Implement Stack Using Queue in DSA Go - Time & Space Complexity

Time Complexity: Implement Stack Using Queue

O(n)

Understanding Time Complexity

We want to understand how fast the stack operations run when we build a stack using a queue.

Specifically, how does the time to add or remove items change as we add more items?

Scenario Under Consideration

Analyze the time complexity of the Push operation in this stack using one queue.

func (s *Stack) Push(x int) {
    s.q.Enqueue(x)
    for i := 0; i < s.q.Size()-1; i++ {
        val := s.q.Dequeue()
        s.q.Enqueue(val)
    }
}
    

This code adds an item to the stack by first adding it to the queue, then moving all other items behind it.

Identify Repeating Operations

Look at what repeats when we push an item.

• Primary operation: Loop that dequeues and enqueues all existing items in the queue.
• How many times: This loop runs once for each item already in the queue, so it runs n-1 times if there are n items.

How Execution Grows With Input

As the number of items grows, the push operation takes longer because it moves all existing items each time.

Input Size (n)	Approx. Operations
10	About 9 moves inside the loop
100	About 99 moves inside the loop
1000	About 999 moves inside the loop

Pattern observation: The work grows directly with the number of items already in the stack.

Final Time Complexity

Time Complexity: O(n)

This means pushing an item takes longer as the stack grows, roughly proportional to the number of items.

Common Mistake

[X] Wrong: "Push operation is always O(1) because we just add one item."

[OK] Correct: Here, push moves all existing items to keep stack order, so it takes time proportional to the stack size.

Interview Connect

Understanding how to build one data structure using another and analyzing its speed is a great skill that shows your problem-solving and reasoning ability.

Self-Check

What if we used two queues instead of one? How would the time complexity of push and pop change?

Practice

1. What is the main idea behind implementing a stack using a queue?

easy

A. Use one or two queues to simulate Last-In-First-Out (LIFO) behavior

B. Use a queue to store elements in sorted order

C. Use a stack to simulate First-In-First-Out (FIFO) behavior

D. Use two stacks to implement a queue

Solution

1. Step 1: Understand stack and queue behaviors

   A stack follows Last-In-First-Out (LIFO), while a queue follows First-In-First-Out (FIFO).

2. Step 2: Simulate stack using queue(s)

   To implement a stack using queue(s), we use one or two queues to reorder elements so the last pushed element is always dequeued first.

3. Final Answer:

   Use one or two queues to simulate Last-In-First-Out (LIFO) behavior -> Option A

4. Quick Check:

   Stack = LIFO simulated by queue(s) [OK]

Hint: Remember stack is LIFO; queue is FIFO, reorder to simulate stack [OK]

Common Mistakes:

• Confusing stack and queue order
• Trying to sort elements instead of reordering
• Using stack to implement queue instead

2. Which of the following is the correct Go syntax to declare a queue using a slice of integers?

easy

A. var queue []int

B. queue := make(map[int]int)

C. queue := []string{}

D. var queue int

Solution

1. Step 1: Identify queue data structure type

   A queue can be implemented using a slice of integers for integer elements.

2. Step 2: Check Go syntax for slice declaration

   "var queue []int" declares a slice of integers, suitable for queue implementation.

3. Final Answer:

   var queue []int -> Option A

4. Quick Check:

   Slice of int = queue [OK]

Hint: Use slice []int for queue of integers in Go [OK]

Common Mistakes:

• Using map instead of slice for queue
• Declaring wrong data type for queue
• Using string slice for integer queue

3. Consider this Go code snippet implementing push operation of stack using one queue:

func (s *Stack) Push(x int) {
    s.q = append(s.q, x)
    for i := 0; i < len(s.q)-1; i++ {
        s.q = append(s.q, s.q[0])
        s.q = s.q[1:]
    }
}

If initially s.q = [1, 2], what is s.q after calling Push(3)?

medium

A. [3, 2, 1]

B. [1, 2, 3]

C. [3, 1, 2]

D. [2, 1, 3]

Solution

1. Step 1: Append new element 3 to queue

   Initial queue: [1, 2]. After append: [1, 2, 3].

2. Step 2: Rotate queue elements to move 3 to front

   Loop runs len(s.q)-1 = 2 times: - Iteration 1: append s.q[0]=1, queue becomes [1, 2, 3, 1], then remove first element: [2, 3, 1] - Iteration 2: append s.q[0]=2, queue becomes [2, 3, 1, 2], then remove first element: [3, 1, 2]

3. Final Answer:

   [3, 1, 2] -> Option C

4. Quick Check:

   Push moves new element front: [3,1,2] [OK]

Hint: New push rotates queue to front [OK]

Common Mistakes:

• Not rotating queue after append
• Removing wrong elements during rotation
• Confusing order of elements after push

4. This Go code tries to implement Pop operation for stack using one queue:

func (s *Stack) Pop() int {
    if len(s.q) == 0 {
        return -1
    }
    val := s.q[0]
    s.q = s.q[1:]
    return val
}

What is the bug in this Pop method?

medium

A. It returns -1 when queue is empty, should panic instead

B. It removes the last element instead of the first

C. It does not check if queue is nil before accessing

D. It removes the front element, but stack pop should remove the last pushed element

Solution

1. Step 1: Understand Pop operation in stack

   Pop removes the last pushed element (top of stack).

2. Step 2: Analyze given Pop code

   Code removes s.q[0], the front of queue, which is incorrect because stack top is at front only if queue is rotated properly.

3. Final Answer:

   It removes the front element, but stack pop should remove the last pushed element -> Option D

4. Quick Check:

   Pop must remove last pushed, not front [OK]

Hint: Pop must remove top, not front element [OK]

Common Mistakes:

• Removing front element without rotation
• Ignoring empty queue check
• Returning wrong default value

5. You want to implement a stack using two queues q1 and q2 in Go. Which approach correctly implements the Push operation to keep the stack order?

hard

A. Dequeue all elements from q1 to q2, then enqueue new element to q1

B. Enqueue new element to q2, then dequeue all from q1 to q2, swap q1 and q2

C. Enqueue new element to q1 only, no other operations

D. Enqueue new element to q1, then dequeue one element from q2

Solution

1. Step 1: Understand two-queue stack push logic

   To keep stack order, new element must be at front of main queue after push.

2. Step 2: Push by enqueue to empty q2, move all from q1 to q2, then swap

   Enqueue new element to q2, move all q1 elements to q2, then swap q1 and q2 so q1 has new order with new element front.

3. Final Answer:

   Enqueue new element to q2, then dequeue all from q1 to q2, swap q1 and q2 -> Option B

4. Quick Check:

   Push uses q2 enqueue + q1 transfer + swap [OK]

Hint: Push to empty q2, move q1 to q2, swap queues [OK]

Common Mistakes:

• Not swapping queues after transfer
• Enqueueing new element to wrong queue
• Forgetting to move all elements