Teleporting Ants & Dynamic Programming  

for all I know (I have rating of 2194 on cf) for any solution to this problem, we would need to find first x[j] greater than y[i] for all i=1..n and we would either need to sort y, or do n binary searches. We can't do n binary searches in O(n) time, and we can't sort general array in less than O(n log n), but what we can do is use constraint that y[i] < 1e9 and use radix sort(basically 2 count sorts in this case, but for numbers not up to 1e9 but 1e4.5, so linear time since n is up to 2e5), but this feels very unsatisfactory and leaves savory taste in mouth, so I would leave this problem at being O(n log n)

No one can explain a 2200-rated problem better than this... definitely made me a bit wiser.

This video actually looks like one of those 4m subscriber channels, great job man!

I solved the problem much differently and pure mathematically before watching the video. it is very hard to explain a solution in text form but the main idea was to see the portals as binary numbers. I split the problem into two. evaluating how many times the player has entered a portal (to teleport not passover) and then calculating the total distance traveled. The second problem's solution was easy I just had to calculate the difference in position between the portal and the teleportation target to calculate the "cost" of the teleportation. the total length of the line plus all the costs were equal to the total distance traveled. now for the solution to the first part, look at the first example with 4 portals (red, orange, yellow, green). in the solution of this part ignore all the distances. if you look at the first red portal it teleports the player right behind itself and is thus equivalent to a binary number with one digit. orange also is a one digit number but yellow perfectly encapsulates orange but no other portal so they collectively create two digit binary number. green is encapsulating the entire one digit and two digit numbers. green is not symbolized as 3 digit but (1 + 2)+1 digits. plus operation is not summation in this notation but symbolizes that those two numbers have to be calculated independently. so collectively the table for the puzzle is 1+2+(1+2+1). we have to take the initial positions of the portals into account. the first one digit portal is closed so it has the value of 1. notice 1 is the maximum value a one digit binary number can hold so the next stop over it will pass it. orange and yellow collectively two digit number seems to have the value (0x01=1) but you can observe they act as inverted. so actually they have the value (0x10 = 2) of 2. and lastly green was symbolizing the entire copy of the puzzle plus itself but when the player passes it, all the portals will be open. so the value for green is (0 + 0 + 0) binary digit table = 1+2+(1+2+1) initial value table = 1+2+(0+0+0) we are close to be able to calculate total teleportation count. just remember that one digit binary numbers can hold maximum of 1 and leak for the value of 2. similarly two digit numbers can hold the maximum of 3 and leaks for the value of 4. binary digit table = 1+2+(1+2+1) initial value table = 1+2+(0+0+0) max value table = 1+3+(1+3+1) difference table = 0+1+(1+3+1) when we sum the difference table 1+1+3+1 = 6 you can see that player will teleport 6 times. conveying the idea behind these tables isn't that easy in text form so if they are not clear try to read it again and calculate them by yourself this is the best that I can do. lastls we have to calculate the costs. as I said calculating the cost for a portal is easy we have to calculate the distance between the portal and the target but when we unite two portals into a multi digit number since the two portals creating that multi digit (in this example two) number can have different costs we have to symbolize them independently. for this example in the case of orange and yellow portals, their costs are 1 and 3 so we will notate them as (1+3) for the 2 digit number. this says that if the player enters the first digit the cost is 1 and if it enters the second digit the cost is 3. we need to somehow evaluate how many times which digit will flip while counting in binary. I will calculate that for the general case later but for two digit binary (00 10 01 11) the first digit will flip every time and the second digit will flip every two times. collectively first 4 times and second 2 times. let me write the tables again this time with the cost table too. cost table = 1+(1+3)+1+(1+3)+7 binary digit table = 1 + 2 + (1 + 2 + 1) initial value table = 1 + 2 + (0 + 0 + 0) difference table = 0 + 1 + (1 + 3 + 1) total cost table = 0+(1+0)+1+(1+3+1)+7 = 14 total distance travelled = cost + distance = 14 + 9 = 23 the reason for the cost for the second two digit number to be not (1+3) but (1+3+1) is because while counting 00 10 01 11 first digit is flipped from 0 to 1 twice and second digit once. thus when the binary digit table is constructed total distance traveled can be evaluated mathematically with near zero computational cost. I have spent around 2 hours for the solution and an additional 2 hours watching the video and writing this comment. I'm very happy if you have read it to here and I hope you like the solution. have a nice day.

