Infix to Postfix Conversion Using Stack in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to push an operator onto the stack.

DSA Rust

stack.[1](ch);

Drag options to blanks, or click blank then click option'

Apeek

Bpop

Cpush

Dremove

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop() instead of push() removes elements.

Using peek() only looks at the top without adding.

We use push() to add an element to the top of the stack.

2fill in blank

medium

Complete the code to check if the stack is empty.

DSA Rust

if stack.[1]() {

Drag options to blanks, or click blank then click option'

Alen

Bis_empty

Cis_full

Dcontains

Attempts:

3 left

💡 Hint

Common Mistakes

Using len() without comparing to zero.

Using is_full() which checks if stack is full.

is_empty() returns true if the stack has no elements.

3fill in blank

hard

Fix the error in the code to pop the top element from the stack.

DSA Rust

let top = stack.[1]().unwrap();

Drag options to blanks, or click blank then click option'

Apop

Bpeek

Cpush

Dremove

Attempts:

3 left

💡 Hint

Common Mistakes

Using peek() which does not remove the element.

Using push() which adds elements instead.

pop() removes and returns the top element; unwrap() gets the value.

4fill in blank

hard

Fill both blanks to compare operator precedence correctly.

DSA Rust

while !stack.[1]() && precedence(stack.[2]().unwrap()) >= precedence(ch) {

Drag options to blanks, or click blank then click option'

Ais_empty

Bpop

Cpeek

Dpush

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop() inside while condition removes elements prematurely.

Using push() instead of peek() to check top element.

Check stack is not empty with is_empty(), then peek top element for precedence.

5fill in blank

hard

Fill all three blanks to build the postfix string correctly.

DSA Rust

postfix.push(stack.[1]().unwrap());
stack.[2](ch);
while !stack.[3]() {

Drag options to blanks, or click blank then click option'

Apop

Bpush

Cis_empty

Dpeek

Attempts:

3 left

💡 Hint

Common Mistakes

Using peek() instead of pop() when adding to postfix.

Checking is_empty() before pushing new operator.

pop() removes top to add to postfix, push() adds current operator, is_empty() checks stack status.

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