Link: https://leetcode.com/problems/implement-queue-using-stacks/
Solution:
Topics: amortized
Intuition
This is a pretty interesting problem. Took me a minute to grasp this one because I was not familiar with the concept of amortized complexity.
We know that pushing to the front of a list is o(n) so obviously we want to avoid this. The key insight is to use two lists! Append elements to stack1. Pop elements from stack2. If stack 2 is empty, pop all elements off stack1 and append them to stack2. This will result in the bottom of stack1 being the top of stack2! If stack2 is not empty, pop of it.
This is analogous to pouring out the contents of stack2 into stack2 while preserving the relative order of the contents.
Implementation
class MyQueue:
def __init__(self):
self.stack1 = []
self.stack2 = []
def push(self, x):
self.stack1.append(x)
def pop(self):
if not self.stack2:
while self.stack1:
self.stack2.append(self.stack1.pop())
return self.stack2.pop()
def peek(self):
if not self.stack2:
return self.stack1[0]
else:
return self.stack2[-1]
def empty(self):
return not self.stack1 and not self.stack2
#time: o(1) amortized
#memory: o(1) Mnemonic
Imagine you have a green cup and a red cup. You fill up the green cup, and drink only from the red cup. If the red cup is empty, fill it up by pour out all the water from the greed cup.
Review 1
Classic problem. Nailed it.