Why Queue Using Two Stacks in DSA C?

The Big Idea

What if you could turn a stack into a queue using just one extra stack? Let's see how!

The Scenario

Imagine you have a line of people waiting to buy tickets, but you only have two stacks of trays to hold their tickets. You want to serve them in the order they arrived, but the trays only let you add or remove tickets from the top. How do you keep the order correct?

The Problem

Trying to serve people in the exact order using just stacks is tricky because stacks work like a pile where you can only take the top item. If you try to serve the first person who came, you might have to move many tickets around, which is slow and confusing.

The Solution

Using two stacks together cleverly solves this problem. One stack holds new tickets, and the other stack helps reverse the order when serving. This way, you can add tickets quickly and serve them in the right order without moving everything every time.

Before vs After

✗ Before

void enqueue(int x) {
    stack[top++] = x; // just push
}

int dequeue() {
    // no direct way to get first element
}

✓ After

void enqueue(int x) {
    push(stack1, x);
}

int dequeue() {
    if (isEmpty(stack2)) {
        while (!isEmpty(stack1)) {
            push(stack2, pop(stack1));
        }
    }
    return pop(stack2);
}

What It Enables

This method lets you build a queue using only stacks, keeping the order correct and operations efficient.

Real Life Example

Think of a cafeteria where food trays are stacked in two piles. One pile is for new trays, and when serving, trays are moved to the other pile to serve in the order they arrived.

Key Takeaways

Stacks alone can't serve in first-in-first-out order easily.

Two stacks together can simulate a queue efficiently.

This technique helps when only stack operations are allowed but queue behavior is needed.

Practice

1. What is the main idea behind implementing a Queue Using Two Stacks?

easy

A. Use two stacks to store elements in sorted order

B. Use two stacks to reverse the order twice to maintain queue order

C. Use one stack for enqueue and another for dequeue operations

D. Use two stacks to perform only enqueue operations

Solution

1. Step 1: Understand queue order

   A queue follows FIFO (First In First Out) order, meaning the first element added is the first to be removed.

2. Step 2: Use two stacks to reverse order twice

   One stack reverses the order when pushing elements, and the second stack reverses it again when popping, restoring FIFO order.

3. Final Answer:

   Use two stacks to reverse the order twice to maintain queue order -> Option B

4. Quick Check:

   Two reversals keep FIFO order [OK]

Hint: Two stacks reverse order twice to simulate queue [OK]

Common Mistakes:

• Thinking both stacks store elements in sorted order
• Using stacks only for enqueue or dequeue, not both
• Confusing stack order with queue order

2. Which of the following is the correct syntax to push an element x onto a stack named stack1 in C?

easy

A. push(stack1, x);

B. stack1.push(x); // C syntax

C. stack1.push(x);

D. push(x, stack1);

Solution

1. Step 1: Recall stack push function in C

   In C, stacks are usually implemented with functions like push(stack, element).

2. Step 2: Identify correct argument order

   The stack is passed first, then the element to push, so push(stack1, x) is correct.

3. Final Answer:

   push(stack1, x); -> Option A

4. Quick Check:

   Function call with stack first, element second [OK]

Hint: In C, push(stack, element) is standard [OK]

Common Mistakes:

• Using object.method() syntax which is not valid in C
• Swapping argument order in push function
• Assuming push is a built-in method

3. Consider the following code snippet implementing a queue using two stacks s1 and s2. What will be the output after the operations?

push(s1, 10);
push(s1, 20);
pop(s2);
if (isEmpty(s2)) {
  while (!isEmpty(s1)) {
    push(s2, top(s1));
pop(s1);
  }
}
pop(s2);
printQueue(s1, s2);

medium

A. Queue elements: 10 -> null

B. Queue elements: null

C. Queue elements: 20 -> null

D. Queue elements: 10 -> 20 -> null

Solution

1. Step 1: Push 10 and 20 onto s1

   s1 = [10 (bottom), 20 (top)], s2 = []

2. Step 2: Pop from s2 (empty), then transfer all from s1 to s2

   Transfer reverses order: s2 = [20 (bottom), 10 (top)], s1 = []

3. Step 3: Pop from s2 removes 10 (top)

   s2 = [20]

4. Step 4: printQueue shows elements in queue order

   Queue order is s2 top to bottom then s1 bottom to top, so only 20 remains.

5. Final Answer:

   Queue elements: 20 -> null -> Option C

6. Quick Check:

   First popped element was 10, next is 20 [OK]

Hint: Transfer reverses stack order to simulate queue dequeue [OK]

Common Mistakes:

• Not transferring elements from s1 to s2 when s2 is empty
• Confusing top and bottom elements during transfer
• Assuming pop removes from s1 directly

4. The following code attempts to implement dequeue operation in a queue using two stacks s1 and s2. What is the bug?

int dequeue() {
  if (isEmpty(s2)) {
    while (!isEmpty(s1)) {
      push(s2, top(s1));
      pop(s1);
    }
  }
  return pop(s2);
}

medium

A. Missing check if both stacks are empty before dequeue

B. pop(s2) returns void, should use top(s2) before pop

C. Should push from s2 to s1 instead of s1 to s2

D. pop(s1) should be called after push(s2, top(s1))

Solution

1. Step 1: Check for empty queue condition

   The code does not check if both s1 and s2 are empty before dequeue, which can cause errors.

2. Step 2: Identify missing empty check

   Without this check, pop(s2) may be called on empty stack causing runtime error.

3. Final Answer:

   Missing check if both stacks are empty before dequeue -> Option A

4. Quick Check:

   Empty queue must be handled before popping [OK]

Hint: Always check if queue is empty before dequeue [OK]

Common Mistakes:

• Assuming pop returns the popped value in C without checking
• Ignoring empty queue condition
• Confusing push direction between stacks

5. You want to implement a queue using two stacks s1 and s2. You perform the following operations:

enqueue(1);
enqueue(2);
dequeue();
enqueue(3);
dequeue();
dequeue();

What will be the output sequence of the dequeued elements?

hard

A. [3, 2, 1]

B. [1, 3, 2]

C. [2, 1, 3]

D. [1, 2, 3]

Solution

1. Step 1: Enqueue 1 and 2

   s1 = [1 (bottom), 2 (top)], s2 = []

2. Step 2: Dequeue first element

   s2 empty, transfer all from s1 to s2: s2 = [2 (bottom), 1 (top)], s1 = []
   Pop from s2 removes 1

3. Step 3: Enqueue 3

   Push 3 to s1: s1 = [3], s2 = [2]

4. Step 4: Dequeue second element

   s2 not empty, pop from s2 removes 2

5. Step 5: Dequeue third element

   s2 empty, transfer s1 to s2: s2 = [3], s1 = []
   Pop from s2 removes 3

6. Final Answer:

   [1, 2, 3] -> Option D

7. Quick Check:

   FIFO order maintained by two stacks [OK]

Hint: Dequeue triggers transfer only if second stack empty [OK]

Common Mistakes:

• Not transferring elements when s2 is empty
• Assuming enqueue adds to s2 instead of s1
• Mixing order of popped elements