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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to push an operator onto the stack.

DSA Java

stack.[1](ch);

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

Apeek

Bpop

Cpush

Dempty

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop instead of push removes an element instead of adding.

Using peek does not add anything to the stack.

The push method adds an element to the top of the stack.

2fill in blank

medium

Complete the code to check if the stack is empty.

DSA Java

if(stack.[1]()) {

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

Aempty

Bpop

Cpush

Dpeek

Attempts:

3 left

💡 Hint

Common Mistakes

Using peek() to check emptiness causes errors if stack is empty.

Using pop() removes elements, not checks emptiness.

The empty() method returns true if the stack has no elements.

3fill in blank

hard

Fix the error in the code to get the precedence of an operator.

DSA Java

int precedence = [1](op);

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

AgetPrecedence

Bprecedence

Cpriority

DgetPriority

Attempts:

3 left

💡 Hint

Common Mistakes

Using a variable name instead of a method call causes errors.

Using incorrect method names leads to compilation errors.

The method to get operator precedence is named getPrecedence in this code.

4fill in blank

hard

Fill both blanks to correctly pop from stack and append to result.

DSA Java

result += stack.[1]();
stack.[2]();

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

Apeek

Bpop

Cpush

Dempty

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop() twice removes two elements instead of one.

Using push() instead of pop() adds elements incorrectly.

pop() removes the top element from the stack and returns it. Using peek() followed by pop() is redundant here.

5fill in blank

hard

Fill all three blanks to correctly check operator precedence and pop from stack.

DSA Java

while(!stack.[1]() && getPrecedence(ch) <= getPrecedence(stack.[2]())) {
    result += stack.[3]();
}

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

Aempty

Bpeek

Cpop

Dpush

Attempts:

3 left

💡 Hint

Common Mistakes

Not checking empty() causes errors when stack is empty.

Using push() instead of pop() causes infinite loops.

empty() checks if stack is empty, peek() gets top element, pop() removes it.

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 infix to postfix

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

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

   Operands are output immediately, not stored in stack. Stack is not used for counting or reversing operands.

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 to manage precedence during conversion [OK]

Common Mistakes:

• Thinking stack stores operands
• Confusing stack with queue
• Believing stack reverses operands

2. Which of the following is the correct way to push an operator onto the stack in Java?

easy

A. stack.push(operator);

B. stack.insert(operator);

C. stack.add(operator);

D. stack.put(operator);

Solution

1. Step 1: Recall Java Stack method for adding elements

   Java's Stack class uses the push() method to add elements to the top of the stack.

2. Step 2: Identify incorrect methods

   Methods add(), insert(), and put() are not standard for Stack push operation in Java.

3. Final Answer:

   stack.push(operator); -> Option A

4. Quick Check:

   Use push() to add to stack [OK]

Hint: Use push() to add elements to Java stack [OK]

Common Mistakes:

• Using add() instead of push()
• Using non-existent methods like insert()
• Confusing stack with other collections

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

medium

A. AB+C*

B. ABC*+

C. A+BC*

D. *ABC+

Solution

1. Step 1: Apply operator precedence in infix to postfix

   Multiplication (*) has higher precedence than addition (+), so B*C is evaluated first.

2. Step 2: Convert expression step-by-step

   Postfix: operands B and C followed by *, then operand A followed by + gives "ABC*+".

3. Final Answer:

   ABC*+ -> Option B

4. Quick Check:

   Multiplication before addition = ABC*+ [OK]

Hint: Multiply before add: operands then operator [OK]

Common Mistakes:

• Ignoring operator precedence
• Placing operators before operands
• Confusing postfix order

4. Given the following code snippet for infix to postfix conversion, what is the error?

if (ch == '+' || ch == '-' || ch == '*' || ch == '/') {
    while (!stack.isEmpty() && precedence(stack.peek()) >= precedence(ch)) {
        postfix += stack.pop();
    }
    stack.push(ch);
}

medium

A. The precedence comparison should use > instead of >=

B. The stack should be checked for null before peek()

C. Operands are not handled in this snippet

D. The while loop condition is incorrect for operator associativity

Solution

1. Step 1: Understand operator associativity in infix to postfix

   For left-associative operators like '-', '+', '*', '/', pop operators with precedence >= current operator.

2. Step 2: Identify issue with while condition

   Using >= causes right-associative operators like '^' to be handled incorrectly; condition should consider associativity.

3. Final Answer:

   The while loop condition is incorrect for operator associativity -> Option D

4. Quick Check:

   Associativity affects while condition [OK]

Hint: Check operator associativity when comparing precedence [OK]

Common Mistakes:

• Ignoring operator associativity
• Not handling operands separately
• Assuming peek() can be null without check

5. Convert the infix expression (A + B) * (C - D) / E to postfix 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 respecting precedence

   Multiply results: "AB+CD-*" then divide by E: "AB+CD-*E/".

3. Final Answer:

   AB+CD-*E/ -> Option C

4. Quick Check:

   Parentheses handled first, then * and / [OK]

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

Common Mistakes:

• Ignoring parentheses order
• Mixing operator precedence
• Dropping operators or operands