232. Implement Queue using Stacks

Problem Link

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.

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.

 

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

 

Constraints:

• 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.

 

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.

Solution

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C++

class MyQueue {
public:
    stack<int> stack1, stack2;
 
    MyQueue() {
    }
    
    void push(int x) {
        stack1.push(x);
    }
    
    int pop() {
        peek();
 
        int e = stack2.top(); stack2.pop();
        return e;
    }
    
    int peek() {
        if (stack2.size() == 0) {
            while (stack1.size() > 0) {
                int e = stack1.top(); stack1.pop();
                stack2.push(e);
            }
        }
 
        return stack2.top();
    }
    
    bool empty() {
        return stack1.empty() && stack2.empty();
    }
};
 
/**
 * Your MyQueue object will be instantiated and called as such:
 * MyQueue* obj = new MyQueue();
 * obj->push(x);
 * int param_2 = obj->pop();
 * int param_3 = obj->peek();
 * bool param_4 = obj->empty();
 */

Python3

class MyQueue:
 
    def __init__(self):
        self.stack1 = []
        self.stack2 = []
 
    def push(self, x: int) -> None:
        self.stack1.append(x)
 
    def pop(self) -> int:
        self.peek()
        return self.stack2.pop()
 
    def peek(self) -> int:
        if not self.stack2:
            while self.stack1:
                self.stack2.append(self.stack1.pop())
        
        return self.stack2[-1]
 
 
    def empty(self) -> bool:
        return len(self.stack1) + len(self.stack2) == 0
 
 
# Your MyQueue object will be instantiated and called as such:
# obj = MyQueue()
# obj.push(x)
# param_2 = obj.pop()
# param_3 = obj.peek()
# param_4 = obj.empty()