Stack and queue in Python

Stack:

>>> stack = [1,2,3,4,5,6,7]
>>> stack.pop()
7
>>> stack
[1, 2, 3, 4, 5, 6]

​Queue:

>>> queue = [1,2,3,4,5,6,7]
>>> queue.pop(0)
1
>>> queue
[2, 3, 4, 5, 6, 7]

Problem: Implement a Min Stack

​leetcode.com/problems/min-stack/submissions/
​

class MinStack:
    
    def __init__(self):
        self.min = None
        self.data = []
        

    def push(self, x: int) -> None:
        self.data.append(x)
        self.min = x if self.min == None else min(x, self.min) 

    def pop(self) -> None:
        self.data.pop()
        self.min = self.top()
        for x in self.data:
            if x < self.min:
                self.min = x
        
    def top(self) -> int:
        if len(self.data) == 0:
            return None
        return self.data[len(self.data) -1]

    def getMin(self) -> int:
        return self.min

same using reduce and one liners for if-else:

import functools

class MinStack:
    
    def __init__(self):
        self.min = None
        self.data = []
        

    def push(self, x: int) -> None:
        self.data.append(x)
        self.min = x if self.min == None else min(x, self.min) 

    def pop(self) -> None:
        self.data.pop()
        self.min = functools.reduce(lambda x,y: x if x<y else y,self.data,self.top())
        
    def top(self) -> int:
        return None if len(self.data) == 0 else self.data[len(self.data) -1]

    def getMin(self) -> int:
        return self.min

Problem: Implement a stack using two queues

​leetcode.com/problems/implement-stack-using-queues/

Implement a last in first out (LIFO) stack using only two queues. The implemented stack should support all the functions of a normal queue (push, top, pop, and empty).

Implement the MyStack class:

• void push(int x) Pushes element x to the top of the stack.
• int pop() Removes the element on the top of the stack and returns it.
• int top() Returns the element on the top of the stack.
• boolean empty() Returns true if the stack is empty, false otherwise.

Notes:

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

Follow-up: Can you implement the stack 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.

 

Example 1:

Input
["MyStack", "push", "push", "top", "pop", "empty"]
[[], [1], [2], [], [], []]
Output
[null, null, null, 2, 2, false]

Explanation
MyStack myStack = new MyStack();
myStack.push(1);
myStack.push(2);
myStack.top(); // return 2
myStack.pop(); // return 2
myStack.empty(); // return False

 

Constraints:

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

SOLUTION:
​

class MyStack:

    def __init__(self):
        """
        Initialize your data structure here.
        """
        self.main_queue = []
        self.temp_queue = []

    def push(self, x: int) -> None:
        """
        Push element x onto stack.
        """
        self.main_queue.append(x)

    def pop(self) -> int:
        """
        Removes the element on top of the stack and returns that element.
        """
        #move all the items but the last one to temp queue
        self.temp_queue = []
        for i in range(0, len(self.main_queue) - 1):
            self.temp_queue.append(self.main_queue.pop(0))
        to_return = self.main_queue.pop(0) # last item
        self.main_queue = []
        for i in range(0, len(self.temp_queue)):
            self.main_queue.append(self.temp_queue.pop(0))
            
        return to_return

    def top(self) -> int:
        """
        Get the top element.
        """
        x = self.pop()
        self.main_queue.append(x)
        return x

    def empty(self) -> bool:
        """
        Returns whether the stack is empty.
        """
        if not self.main_queue:
            return True


# Your MyStack object will be instantiated and called as such:
# obj = MyStack()
# obj.push(x)
# param_2 = obj.pop()
# param_3 = obj.top()
# param_4 = obj.empty()