Hey there can you help me I want to reinstall my windows 10 and I have to download a new Iso file for it but in my area internet is not working what should I do now plz reply
If i make a subsequence array and it has a tuple of 2 numbers showing the biggest number in the subsequence and the length and then i go through that and if the number i am currently at is bigger than k subsequences in the subsequence array and i change every single one that's smaller than what i am at to the number then that would work the same right?
So the cache in Joma's solution is [{The Longest That End With Current Value}], and it make the inner looping O(n) I think we can make better cache which is [{length}, {min value that have that length}]. For example, my cache for [0, 8, 4, 12, 2, 10] would be: [ (1, 0), #for length 1, min val is 0 (2, 2), (3, 10) ] Doing that, we can binary search the cache on min val when looping and it would take O(log n) => The entire thing would be O(n*logn)
at 7:40 shouldn't the result be 5?
These videos are great I love watching them
Which software is he using to make these videos?
Domo arigato Joma-chan ❤️
which software you use for the drawing.
thats really cool thank you
Hey there can you help me I want to reinstall my windows 10 and I have to download a new Iso file for it but in my area internet is not working what should I do now plz reply
If i make a subsequence array and it has a tuple of 2 numbers showing the biggest number in the subsequence and the length
and then i go through that and if the number i am currently at is bigger than k subsequences in the subsequence array and i change every single one that's smaller than what i am at to the number then that would work the same right?
Love the clear explanation!
The longest subsequence on the drawing board should be 5 tho
isn't the magnitude of subsequence for 9 must be 3 (1,8 and 9=3 numbers)
Now try to solve it in O(n * log n) (not O(n ^ 2))
So the cache in Joma's solution is [{The Longest That End With Current Value}], and it make the inner looping O(n)
I think we can make better cache which is [{length}, {min value that have that length}].
For example, my cache for [0, 8, 4, 12, 2, 10] would be:
[
(1, 0), #for length 1, min val is 0
(2, 2),
(3, 10)
]
Doing that, we can binary search the cache on min val when looping and it would take O(log n)
=> The entire thing would be O(n*logn)
can you make a video on O(nlogn) solution?
Class is in session.
7:39
answer should be 5!?
Third comment
Guess what, I'm the First!