Queue Using Two Stacks in DSA C - Execution Trace

Concept Flow - Queue Using Two Stacks

Enqueue Operation

↓

Push element onto Stack1

↓

Done

↓

Dequeue Operation

↓

Is Stack2 empty?

No→Pop element from Stack2

Yes↓

Transfer all elements from Stack1 to Stack2

↓

Pop element from Stack2

↓

Done

Enqueue pushes elements onto Stack1. Dequeue pops from Stack2; if Stack2 is empty, transfer all elements from Stack1 to Stack2 first.

Execution Sample

DSA C

push(Stack1, 1);
push(Stack1, 2);
if (empty(Stack2)) transfer from Stack1 to Stack2;
pop(Stack2);  // dequeues 1
pop(Stack2);  // dequeues 2

This code enqueues 1 and 2, then dequeues 1 (transferring first if needed) and then dequeues 2 using two stacks.

Execution Table

Step	Operation	Stack1 (top->bottom)	Stack2 (top->bottom)	Pointer Changes	Visual State
1	Enqueue 1	1	empty	Stack1 push 1	Stack1: [1] Stack2: []
2	Enqueue 2	2 -> 1	empty	Stack1 push 2	Stack1: [2, 1] Stack2: []
3	Dequeue - Stack2 empty, transfer all from Stack1 then pop	empty	1	Pop all from Stack1, push to Stack2 then pop 1 from Stack2	Stack1: [] Stack2: []
4	Dequeue - Pop from Stack2	empty	empty	Stack2 pop 2	Stack1: [] Stack2: []
5	Dequeue - Both stacks empty	empty	empty	No elements to dequeue	Stack1: [] Stack2: []
6	Dequeue - Both stacks empty	empty	empty	No elements to dequeue	Stack1: [] Stack2: []

💡 Execution stops when both stacks are empty and dequeue is requested.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6
Stack1	empty	[1]	[2, 1]	empty	empty	empty	empty
Stack2	empty	empty	empty	[2]	empty	empty	empty

Key Moments - 3 Insights

Why do we transfer all elements from Stack1 to Stack2 during dequeue?

Because Stack2 holds elements in reverse order for dequeue. When Stack2 is empty, transferring reverses Stack1's order to maintain queue order. See Step 3 in execution_table.

What happens if we try to dequeue when both stacks are empty?

There are no elements to dequeue, so the operation cannot proceed. This is shown in Step 6 where both stacks are empty.

Why do we push new elements only onto Stack1 during enqueue?

Stack1 collects new elements in order. We only transfer to Stack2 when dequeuing to reverse order. This keeps enqueue simple and efficient. See Steps 1 and 2.

Visual Quiz - 3 Questions

Look at the execution_table, what is the state of Stack2 after Step 3?

A empty

B [2, 1]

C [1, 2]

D [2]

💡 Hint

Check the 'Stack2 (top->bottom)' column in Step 3.

At which step does Stack1 become empty for the first time?

A Step 2

B Step 4

C Step 3

D Step 5

💡 Hint

Look at the 'Stack1 (top->bottom)' column and find when it changes to empty.

If we enqueue 3 after Step 5, what will be the state of Stack1?

A [3, 2]

B [3]

C empty

D [2, 3]

💡 Hint

Enqueue pushes onto Stack1; after Step 5 Stack1 is empty.

Concept Snapshot

Queue Using Two Stacks:
- Enqueue: push element onto Stack1
- Dequeue: pop from Stack2 if not empty
- If Stack2 empty, transfer all from Stack1 to Stack2
- This reverses order to maintain FIFO
- Efficient amortized enqueue and dequeue

Full Transcript

This visualization shows how a queue can be implemented using two stacks. Enqueue operations push elements onto Stack1. Dequeue operations pop elements from Stack2. If Stack2 is empty during dequeue, all elements from Stack1 are moved to Stack2, reversing their order to maintain the queue's FIFO behavior. The execution table tracks each step, showing the contents of both stacks and pointer changes. Key moments clarify why transfer is needed and what happens when stacks are empty. The visual quiz tests understanding of stack states at different steps. This method efficiently simulates a queue using stack operations.

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