232. Implement Queue using Stacks

1. Description

Implement a first in first out (FIFO) queue using only two stacks. The implemented queue should support all the functions of a normal queue (push, peek, pop, and empty).
Implement the MyQueue class:

• void push(int x) Pushes element x to the back of the queue.
• int pop() Removes the element from the front of the queue and returns it.
• int peek() Returns the element at the front of the queue.
• boolean empty() Returns true if the queue is empty, false otherwise.

2. Notes

• You must use only standard operations of a stack, which means only push to top, peek/pop from top, size, and is empty operations are valid.
• Depending on your language, the stack may not be supported natively. You may simulate a stack using a list or deque (double-ended queue) as long as you use only a stack’s standard operations.

3. Follow Up

Can you implement the queue such that each operation is amortized O(1) time complexity? In other words, performing n operations will take overall O(n) time even if one of those operations may take longer.

4. Example

Example 1:
Input
[“MyQueue”, “push”, “push”, “peek”, “pop”, “empty”]
[[], [1], [2], [], [], []]
Output
[null, null, null, 1, 1, false]

Explanation
MyQueue myQueue = new MyQueue();
myQueue.push(1); // queue is: [1]
myQueue.push(2); // queue is: [1, 2] (leftmost is front of the queue)
myQueue.peek(); // return 1
myQueue.pop(); // return 1, queue is [2]
myQueue.empty(); // return false

5. Note

• 1 <= x <= 9
• At most 100 calls will be made to push, pop, peek, and empty.
• All the calls to pop and peek are valid.

6. Solutions

My Accepted Solution(Follow Up)

n is count of numbers in the stack
Time complexity: O(n)
Space complexity: O(n)

// if the push operation is more frequent, we could design the queue like mine
// if the pop or top operation is more freuent, we could design the stack by letting pop and top operation to O(1), letting push operation to O(n)
// we could make this by changing the push operation, we push all the value to the operation stack, then push all numbers in the opeartion stack into the storage stack

class MyQueue 
{
private:
    stack<int> storage;
    stack<int> operation;
public:
    MyQueue() {}
    
    void push(int x) 
    {
        storage.push(x);
    }
    
    int pop() 
    {
        int lastValue;
        while(storage.size() > 1)
        {
            operation.push(storage.top());
            storage.pop();
        }
        
        lastValue = storage.top();
        storage.pop();
        
        while(!operation.empty())
        {
            storage.push(operation.top());
            operation.pop();
        }
        
        return lastValue;
    }
    
    int peek() 
    {
        
        while(!storage.empty())
        {
            operation.push(storage.top());
            storage.pop();
        }
        
        int lastValue = operation.top();
        
        while(!operation.empty())
        {
            storage.push(operation.top());
            operation.pop();
        }
        
        return lastValue;
    }
    
    bool empty() 
    {
        return storage.empty();
    }
};