How do you solve Implement Stack using Queues on LeetCode?

Implement Stack using Queues (LeetCode #225) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using two queues can simulate stack behavior effectively.

What is the time complexity of Implement Stack using Queues?

The optimal solution for Implement Stack using Queues runs in O(n) time and uses O(n) space. The time complexity is O(n) for pop and top operations because we may need to transfer n-1 elements between queues. However, push is O(1).

What is the brute force solution for Implement Stack using Queues?

The brute force approach for Implement Stack using Queues is Brute Force, with O(n²) time complexity and O(1) space complexity. We can simulate stack behavior by using two queues. The simplest approach is to use one queue for pushing elements and another for popping, which can be inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Implement Stack using Queues?

Implement Stack using Queues uses the following concepts: Stack, Design, Queue. The recommended patterns to study are: Queue, Data Structure Design.

What are the interview tips for Implement Stack using Queues?

Explain your thought process clearly while coding. Discuss the trade-offs of your approach. Practice implementing both brute-force and optimal solutions.

What are common mistakes when solving Implement Stack using Queues?

Forgetting to swap queues after pop/top operations. Not handling edge cases like popping from an empty stack.

#225

Implement Stack using Queues

Easy

StackDesignQueueQueueData Structure Design

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

We can optimize the stack operations by maintaining the stack's top element in the first queue, allowing for constant time access to the top element and more efficient push and pop operations.

⚙️

Algorithm

4 steps

1Step 1: For push, simply enqueue the new element to the first queue.
2Step 2: For pop, dequeue all elements except the last one, which is the top of the stack. Swap the queues afterward.
3Step 3: For top, perform the same transfer as in pop but keep the last element in the first queue.
4Step 4: For empty, check if the first queue is empty.

solution.py27 lines

1from collections import deque
2
3class MyStack:
4    def __init__(self):
5        self.q1 = deque()
6        self.q2 = deque()
7
8    def push(self, x):
9        self.q1.append(x)
10
11    def pop(self):
12        while len(self.q1) > 1:
13            self.q2.append(self.q1.popleft())
14        top_element = self.q1.popleft()
15        self.q1, self.q2 = self.q2, self.q1
16        return top_element
17
18    def top(self):
19        while len(self.q1) > 1:
20            self.q2.append(self.q1.popleft())
21        top_element = self.q1[0]
22        self.q2.append(self.q1.popleft())
23        self.q1, self.q2 = self.q2, self.q1
24        return top_element
25
26    def empty(self):
27        return len(self.q1) == 0

ℹ

Complexity note: The time complexity is O(n) for pop and top operations because we may need to transfer n-1 elements between queues. However, push is O(1).

1Using two queues can simulate stack behavior effectively.
2Understanding the transfer of elements between queues is crucial for efficient operations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.