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  1. Queues: A Tale of Two Stacks
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Queues: A Tale of Two Stacks

  • brodan
     + 3 comments

    Here is my Python3 solution (100%, no timeouts):

    class MyQueue(object):
    def __init__(self):
        self.stack1 = []
        self.stack2 = []
    
    def peek(self):
        if len(self.stack2) > 0:
            top = self.stack2.pop()
            self.stack2.append(top)
        else:
            while(len(self.stack1) > 1): 
                self.stack2.append(self.stack1.pop())
            top = self.stack1.pop()
            self.stack2.append(top)
        return top
        
    def pop(self):
        if len(self.stack2) > 0:
            self.stack2.pop()
        else:
            while(len(self.stack1) > 1): 
                self.stack2.append(self.stack1.pop())
            self.stack1.pop() #stack1 is now empty
        
    def put(self, value):
            self.stack1.append(value)

queue = MyQueue()
t = int(input())
for line in range(t):
    values = map(int, input().split())
    values = list(values)
    if values[0] == 1:
        queue.put(values[1])        
    elif values[0] == 2:
        queue.pop()
    else:
        print(queue.peek())