The sum on contiguous elements idea was genius, I never would have thought of that but it's so obvious in hindsight

Probably the best explanation to a problem I've ever seen

I'm shocked that this is the first video on this channel, the quality is something I would expect it to take months or even years to achieve. Can't wait for the next one!!!

i really liked the animations, especially the last part where the lines of code are highlighted each step and you can see where the results move to.

Absolute gold. Keep it up. Wish to see several algorithm implementations and problems in this format.

Absolutely incredible. Amazing content right out the gate! Keep making awesome vids

You are an awesome storyteller and a talented movie maker, and not to mention an excellent teacher whatsoever.

Just great! Can’t wait for the next video. And any video made using manim have to be great:)

So I might have taken your challenge to solve this problem in O(n) a bit too seriously. It took a variant of counting sort to allow building a hash map from each exit to the closest entrance in O(n), so we can skip the binary search and replace it with an access to the hash map in O(1), thus resulting in a total complexity of O(n) (where n is the length of the road rather than the portal count, though).

Incredible video and excellent teaching. Like many others, I’m amazed this is your first video! You earned a sub - can’t wait to see what else you have for the future!

Loving the animations. Really made the video so much more understandable. Really great job!

Checked out your channel to see more vids and realised thats the only one. The music/animation was honestly amazing

Great video! We definitely need more of that. Keep it up man

Love this video and the animations. Keep making more of these videos please

Awesome video. I noticed this is your first on RU-vid, I hope you keep it up! Really tough problem but you explained it well. Always wished there was a 3b1b for data structure and algs. Could be you.

This is unbelievably good explanation, please keep up and post more!

The animations were really nice, particularly for the DP graph visualization. Very creative.

wow such a high quality video for a competitive programming problem. Great explanation and visualization. I just love it.

the motion graphics illustrating the algorithm are absolutely delightful!

Just soo good. Please, never stop making these ❤❤

The production value, explanation, pacing, audio, visuals, everything about this screams top-notch quality. Can't wait to see more!

As a green coder (1200-1400 CF) I can say that I understood the 2200 rated problems solution proving that this guy have put a great work into his explanation!

GOAT explanation skills, amazing work! Hoping for a lot more great videos in the near future.

Please make more videos, the wonderful quality and entertainment value this has is beautiful, thank you.

Loved it! Thank you for this and all upcoming videos, orz

Thank you for explaining this so nicely, I gave this problem a try before looking at the solution and when I saw my solution matched up with the one you presented I was very happy, this is a great explanation of dynamic programming, great video!!

Amazing effort for a video about one dp problem! Hope to see another entry this year for some3!

Very Video, much good, such wow. First video, and despite some audio that's clearly form different recording sessions its mostly flawless. Keep on rocking dude.

Fantastic video! Use of prefix sums is pretty genius.

Incredible video! When I first studied dynamic programing I had lots of troubles wrapping my head around it and seeing it explained so clearly was really nice! I wish your video came out earlier 🤣

Excellent video. Very well-explained and demonstrated. I hope you post more.

Beautifully done. Simply awesome

Thank you for the hard work and for sharing it!

Banger video

Fantastic visuals. I wonder how many followed it all while watching the video without pausing it. I sure didn't. But still I'm impressed.

you probably are the best teacher ever for dsa. I have never seen a better explanation

Fantastic. What a superb first video. I hope there will be more, but even if not, wow.

Thank you for making such a great video!

This is the best video on the internet, until you upload the next one

this video is really impressive, please make more!

Amazing content, great animations. Really hope you still have plans of making more videos!

I really liked how you showed how much DP speeds things up at 7:18. I also like how you cut off the music at 6:37 when making your conclusion. One criticism I have is you sometimes show graphics on the screen only while you talk about it, so it's hard to visually digest what you're showing and listen in that short time (ie sometimes it's a bit fast such as at 5:30)

This is amazing! Calling now that this will be in the winners video :D

Damn thats some quality work! Your Channel needs to blow up. You have my subscription at least :)

that was incredible man, please keep on going :)

Great first video. Loved it. Subscribed.

I just subscribed, I love competitive programming and I think this channel will help me on my journey.

Excellent video, this is 3b1b-quality explaining!

This is a really awesome video with great production❤

When I saw your video mentioned in 3B1B's SOME2 results video, I was dead sure you were gonna win. Welp, like he said, every video is the best for different demographics of people, and this is the one for me. Fucking incredible video, and thank you for making this!

Oh man.. you took it to the next level. Also how many hours did you spend making this?

I'm a beginner EASY coder and this video popped up in my feed, my Feedback :- Good Work 👍🏻, nicely explained, with little bit of pace management, even beginner's like me can understand the Problem Solving aspect. Thank You.

