Implement Queue using Stacks - Leetcode 232 - Python

0:00 - Read the problem
0:36 - Drawing Explanation
12:11 - Coding Explanation

leetcode 232

#neetcode #leetcode #python

[NeetCodeIO]

Even though this is an easy, I spent some extra time to really explain why the solution works.

I think it's tricky enough to warrant that, but let me know if it felt too long though.

[servantofthelord]

I used to watch your videos on 1.5x speed, yet I always had to rewind because I kept missing the logic. Once I humbled myself and decided to slow it down to 1x speed - I have a much more thorough understanding of your content. THANK YOU!

[garsidrag]

i think you should really consider changing it up and once in a while instead of saying its pretty efficient, you should say its hella efficient. would be very satisfying to hear.

[Amy-]

Literally just solved it today after 4 years. I think it was daily challenge or something. I remembered to push onto in-stack and pop from outstack, checking if it’s empty and doing an outstack.push( in stack.pop()). Lol 😂. But I liked your “ amortized” explanation, didn’t think of “ amortized”. - Amy

[adventurer]

For peek, you don't need to perform the copy operation all over again. Just check if s2 has items, and if so return from position -1; else if s1 has items, return from position 0:
if s2:
return s2[-1]
elif s1:
return s1[0]

[garth]

Great solution and explanation but this shouldn't be an *easy* question :(

[utkarshchaturvedi]

Found out today that you solve daily problems, will be following these videos from today♥

[SoRandominfo]

Plz add this question to neetcode all on ur website it will be really helpful.

[Zefnx]

Man fuked up google’s Interview. Will be grinding your sheet and doing LT contests. Thanks for your explanation! ❤

[mikerico]

@NeetcodeIO, you should create a private method to avoid duplucate code lines 11 and 17

[sourabpramanik]

Do you think it is possible to do it using one stack only (if there is a follow up question)? I think it is possible but it won't be memory efficient I guess

[scoobyy]

I was asked this question in my OOD round at Amazon. Messed it up :/

[shaanvijaishankar]

Why is this video not listed in 'Neetcode All'?

[neVessential]

is this actually correct? i think you need to move everything from the second stack to the first before pushing.

[brucem]

Queues are FIFO. Stacks are LIFO.

Queue, original order: user1, user2, user3

Queue, popping: user3, user2, user1...

Queue popping reverses order.

This is why 2 (!) queues are necessary - reversing reversal preserves the original order

🙂

[Jason-lblu]

This is definitely not an "easy" for me, the concept is pretty damn hard to come up with on the spot

[ooouuuccchhh]

in the peek method, you can just return the first element of the stack1 rather than again popping and appending the elements.
it can go like

if not self.s2:
return self.s1[0]

[dumbfailurekms]

When I first started watching your videos I would probably look at this problem and take some time to do the O(n) solution. After about a year of practicing, there was literally not a microsecond that I even considered to solve it in O(n) because I instantly realized that would be trivial and the only real problem is doing this in amortized O(1). point is you helped me a lot to grow