How do you solve Implement Queue using Stacks on LeetCode?

Implement Queue using Stacks (LeetCode #232) 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 stacks allows us to maintain FIFO order by reversing the order of elements when necessary.

What is the brute force solution for Implement Queue using Stacks?

The brute force approach for Implement Queue using Stacks is Brute Force, with O(n²) time complexity and O(1) space complexity. We can use two stacks to simulate a queue by moving elements between them. When we want to pop or peek, we transfer elements from one stack to the other, reversing their order. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Implement Queue using Stacks?

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

What are the interview tips for Implement Queue using Stacks?

Explain your thought process clearly as you implement the solution. Discuss the trade-offs of your approach, especially regarding time complexity. Practice explaining the difference between worst-case and amortized time complexity.

What are common mistakes when solving Implement Queue using Stacks?

Forgetting to check if stack2 is empty before popping or peeking. Not transferring elements between stacks when needed, leading to incorrect results.

#232

Implement Queue using Stacks

Easy

StackDesignQueueStackQueue

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)

By using two stacks efficiently, we can ensure that each operation is amortized O(1). We only transfer elements when necessary, keeping operations fast.

⚙️

Algorithm

3 steps

1Step 1: For push, simply push the element onto stack1.
2Step 2: For pop and peek, check if stack2 is empty. If it is, transfer all elements from stack1 to stack2 (this reverses their order). Then, pop or peek from stack2.
3Step 3: For empty, check if both stacks are empty.

solution.py20 lines

1class MyQueue:
2    def __init__(self):
3        self.stack1 = []
4        self.stack2 = []
5
6    def push(self, x):
7        self.stack1.append(x)
8
9    def pop(self):
10        self.peek()  # Ensure stack2 has the current front
11        return self.stack2.pop()
12
13    def peek(self):
14        if not self.stack2:
15            while self.stack1:
16                self.stack2.append(self.stack1.pop())
17        return self.stack2[-1]
18
19    def empty(self):
20        return not self.stack1 and not self.stack2

ℹ

Complexity note: The amortized time complexity for each operation is O(1) because we only transfer elements when stack2 is empty, leading to efficient usage of both stacks.

1Using two stacks allows us to maintain FIFO order by reversing the order of elements when necessary.
2Amortized analysis helps us understand that even though some operations can be costly, they are infrequent.

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