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

Concept Flow - Infix to Postfix Conversion Using Stack

Read next token from infix expression

↓

Is token operand?

Yes→Add operand to postfix output

No↓

Is token '(' ?

Yes→Push '(' onto stack

No↓

Is token ')' ?

Yes→Pop and add to postfix until '(' found

No↓

While stack top has higher or equal precedence

↓

Pop from stack to postfix

↓

Push current operator onto stack

↓

Repeat until all tokens processed

↓

Pop remaining stack to postfix

↓

End

This flow shows how each token from the infix expression is processed: operands go directly to output, '(' is pushed, ')' triggers popping until '(', and operators are pushed after popping higher precedence operators.

Execution Sample

DSA Rust

infix = "A+B*C"
stack = []
postfix = ""
for token in infix:
  if operand: postfix += token
  else if token == '(': push
  else if token == ')': pop until '('
  else: pop higher prec then push
pop all stack to postfix

Converts infix expression A+B*C to postfix by scanning tokens, using stack for operators, and building postfix string.

Execution Table

Step	Operation	Token	Stack Before	Stack After	Postfix Output	Visual State
1	Read token	A	[]	[]	A	Postfix: A | Stack: []
2	Read token	+	[]	+	A	Postfix: A | Stack: [+]
3	Read token	B	+	+	AB	Postfix: AB | Stack: [+]
4	Read token	*	+	+, *	AB	Postfix: AB | Stack: [+ *]
5	Read token	C	+, *	+, *	ABC	Postfix: ABC | Stack: [+ *]
6	End of input	None	+, *	+	ABC*	Pop * to postfix
7	End of input	None	+		ABC*+	Pop + to postfix
8	Done	None			ABC*+	Final postfix expression

💡 All tokens processed and stack emptied to postfix output

Variable Tracker

Variable	Start	After 1	After 2	After 3	After 4	After 5	After 6	After 7	Final
token	None	A	+	B	*	C	None	None	None
stack	[]	[]	+	+	+, *	+, *	+
postfix	""	A	A	AB	AB	ABC	ABC*	ABC*+	ABC*+

Key Moments - 3 Insights

Why do we push '(' onto the stack instead of adding it to postfix?

Because '(' is not an operand or operator to output; it marks a boundary for operators inside parentheses. See step 3 in execution_table where '(' would be pushed to stack to wait for matching ')'.

Why do we pop operators from the stack when we see a new operator with lower precedence?

To maintain correct order in postfix, operators with higher or equal precedence on stack must be output before pushing the new operator. See step 6 where '*' is popped before '+' to postfix.

What happens if the stack is empty when we try to pop for a ')'?

This means mismatched parentheses in expression. The algorithm expects '(' to be found before popping stops. This is a syntax error not shown in this example.

Visual Quiz - 3 Questions

Look at the execution_table at step 4, what is the stack after processing token '*'?

A ['+']

B ['*']

C ['+', '*']

D []

💡 Hint

Check the 'Stack After' column at step 4 in execution_table

At which step does the postfix output first include the operator '*'?

A Step 5

B Step 6

C Step 7

D Step 8

💡 Hint

Look at the 'Postfix Output' column and see when '*' appears

If the input expression was 'A*(B+C)', what would happen when ')' is read?

A Pop operators until '(' is found and discard '('

B Push ')' onto stack

C Add ')' to postfix output

D Ignore ')' and continue

💡 Hint

Refer to concept_flow step for handling ')' token

Concept Snapshot

Infix to Postfix Conversion Using Stack:
- Scan tokens left to right
- Operands go directly to output
- '(' pushed to stack
- ')' triggers pop until '('
- Operators pop higher/equal precedence then push
- At end, pop all stack to output

Full Transcript

This visual trace shows how to convert an infix expression like A+B*C into postfix using a stack. Each token is read: operands are added to output immediately; '(' is pushed to stack; ')' causes popping until '('; operators pop higher or equal precedence operators before pushing themselves. The stack and postfix output change step by step until all tokens are processed and stack is empty. This ensures correct operator order in postfix form.

Practice

1. What is the main purpose of using a stack in converting an infix expression to postfix?

easy

A. To temporarily hold operators and manage their 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 operator handling

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

2. Step 2: Clarify why operands are not stored in stack

   Operands are directly added to the postfix output; only operators need temporary storage to ensure correct order.

3. Final Answer:

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

4. Quick Check:

   Stack holds operators = A [OK]

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

Common Mistakes:

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

2. Which of the following Rust code snippets correctly initializes an empty stack for characters used in infix to postfix conversion?

easy

A. let stack: Vec<char> = new Vec();

B. let mut stack: Vec<char> = Vec::new();

C. let stack = Vec<char>();

D. let mut stack = Vec::empty();

Solution

1. Step 1: Identify correct Rust syntax for empty vector

   In Rust, an empty vector is created with Vec::new(), and mutability requires 'mut'.

2. Step 2: Check each option for syntax correctness

   let mut stack: Vec<char> = Vec::new(); uses correct syntax: mutable vector of chars with Vec::new(). Others have syntax errors or invalid methods.

3. Final Answer:

   let mut stack: Vec<char> = Vec::new(); -> Option B

4. Quick Check:

   Vec::new() + mut = C [OK]

Hint: Use 'let mut stack: Vec<char> = Vec::new();' to create empty stack [OK]

Common Mistakes:

• Omitting 'mut' for mutable stack
• Using invalid Vec constructors
• Confusing Vec::new() with Vec::empty()

3. Given the infix expression a + b * c, what is the correct postfix expression after conversion using a stack?

medium

A. a b + c *

B. a * b + c

C. a b c * +

D. a + b c *

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 respecting precedence

   Postfix becomes 'a b c * +' where 'b c *' is first, then 'a' and '+' applied last.

3. Final Answer:

   a b c * + -> Option C

4. Quick Check:

   Multiplication before addition = a b c * + [OK]

Hint: Multiply before add: postfix is a b c * + [OK]

Common Mistakes:

• Ignoring operator precedence
• Placing '+' before '*' in postfix
• Mixing operand order

4. Consider this Rust snippet for infix to postfix conversion stack push operation: stack.push(token); If the stack is not declared mutable, what error will occur?

medium

A. No error, code runs fine

B. Stack overflow error

C. Syntax error: missing semicolon

D. Cannot borrow immutable variable `stack` as mutable

Solution

1. Step 1: Understand Rust mutability rules

   In Rust, to modify a variable like pushing to a vector, it must be declared mutable with 'mut'.

2. Step 2: Identify error from pushing to immutable stack

   Trying to push to an immutable vector causes a compile-time error: cannot borrow immutable variable as mutable.

3. Final Answer:

   Cannot borrow immutable variable `stack` as mutable -> Option D

4. Quick Check:

   Push needs mutable stack = D [OK]

Hint: Declare stack mutable to push: 'let mut stack' [OK]

Common Mistakes:

• Ignoring mut keyword
• Assuming runtime error instead of compile error
• Confusing syntax errors with mutability errors

5. Given the infix expression (a + b) * (c - d) / e, what is the correct postfix expression after conversion using a stack?

hard

A. a b + c d - * e /

B. a b c d - + * e /

C. a b + c d - e / *

D. a b + * c d - 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 stepwise to postfix

   Postfix of (a + b) is 'a b +', (c - d) is 'c d -', then multiply '*', then divide by 'e' with '/'.

3. Final Answer:

   a b + c d - * e / -> Option A

4. Quick Check:

   Parentheses respected: a b + c d - * e / [OK]

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

Common Mistakes:

• Ignoring parentheses order
• Misplacing operators in postfix
• Mixing operand order