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Really good explanation!

You can do it in O(1) time O(N) space using a queue.

This feeling then you are a completely noob in Python, went to TH-cam because you code didn't work... and see this video and think about pop(0) instead of the second stack.

great

beautifullll

Thank you

Great explanation! Question: are we using the size as the tail of the linked list?

Your explanation is by far the easiest to understand. thank you

thanksss,

Thank you! :)

Good explanation with clean code. Thank you _/\_

i just never seen a good explaination like this for dfs, you are just the best teacher in the universe. I think a baby can understand your explanation

Your explanation is top notch 🛐

Worst solution seen so far

Nicely explained

how does it work if it's in a separate file?

Amazing explanation

Thank you <3

how does this code guarantee the shortest distance?

Your brains are on another leve.

This is a very interesting and unique way to approach the problem, thanks for sharing I’ve learned a new way how to do it

beautiful

Very good Explanation for Beginners

Thank you so much

such a awesome explanation ...

Thank you so much sir... I loved it!

much easier approach

great explanation : )

This reminds me of the towers of hanoi solution, thank you for this explanation

This was one of the best explanations I have seen! P.S. Your linkedin link doesn't work.

very clear explanation, thanks

Thank you so much such a wonderful explanation

Very easy and Understandable solution thank you.

Thanks a lot bro your explanation is very good

Holy Thanks mate

I got memory limit exceeded for the same code but thanks for the video

An alternative solution might be to have one stack hold input values, and the other stack holds the output values. You just add to the input stack when pushing. And when you need to peek or pop, you just check if the output stack is nonempty and return the poped/peeked value from that stack, otherwise, you flip the values from the input to the output and then pop/peek. Of course, it'll depend on where you want most of the computing to be spent, when writing data or when reading data

Hey man great video but you should focus more on the solution as opposed to explaining what parts of the code do syntatically. Anyone who is attempting leetcode mediums should already understand things like accessing an element in a 2D array.

Thanks for the tutorial! This helped me understand.

.push() is O(n) , .pop() is O(1)

thank you!!!!

Thank you so much, I was really lost since I've just started learning about data structures! but your explanation was amazing!

Time complexity analysis?

Whats the Complexity of this solution?

Thank you for the explanation. While this is algorithmically correct, it may fail in large test case. Specifically, if the distance of a cell is previously computed as 35 (exactly the ascii value of ‘#’), your code will mistakenly flag it as not being computed and therefore incrementing it again (and potentially more than once due to the same reason). I suggest you change ‘#’ to either some negative number or maximum of int so this corner case will be fine. Otherwise, love the short and concise explanation!

You don't wanna be recalculating the row and col len with every recursive call.

hi, please why is it -self.size

Not only do you explained it quite well, but your voice is also quite soothing, which makes the lesson enjoyable.

Thanks a lot : )

thank you