Видео

thanks for discussing all the possible solutions. it helps to see the problems from different sides.

goated

Thank youu

The solution is straightforward, but I could never have come up with it myself.😢

nice explanation

Very good explanation. I was having a hard time understanding the prefix sum thing, thank you. I love that you use Excalidraw, it makes it very beautiful and readable. One note: mod still works fine for negative numbers. k mod p = k + n*p mod p, so adding or subtracting p is essentially the same as adding 0.

Explained this code very well

loved the explanation 👍

awesome video :)

thanks dude

Way better solution than neetcode!

the solution is amazing , you are doing a great job . I am following you for a while.🙌

dude voice too low.

Thank you, this is very clever implementation. PS Using the Integer[] cache will result in an impressive 6ms execution time.

Great explanation man 👌, Loved it!

This is the best and most cleaned and well-explained, Thank you so much for the excellent content. Subscribed

it was really impressive solution

The dedication is impressive. Keep up the good work.

Great explanation

doing a great job man , helping a. noobie like me .

Subscribed. This is brutal! I can barely understand and follow the algorithm, even with chatgpt, the solution and a video explanation I'm still having difficulties understanding how to apply this approach to another problem like counting all occurrences of lowercase characters (a-z) in a given string. Masking is extremely powerful and is a must-have tool, so thank you for the video!

Man, you rock! Thanks for all you're doing!

nice solution!

Thank you, you are a legend.

Can you explain contest questions aswell?

I just realised, you use Ai to generate your thumbnails right?

I had an interesting problem where I accidentally derived this algorithm. The question was to place each number of an input array into one of 2 arrays as you go, depending on which of the 2 arrays had more elements strictly greater than the current number at the time. I tracked a sorted version of each array, comparing the bisect.bisect_right index relative to the size of the array to get the strictly greater than count and then using bisect.insort to insert into the appropriate sorted array. The funny thing is I did that recently and didn't see the application it had to this problem haha

Love your videos, please make videos daily

Arriving at that intuition is key. I started with a subset based solution, got TLE, did early pruning and returns with memoisation if we already have a result. Still got TLE. I guess this is more like a math observation question

Nice video

Voice too low. Can't even hear on full volume

I created a if-else hell for directions it works just as the simulation says, but I wonder how what can we do if the inputs were large ? Binary search or something ?

What's the proof of it if we replace one by one every -1 weight to positive weight and at some point its less than equal to target is our result?

12:15 for this subproblem shouldn't the result of this be 3 (just write e's then replace with T and D)? If you remove the first e then split it into TE and DE woudln't that return 2 + 2 which is 4? please correct me if i am wrong

Why are we turning on only one at a time?

Great job

What if do something like this : We convert all -1 nodes into 1 and we try to save the shortes path. When we get the shortes path, we will travers it backwards and check if any of the edges is -1. We can add the difference between the shortest path and target to the first edge that was originally -1 For the rest we can update them with the MAX

You can also use for loop and scanner class to solve it... If it is possible

Thanks, Good explanation

Had some background noise towards the end, I'll fix that up in the future.

Thanks for the video, but You can use grid2 instead of creating new visited array

Excellent explanation !! Hats off to your consistency.

Thanks for explaning why we need 3*3 grid, because once you figure that out its basically number of island problem.

Just my intuition from what you have explained so far correct me if I am wrong. So for printer(i, j) it returns the minimal number of operations needed to print characters from [i, j]. In the for loop you are splitting whenever we see a matching character with our start character in the remaining string form [i + 1, j]. In that case we know that we could possibly achieve a lower number of operations for printing the first part [i, k]. The reason why we pass in dp(i, k - 1) is because when we have matching characters we are assuming that at index i we are already printing a sequence of similar character from i until index k inclusive. (i.e. "aba", when 0 == 2, at that instance we assume that we printed 'aaa' already, we actually only really consider the cost of printing 'ab' -> when we print 'a' -> 'aaa' and then when we print 'b' we are actually replacing the middle 'a' -> 'aba'.

I think pre/in/post traversal just clicked in my mind haha I wasn't sure why I always used pre for dfs xD

I think it would be a good add on to show why a greedy method won't work here. That being just match the first occurrence with the last occurrence of the character trying to cover as many same characters as possible. Apart from that, excellent explanation

AlphaCode!!!

Very hard to come up with a solution if you have never seen this problem before

Quite a bit of edge cases in this problem, I was able to get to the reversing the first n/2 digits and subtracting and adding 1 to the significant bits but eventually gave up on trying to search for measures that would take care of these 101 and 99 cases. Glad I was able to come across your video! Great explanation, I wish you all the best in everything that you are doing. Thank you tons for posting content that helps people like us!

Nice solution These types of problems can be so annoying