It's a god level explanation of monotonic stack. A request from your rest subscriber: Don't have that running background in the tutorial it's irritating .
Thank you for the monotonic stack explainer. It may be helpful to also show an image of the heights array above the image of the answer array to help visualize the relationship between the two better while popping and inserting.
Great video thanks. Trying to understand how the monotonic stack approach would be of O(n) time. We traverse the array only once for sure, but what about the pop operations? For a few elements of array, we are popping out of stack multiple times until we find a stack element that is greater than the current element, so are we not counting this towards the time complexity? If length of stack is k, wouldn't the overall time be O(nk)? Thanks again!
Every element in the array is processed exactly once. When an element is popped from the stack, it does not re-enter. Consequently, the maximum number of operations is 2n, accounting for each element being pushed and popped once.
Really good explanation. I'll agree with everyone else and say that the background is extremely distracting, and it actually makes the bitrate of the whole video go down, making the quality poor whether I select 144p or 1080p. I also feel like you didn't explain the algorithm before going through the problem with the heights of people. You started going through the example, but I didn't know precisely what it was you were doing or why you were making the decisions that you were making. I think it would be nice to have a plain explanation of the problem and how the algorithm solves it before going through the problem step by step.
I kindly request that you should make videos avoiding moving backgrounds, as they can be distracting while concentrating . By the way, great explanation!
Why is the time complexity O(n) if we have two nested loops? shouldn't it be o(n*m) where n is the number of elements in the array and m is the number of elements in the stack?
why brute force is O(n^2)? Each element doesn't start from the beginning of the array. Isn't it O(n*m) where m is i - k - 1? where k is the current element.
Couldn’t you also go from left to right? You could store the indexes in the stack and as soon as you add an element greater than the element at the top of the stack, you keep removing the top until it’s equal or greater to the new element, while updating their values (sorry for the bad explanation)
![ROSÉ(로제) - APT. (Band VER.) (With. 이영지) [더 시즌즈-이영지의 레인보우] | KBS 241129 방송](http://i.ytimg.com/vi/IW8qaemx74g/mqdefault.jpg)
Very well explained, would be easier to learn without the moving background.
I loved the moving background. It helped me stay focused
Everything was fine before i read your comment. Now i cant ignore it .
@@bluesteel1 😆
@@bluesteel1 😂
@@JackHou-vw7hs my man
It's a god level explanation of monotonic stack.
A request from your rest subscriber: Don't have that running background in the tutorial it's irritating .
first time heard about Monotonic Stack, thanks for explaining it so properly
Our pleasure!
the background of this video is greatly designed. its very easy to focus with this background in motion
can't tell if it's positive or sarcasm haha but yeah glad you like it!
@@algo.monster the background in motion acts as a boundary to the actual content, so I think it really works.
I am a complete opposite to this idea, I feel like this 🤮 when I see it. It is not optimal for every 🧠
phenomenal bro, u got a talent for this
This helped me solve Next Greater element without looking at your code, thanks!
Wow! Thanks for walking through the example, it was very insightful
Thank you for the monotonic stack explainer. It may be helpful to also show an image of the heights array above the image of the answer array to help visualize the relationship between the two better while popping and inserting.
Great video thanks. Trying to understand how the monotonic stack approach would be of O(n) time. We traverse the array only once for sure, but what about the pop operations? For a few elements of array, we are popping out of stack multiple times until we find a stack element that is greater than the current element, so are we not counting this towards the time complexity? If length of stack is k, wouldn't the overall time be O(nk)? Thanks again!
Every element in the array is processed exactly once. When an element is popped from the stack, it does not re-enter. Consequently, the maximum number of operations is 2n, accounting for each element being pushed and popped once.
Perfect explanation, thank you !
Glad it was helpful!
Really good explanation. I'll agree with everyone else and say that the background is extremely distracting, and it actually makes the bitrate of the whole video go down, making the quality poor whether I select 144p or 1080p. I also feel like you didn't explain the algorithm before going through the problem with the heights of people. You started going through the example, but I didn't know precisely what it was you were doing or why you were making the decisions that you were making. I think it would be nice to have a plain explanation of the problem and how the algorithm solves it before going through the problem step by step.
Very nice explanation, great job.
Thank you!
Awesome video! Easy to understand
wow very good explanation. thank you..
Great!
Wonderful Explanation and PPT's
Glad you liked it!
I kindly request that you should make videos avoiding moving backgrounds, as they can be distracting while concentrating . By the way, great explanation!
Noted thanks!
Awesome explaination
you made it very easy by making visualizing approach. Thanks
Glad it was helpful!
Great video, thank you a lot!
Glad it helped!
would you be kind enough to share what is the application that you are using for the walk through of the solution :) Thank you in advance.
The background is too distracting
Why is the time complexity O(n) if we have two nested loops? shouldn't it be o(n*m) where n is the number of elements in the array and m is the number of elements in the stack?
The code can be indeed deceiving. But remember each element enters or exits at most once. This is why monotonic stack is a efficient data structure.
Thank you so much
You're most welcome
So helpful 😭🖤
Glad it helped!
thanks
bro is god
You are Osm bro..
Thanks ✌️
Before learning this, I'd just use a segment tree for the first question xD.
Why the solution iterative the array from right side to left side, is this order matter?
The order is a choice, you could have used a strictly increasing function instead
New Topic Learned!!
Sep'11, 2023 06:36 pm
Nice
Thanks for the video! Feedback on the moving background : VERY DISTURBING
Thanks and Noted!
why brute force is O(n^2)? Each element doesn't start from the beginning of the array.
Isn't it O(n*m) where m is i - k - 1? where k is the current element.
k is not an input parameter. we have to express it in n. after dropping the content it's n^2
its n-1 + n-2 +n-3 ... = n**2
@@douglas5260 its not multiplication it should be addition
@@BurtPredrag Yeah, I've corrected it. Thanks!
Couldn’t you also go from left to right? You could store the indexes in the stack and as soon as you add an element greater than the element at the top of the stack, you keep removing the top until it’s equal or greater to the new element, while updating their values (sorry for the bad explanation)
Literally the worst sample, because not much of exploration... Maybe thats why for you its most confusing... As so its this video...
great explanation but the moving background really makes me dizzy, could not watch it for more than 1 minute
The moving background make it hard to follow.
The moving background is very distracting. I can't even watch the full video
bro remove that background!
very bad background
yup!
It's really helpful to stay focused actually
Slow down, it's not a race. Unless you don't understand what you are doing.