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DSA Typescriptprogramming~5 mins

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

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Time Complexity: Implement Stack Using Queue
O(n) for push, O(1) for pop
Understanding Time Complexity

We want to understand how the time needed changes when we use a queue to build a stack.

Specifically, how fast can we add and remove items when using this method?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.


class StackUsingQueue {
  private queue: number[] = [];

  push(x: number): void {
    this.queue.push(x);
    for (let i = 0; i < this.queue.length - 1; i++) {
      this.queue.push(this.queue.shift()!);
    }
  }

  pop(): number | undefined {
    return this.queue.shift();
  }
}
    

This code uses one queue to simulate a stack by rotating elements after each push.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

  • Primary operation: The for-loop inside the push method that moves elements from front to back.
  • How many times: It runs once for each element already in the queue, so roughly n times when pushing the nth element.
How Execution Grows With Input

Each push moves all existing elements to the back, so the work grows as the stack grows.

Input Size (n)Approx. Operations in push
10About 9 moves
100About 99 moves
1000About 999 moves

Pattern observation: The number of moves grows linearly with the number of elements.

Final Time Complexity

Time Complexity: O(n) for push, O(1) for pop

This means pushing takes longer as the stack grows, but popping is always quick.

Common Mistake

[X] Wrong: "Push and pop both take constant time because queues are fast."

[OK] Correct: Because push rotates all elements to keep stack order, it takes time proportional to the number of elements.

Interview Connect

Understanding this helps you explain how data structures can be combined and what trade-offs happen in time costs.

Self-Check

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

Practice

(1/5)
1. What is the main idea behind implementing a stack using a queue?
easy
A. Use a queue to sort elements in ascending order
B. Use two stacks to implement a queue
C. Use a stack to implement queue operations
D. Use one or two queues to simulate stack operations like push and pop

Solution

  1. Step 1: Understand stack and queue operations

    A stack follows Last-In-First-Out (LIFO) and a queue follows First-In-First-Out (FIFO).
  2. Step 2: Simulate stack using queue(s)

    By using one or two queues cleverly, we can reorder elements to behave like a stack.
  3. Final Answer:

    Use one or two queues to simulate stack operations like push and pop -> Option D
  4. Quick Check:

    Stack via queue = simulate LIFO with FIFO [OK]
Hint: Remember stack is LIFO, queue is FIFO; simulate LIFO with queue(s) [OK]
Common Mistakes:
  • Confusing stack and queue order
  • Thinking queue sorts elements automatically
  • Using stack to implement queue instead
2. Which of the following TypeScript code snippets correctly declares a queue using an array?
easy
A. const queue: number[] = [];
B. let queue = new Stack<number>();
C. const queue = {};
D. let queue: Queue = new Array();

Solution

  1. Step 1: Check correct TypeScript array declaration

    To declare a queue as an array of numbers, use const queue: number[] = [];
  2. Step 2: Identify incorrect declarations

    let queue = new Stack<number>(); uses Stack, not queue; C is an empty object; A uses undefined type Queue.
  3. Final Answer:

    const queue: number[] = []; -> Option A
  4. Quick Check:

    Array queue declaration = const queue: number[] = [] [OK]
Hint: Use array with type annotation for queue: const queue: number[] = [] [OK]
Common Mistakes:
  • Using Stack instead of array for queue
  • Declaring queue as empty object {}
  • Using undefined types without import
3. Given the following TypeScript code implementing a stack using one queue, what will be the output after the operations?
class StackUsingQueue {
  private queue: number[] = [];

  push(x: number) {
    this.queue.push(x);
    for (let i = 0; i < this.queue.length - 1; i++) {
      this.queue.push(this.queue.shift()!);
    }
  }

  pop(): number | undefined {
    return this.queue.shift();
  }
}

const stack = new StackUsingQueue();
stack.push(1);
stack.push(2);
stack.push(3);
console.log(stack.pop());
console.log(stack.pop());
medium
A. 3 2
B. 1 2
C. 2 3
D. Error at runtime

Solution

  1. Step 1: Trace push operations

    Push 1: queue=[1]
    Push 2: queue=[2,1] after rotation
    Push 3: queue=[3,2,1] after rotation
  2. Step 2: Trace pop operations

    Pop returns queue.shift(): 3, then next pop returns 2.
  3. Final Answer:

    3 2 -> Option A
  4. Quick Check:

    Stack pop order = 3 then 2 [OK]
Hint: Push rotates queue to front; pop removes front element [OK]
Common Mistakes:
  • Not rotating queue after push
  • Assuming pop removes last element of array
  • Confusing shift() with pop()
4. Identify the bug in this TypeScript code for stack using two queues and suggest the fix:
class StackUsingTwoQueues {
  private q1: number[] = [];
  private q2: number[] = [];

  push(x: number) {
    this.q1.push(x);
  }

  pop(): number | undefined {
    while (this.q1.length > 1) {
      this.q2.push(this.q1.shift()!);
    }
    return this.q1.shift();
  }
}
medium
A. pop() should return q2.shift() instead of q1.shift()
B. push() should push to q2 instead of q1
C. pop() does not swap q1 and q2 after popping; add swap after pop
D. No bug; code works correctly

Solution

  1. Step 1: Analyze pop method behavior

    Pop moves all but last element from q1 to q2, then removes last from q1.
  2. Step 2: Identify missing queue swap

    After pop, q1 is empty and q2 has elements, but q1 and q2 are not swapped, so next push/pop fail.
  3. Final Answer:

    pop() does not swap q1 and q2 after popping; add swap after pop -> Option C
  4. Quick Check:

    Swap queues after pop to maintain correct state [OK]
Hint: Remember to swap queues after pop in two-queue stack [OK]
Common Mistakes:
  • Forgetting to swap queues after pop
  • Pushing to wrong queue
  • Returning wrong queue element
5. You want to implement a stack using a single queue in TypeScript. Which approach correctly ensures that the pop() operation returns the last pushed element efficiently?
class StackUsingOneQueue {
  private queue: number[] = [];

  push(x: number) {
    this.queue.push(x);
    // What should be done here?
  }

  pop(): number | undefined {
    return this.queue.shift();
  }
}
hard
A. After push, sort the queue in descending order
B. After push, rotate the queue by moving all elements except the last to the back
C. After push, reverse the queue array
D. No extra operation needed after push

Solution

  1. Step 1: Understand stack behavior with one queue

    To simulate LIFO, after pushing, rotate queue so new element is at front.
  2. Step 2: Rotate queue after push

    Move all elements except last pushed to back, so pop (shift) removes last pushed.
  3. Final Answer:

    After push, rotate the queue by moving all elements except the last to the back -> Option B
  4. Quick Check:

    Rotate queue after push to simulate stack pop [OK]
Hint: Rotate queue after push to keep last element at front [OK]
Common Mistakes:
  • Sorting queue instead of rotating
  • Reversing queue array (inefficient)
  • Not rotating queue after push
Recommended

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10 questions~5 mineasy hard