A

4. Quick Check:

   Stack holds operators = A [OK]

Hint: Stack holds operators to manage order and precedence [OK]

Common Mistakes:

• Thinking stack stores operands
• Assuming stack reverses operands
• Believing stack counts operators

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

easy

A. if (stack == NULL)

B. if (top == -1)

C. if (stack->top == 0)

D. if (stack.size == 0)

Solution

1. Step 1: Understand stack implementation in C

   In array-based stack, 'top' is the index of the last pushed element. When empty, top is -1.

2. Step 2: Identify correct empty condition

   Checking if top == -1 correctly indicates the stack is empty.

3. Final Answer:

   if (top == -1) -> Option B

4. Quick Check:

   top == -1 means empty stack [OK]

Hint: Empty stack means top index is -1 [OK]

Common Mistakes:

• Checking stack pointer NULL instead of top
• Using top == 0 which means one element present
• Using struct size field incorrectly

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

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

   Postfix places operands first, then operator: B C * then A + gives ABC*+

3. Final Answer:

   ABC*+ -> Option A

4. Quick Check:

   Infix A + (B * C) = 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. Given the following code snippet for infix to postfix conversion, what is the error?

if (isOperator(ch)) {
  while (!isEmpty(stack) && precedence(peek(stack)) >= precedence(ch)) {
    postfix[j++] = pop(stack);
  }
  push(stack, ch);
}

medium

A. The pop operation should not be inside the while loop

B. The push operation should be outside the if block

C. The condition should use > instead of >= for precedence comparison

D. The isEmpty function is called incorrectly

Solution

1. Step 1: Understand operator precedence handling

   For correct associativity, when precedence is equal and operator is right-associative (like '^'), we should not pop.

2. Step 2: Identify correct comparison operator

   Using >= causes popping on equal precedence, which breaks right-associativity. Using > fixes this.

3. Final Answer:

   The condition should use > instead of >= for precedence comparison -> Option C

4. Quick Check:

   Use > to handle associativity correctly [OK]

Hint: Use >, not >=, to handle right-associative operators [OK]

Common Mistakes:

• Using >= causes wrong operator order
• Pushing outside operator check
• Misusing isEmpty function

5. Convert the infix expression (A + B) * (C - D) / E to postfix notation using stack. What is the correct postfix expression?

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 to postfix

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

2. Step 2: Combine with multiplication and division

   Expression becomes (AB+) * (CD-) / E. Multiplication and division are left associative, so postfix is AB+ CD- * E /.

3. Final Answer:

   AB+CD-*E/ -> Option D

4. Quick Check:

   Postfix of (A+B)*(C-D)/E = AB+CD-*E/ [OK]

Hint: Convert parentheses first, then apply operators left to right [OK]

Common Mistakes:

• Ignoring parentheses grouping
• Mixing operator order
• Misplacing division operator