Queue Using Two Stacks in DSA Javascript - Time & Space Complexity

Time Complexity: Queue Using Two Stacks

O(1)

Understanding Time Complexity

We want to understand how fast operations run when we build a queue using two stacks.

Specifically, how does the time to add or remove items grow as the queue gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

class QueueUsingStacks {
  constructor() {
    this.stackIn = [];
    this.stackOut = [];
  }

  enqueue(value) {
    this.stackIn.push(value);
  }

  dequeue() {
    if (this.stackOut.length === 0) {
      while (this.stackIn.length > 0) {
        this.stackOut.push(this.stackIn.pop());
      }
    }
    return this.stackOut.pop();
  }
}
    

This code uses two stacks to add items to the queue and remove them in the correct order.

Identify Repeating Operations

• Primary operation: Moving all items from stackIn to stackOut when stackOut is empty.
• How many times: This move happens only when stackOut is empty, so not every dequeue triggers it.
• Other operations: Pushing to stackIn and popping from stackOut happen once per enqueue or dequeue.

How Execution Grows With Input

When we add items, each enqueue is quick and takes the same time no matter how many items are in the queue.

When we remove items, sometimes we move many items at once, but this happens rarely.

Input Size (n)	Approx. Operations
10	About 10 pushes and pops, with occasional moves of up to 10 items.
100	About 100 pushes and pops, with occasional moves of up to 100 items.
1000	About 1000 pushes and pops, with occasional moves of up to 1000 items.

Pattern observation: Most operations are quick, but some take longer when moving items. Overall, the average time per operation stays steady.

Final Time Complexity

Time Complexity: O(1) amortized per operation

This means on average, each add or remove action takes a constant amount of time, even if some removals take longer.

Common Mistake

[X] Wrong: "Each dequeue always takes time proportional to the number of items in the queue because of moving stacks."

[OK] Correct: The moving happens only when the output stack is empty, so most dequeues are fast. The cost spreads out over many operations.

Interview Connect

Understanding this helps you explain how clever data structures can keep operations fast on average, even if some steps look slow.

Self-Check

What if we moved items from stackIn to stackOut on every dequeue, not only when stackOut is empty? How would the time complexity change?

Practice

1. What is the main idea behind implementing a Queue using two stacks?

easy

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

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

C. Use two stacks to store elements in sorted order

D. Use one stack to hold even numbers and another for odd numbers

Solution

1. Step 1: Understand queue order

   A queue follows FIFO (First In First Out), so the first element added should be the first 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 back when popping, preserving FIFO.

3. Final Answer:

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

4. Quick Check:

   Two stacks reverse order twice = Queue order [OK]

Hint: Two reversals with stacks keep queue order intact [OK]

Common Mistakes:

• Thinking one stack alone can maintain queue order
• Confusing stack sorting with queue behavior
• Using stacks for odd/even separation incorrectly

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

easy

A. stack1.push(x);

B. stack1.add(x);

C. stack1.insert(x);

D. stack1.enqueue(x);

Solution

1. Step 1: Recall JavaScript stack push method

   JavaScript arrays use push() to add elements at the end, simulating stack push.

2. Step 2: Verify other options

   add(), insert(), and enqueue() are not valid array methods for pushing.

3. Final Answer:

   stack1.push(x); -> Option A

4. Quick Check:

   Array push method = stack1.push(x) [OK]

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

Common Mistakes:

• Using enqueue() which is for queues, not stacks
• Using add() or insert() which are not JavaScript array methods
• Confusing stack push with queue enqueue

3. What will be the output of the following code snippet implementing a queue using two stacks?

class QueueTwoStacks {
  constructor() {
    this.stack1 = [];
    this.stack2 = [];
  }
  enqueue(x) {
    this.stack1.push(x);
  }
  dequeue() {
    if (this.stack2.length === 0) {
      while (this.stack1.length > 0) {
        this.stack2.push(this.stack1.pop());
      }
    }
    return this.stack2.pop();
  }
}

const q = new QueueTwoStacks();
q.enqueue(10);
q.enqueue(20);
q.enqueue(30);
console.log(q.dequeue());
q.enqueue(40);
console.log(q.dequeue());

medium

A. 20
30

B. 30
40

C. 10
30

D. 10
20

Solution

1. Step 1: Trace enqueue operations

   Elements 10, 20, 30 are pushed into stack1 in order: [10, 20, 30].

2. Step 2: Trace first dequeue

   stack2 is empty, so pop all from stack1 to stack2: stack2 becomes [30, 20, 10]. Then pop from stack2 returns 10.

3. Step 3: Enqueue 40 and second dequeue

   40 is pushed to stack1: [40]. stack2 still has [30, 20]. Dequeue pops from stack2 returning 20.

4. Final Answer:

   10 20 -> Option D

5. Quick Check:

   FIFO order maintained: 10 then 20 dequeued [OK]

Hint: Dequeue empties stack2 first, then pops top [OK]

Common Mistakes:

• Assuming dequeue pops directly from stack1
• Not transferring all elements to stack2 before dequeue
• Mixing order of elements after transfer

4. Identify the bug in this dequeue method of a queue implemented using two stacks:

dequeue() {
  while (this.stack1.length > 0) {
    this.stack2.push(this.stack1.pop());
  }
  return this.stack2.pop();
}

medium

A. It pops from stack1 instead of stack2

B. It never transfers elements from stack1 to stack2

C. It empties stack1 every dequeue without checking if stack2 is empty

D. It returns undefined always

Solution

1. Step 1: Analyze the transfer logic

   The method transfers all elements from stack1 to stack2 every time dequeue is called, without checking if stack2 is empty.

2. Step 2: Understand consequence

   This causes new elements to be appended to stack2 (which may have remaining older elements), violating FIFO as new elements are popped before older ones.

3. Final Answer:

   It empties stack1 every dequeue without checking if stack2 is empty -> Option C

4. Quick Check:

   Transfer only if stack2.length === 0 [OK]

Hint: Transfer only when stack2 is empty to keep elements safe [OK]

Common Mistakes:

• Transferring elements every dequeue call
• Not checking if stack2 is empty before transfer
• Confusing which stack to pop from

5. You want to implement a queue using two stacks where enqueue and dequeue operations are both efficient. Which approach below achieves amortized O(1) time per operation?

hard

A. Push elements alternately to stack1 and stack2 on enqueue; pop from stack1 on dequeue

B. Push all elements to stack1 on enqueue; transfer to stack2 only when stack2 is empty during dequeue

C. Push elements to stack1 and stack2 simultaneously on enqueue; pop from stack2 on dequeue

D. Push all elements to stack2 on enqueue; pop from stack1 on dequeue

Solution

1. Step 1: Understand amortized O(1) with two stacks

   Enqueue pushes to stack1 only. Dequeue pops from stack2. Transfer happens only when stack2 is empty, moving all elements from stack1 to stack2.

2. Step 2: Analyze other options

   Alternating pushes or pushing to both stacks simultaneously breaks queue order or wastes operations. Pushing all to stack2 and popping from stack1 is incorrect.

3. Final Answer:

   Push all elements to stack1 on enqueue; transfer to stack2 only when stack2 is empty during dequeue -> Option B

4. Quick Check:

   Lazy transfer on dequeue = amortized O(1) [OK]

Hint: Transfer only when needed during dequeue for efficiency [OK]

Common Mistakes:

• Transferring elements on every enqueue or dequeue
• Pushing to both stacks simultaneously
• Ignoring stack2 emptiness check before transfer