Infix to Postfix Conversion Using Stack in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to push an operator onto the stack.

DSA C++

stack.push([1]);

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

Ach

Bpop

Ctop

Dempty

Attempts:

3 left

💡 Hint

Common Mistakes

Using stack.pop() instead of push.

Using stack.top() which only returns the top element.

We push the current operator character 'ch' onto the stack.

2fill in blank

medium

Complete the code to check if the stack is empty before accessing the top element.

DSA C++

if (!stack.[1]()) {

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

Aempty

Bpush

Ctop

Dpop

Attempts:

3 left

💡 Hint

Common Mistakes

Using stack.pop() which removes elements.

Using stack.top() which returns the top element but does not check emptiness.

We use stack.empty() to check if the stack is empty before accessing top.

3fill in blank

hard

Fix the error in the precedence comparison to correctly compare operator precedence.

DSA C++

while (!stack.empty() && precedence(stack.top()) [1] precedence(ch)) {

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

A<

B<=

C>=

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' or '<=' which causes incorrect order.

Using '==' which only checks equality.

We pop from the stack while the top operator has precedence greater than or equal to the current operator.

4fill in blank

hard

Fill both blanks to correctly append operands and pop operators to the postfix string.

DSA C++

if (isalnum(ch)) {
    postfix += [1];
} else {
    postfix += stack.[2]();
    stack.pop();
}

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

Ach

Btop

Cpop

Dempty

Attempts:

3 left

💡 Hint

Common Mistakes

Using stack.pop() to get the operator which returns void.

Appending stack.empty() which is a boolean.

Operands are appended directly using 'ch'. Operators are appended from the top of the stack using stack.top() before popping.

5fill in blank

hard

Fill all three blanks to correctly define the precedence function for operators.

DSA C++

int precedence(char op) {
    if (op == '+' || op == [1]) return 1;
    if (op == '*' || op == [2]) return 2;
    return [3];
}

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

A-

B/

C0

D%

Attempts:

3 left

💡 Hint

Common Mistakes

Using '%' instead of '-' or '/' in precedence.

Returning 1 or 2 for unknown operators.

The function returns 1 for '+' and '-', 2 for '*' and '/', and 0 for others.

Practice

1. What is the main purpose of using a stack in Infix to Postfix Conversion?

easy

A. To count the number of operators

B. To temporarily hold operators and manage their precedence

C. To reverse the order of the expression

D. To store operands permanently

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: Identify what is stored in the stack

   Operands are directly added to the output, while operators are pushed to the stack to manage order and precedence.

3. Final Answer:

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

4. Quick Check:

   Stack holds operators = D [OK]

Hint: Stack holds operators to manage precedence [OK]

Common Mistakes:

• Thinking stack stores operands
• Believing stack reverses the entire expression
• Confusing stack role with counting operators

2. Which of the following is the correct syntax to push an operator onto a stack named st in C++?

easy

A. st.push(operator);

B. push(st, operator);

C. st.add(operator);

D. push.operator(st);

Solution

1. Step 1: Recall C++ stack push syntax

   In C++, the standard stack uses the member function push() to add elements.

2. Step 2: Match correct syntax

   The correct syntax is st.push(operator); where st is the stack object.

3. Final Answer:

   st.push(operator); -> Option A

4. Quick Check:

   Stack push method = st.push() [OK]

Hint: Use st.push(element) to add to stack [OK]

Common Mistakes:

• Using non-member functions like push(st, element)
• Using incorrect method names like add()
• Incorrect dot notation like push.operator()

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

medium

A. A+BC*

B. AB+C*

C. ABC*+

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 +, resulting in ABC*+.

3. Final Answer:

   ABC*+ -> Option C

4. Quick Check:

   Postfix of A + B * C = ABC*+ [OK]

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

Common Mistakes:

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

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

std::stack<char> st;
std::string infix = "A+B*C";
std::string postfix = "";
for (char ch : infix) {
  if (isalnum(ch))
    postfix += ch;
  else {
    while (!st.empty() && st.top() >= ch) {
      postfix += st.top();
      st.pop();
    }
    st.push(ch);
  }
}
while (!st.empty()) {
  postfix += st.top();
  st.pop();
}

medium

A. Operator precedence comparison is incorrect

B. Operands are not added to postfix

C. Stack is not used for operators

D. Postfix string is not initialized

Solution

1. Step 1: Analyze operator precedence check

   The code compares operators using >= directly on characters, which does not correctly reflect operator precedence.

2. Step 2: Identify correct precedence handling

   Operator precedence should be checked using a function that assigns precedence values, not character comparison.

3. Final Answer:

   Operator precedence comparison is incorrect -> Option A

4. Quick Check:

   Operator precedence needs function, not char compare [OK]

Hint: Use precedence function, not char comparison [OK]

Common Mistakes:

• Comparing operators as characters
• Ignoring precedence rules
• Not popping stack correctly

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: Handle parentheses and operator precedence

   Parentheses force A+B and C-D to be evaluated first, then multiply results, then divide by E.

2. Step 2: Convert step-by-step to postfix

   A B + (for A+B), C D - (for C-D), then multiply: AB+CD-*, finally divide by E: 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: Process parentheses first, then operators by precedence [OK]

Common Mistakes:

• Ignoring parentheses order
• Mixing operator order
• Placing division before multiplication