Wow amazing visualization! Good job!

I really appreciated this video. I paused at the 2 minute mark, solved it myself in O(n log n), and then saw that I came to the same conclusions you came to in this video. It felt very validating! Thank you for the great visual and description of your terms. I wouldn't have been sure I had actually had the same solution as you if you hadn't. Also yours is much cleaner and easier to understand than mine.

An academic epiphany of many proportions

This is a great video! But the part near 6:40, explaining how dp was calculated was paced a bit too quick, and I had to rewatch that part to understand it. Otherwise, a great explanation of a hard problem.

amazing video!!! Please continue to create similar video.

Very promising video. Looking forward to more!

This is so well done !!

This is so high quality! I thought I was watching a 3blue1brown video

Fantastic video. Great job!

Great explanation, felt really absorbed during whole video

Amazing work, you earned a subscriber!

amazing work man, keep it up!!!

Every math or CS student should subscribe! Great work man, cant wait to see more of DP or Machine Learning ect ...

Great video. One thing I liked about this is that it didn't break-down the solution, instead it just introduced the ideas and let the viewer understand it by though. I loved this methodology for this problem is particular, but you may run into issues doing this with much more complicated problems.

Wow I really liked this explanation, I want more!

Fantastic video. Keep up the great work.

I've always wanted such content on youtube, and finally here it is. Looking forward for next videos. But personally for me it was very hard to follow the solution, the key part about what we store as dp_i and how we calculate it, is explained too briefly. (although I am 1900 on codeforces).

Great video! It took me some time to figure out some claims when you tell us to convince ourselves. A brief explanation or key idea would make it more digestable.

Great video, subbed from here. I could follow right up until 8:50 then it started getting over my head, I think I'll have to watch the last part a couple of times before I start to get it.

Hats off for the informative video, good stuff! I can't implement binary search off the top of my head for the life of me, not having actually ever done it. Caught that pattern at 2:14, the movement straight up looked like counting in binary with variable offsets. Fascinating.

that was smooth, well done!!

Keep making videos, one day this channel would be huge

Great video. Good start. Keep up the good work.

I don't see 84 subscribers, I see 84M!

The quality of this video is amazing! I really liked how well you made the music sync up to the algorithm at the end with a satisfying ding every time the cost was calculated. The part at 10:30 felt a bit rushed though, I had to rewatch and study your python implementation to realize why exactly you needed to binary search to find the index of the ps array, but other than that it was an excellent video and a delight to watch.

Great Explanation. You sure must upload more

Looking forward to more of these!

Greate video, Thanks for the sorce code of the animations. i am also on codeforces. I am probably learning more there as anywhere else.

Looking forward to the next one! ❤️

I feel like I'm going to love this channel....

Please make more videos This was very well explained

superb work man!

this feels like protagonist in a anime figuring things out

this was superbly presented!

Bruh this content is gold, subscribing and expecting more gold content here 👋

We need more! More videos!

Nice first video, My left Ear enjoyed it.

3000 subscribers? no way, this is an amazing work. And i want to say that in the bisect method we can specify "Lo" and "hi" parametres(the start and end indices in the array where we want to search our element), so if we're in portal X[i] and go out in Y[i], we don't need to search all over X to find a nearest portal, we need to search in X[from 0 to i], so if we specify bisect(X, Y[i], hi=i) time complexity for all calls will be: log(1)+ log(2) + log(3) + ... + log(n) = log(n!) n / 2 * log(n / 2)

i had a very different idea to solve this fast. from a mathematician's perspective. every portal has a length (distance from exit to entry) and a number (how often it teleports the ant). both are calculated easily: 1) distance is trivial 2) give a value of 0 to all closed, and a value of 1 to all open entries. now start from the rightmost entry and add its number (0 or 1) to all portal entries enclosed between its entry and exit. do the same for the second entry from the right (add the updated numbers of course, not the starting value!). and so on. 3) just calculate number*distance for each portal, add everything to the base distance. ready. this works out to give the correct numbers of teleportations. each portal can only affect the numbers to their left, so for the rightmost portal it is trivially true. now thinking it through from right to left, you can see that the 2nd portal is visitied again if and only if the 1st teleports the ant somewhere behind it. all the other portals don't matter. the 3rd is visited again as often as the 2nd or 1st teleports the ant behind it. and so on.

thanks waiting for more such content

I learned so much!!!

Please keep making these!

Great work man... Hat's off

dynamic programming is a weak spot I need to work on for coding interviews. Always good to get more exposure on the subject