Inculcation | Comp Prog Reaching 2000 Rating | Part-153 | STREAM

[00:00:01]And I'm live. That is pretty cool. Today is 21st of July 2026, Tuesday. And with that, let's just begin. Uh what we have to do is just solve some CP31 problems at 1300 rating. That is not that is lower than my current rating. But I'm doing these so that I can get back to I guess the spirit or whatever you want to call it of competitive programming.

[00:00:30]Cool. So this was the problem we were doing uh day before yesterday I guess on Saturday but uh I was not able to solve this but then I read the solution for this and it is it is using some maths.

[00:00:43]So I will first of all write the solution and then we can discuss what it actually does.

[00:00:54]update whim packages and yep sync with the latest packages.

[00:01:02]Cool.

[00:01:05]And here cool. Uh remove this. And here we go.

[00:01:17]Competitive programming. Yep. This was the thing that I was writing earlier.

[00:01:23]And now let's just remove this whole thing and see how the input is being given to us. It is being given to us as a string 13 string s and an end n= s dot size sounds better and I guess is on. No, codium is not on. I changed it. Yep, it disabled. Chium is disabled. problem name is make it alternating make.

[00:01:55]Yep, that is all.

[00:01:58]After taking in the input uh let's have long long response initially is equal to one pool and int M that is the number with which we have to take the modus is this one P cool whatever the answer is would be modulus of this number and here we have to do a very like cool thing L equ= 0 R= R also equals Z. Let me screen increase the size. While R is smaller than N. L= R. After each iteration while R is smaller than N and S L= SR. What we do is increase R. That sounds better. And rest multiplied by equal to the size of this token which is L into R. and then take modulus. Rest is long long. So that is pretty cool. Take modulus of rest with M. Okay, that is just multiplying everything together and minus one would be cool. Minus one would be cool. Yep, minus one would be cool.

[00:03:12]And then we have to take factorial of the number of elements that we are not going to remove which would be let's say here we will have K and C count initial equal to zero. 20 + count increment table. Cool. And here uh the final response would be multiply final response would be rest multiply by factorial of count. And how do we calculate factorial of count? Count can be big. I I guess there is no problem if we genuinely without any pre-programming go and take count steps to find the factorial of count.

[00:03:58]Yep, there would not be any problem.

[00:04:01]We can simply write a function here in or that would return long long.

[00:04:10]In fact, it's taken it takes in an integer I and long long press because initial equals 1 y in i= z with int x= 0 or we can say that this takes in x instead of i= = to for in i= 0 i is smaller than x i ++ or smaller and n equal two and we will not start with zero that does not make sense start with one and rest multiplied by i multiply this by i okay and then take its modulus with m and this might give error because this modulo operation is crazy uh yep so here we can have a this mod this operation takes time so this might give us error like time complexity error Hello. Entering college this year.

[00:05:11]Any advice for competitive programming?

[00:05:14]Uh just keep at it. Have a plan. Do not look for the best resource out there.

[00:05:21]Just pick one thing, one or two things and continue with them. That would be the best advice I would say. And yeah, um go as far as you can in with those few chosen resources.

[00:05:36]Like for me I am going through this CP31 sheet and for problems this and there is this also CSCS sheet. These two are the only things that I use for problems for finding out problems. Yep. Cool. Just have one or two things and follow them regorously. Do not just keep stretching from one thing to another. That would be cool. Okay. And now here if this is greater than or equal to m only then find this modulus.

[00:06:11]Cool. Makes sense. And at the end return this makes sense. Cool. Here cool. Uh get factorial of this thing. Cool. Multiply these two things together and take modulus with M. Cool.

[00:06:34]And here print see uh rest with a new line.

[00:06:40]That makes a lot of sense.

[00:06:44]Okay. Test. And should we have to output two things? Yep, we do have to output two things. And whatever we did output that was also wrong. [ __ ] Uh we have to first of all output CN.

[00:07:00]CNT. No, we do not have to output CNT.

[00:07:03]We do have to output not CNT but something similar to CNT. That would be N minus CNT space and response. Okay, I suppose the first part would be correct. Yep, the first part is correct. But the second part is not too sure.

[00:07:29]Why is this wrong? I guess because minimum or maximum of this or one makes sense.

[00:07:45]Yeah, makes sense.

[00:07:51]That is also wrong somehow.

[00:07:54]Okay.

[00:08:02]Maximum of R minus L is the range one is there. Take modulus.

[00:08:10]Okay. Factorial of C and T.

[00:08:17]Nope. Factorial of C and D. Is that what we want? Yes, that is what we want.

[00:08:22]Factorial of C and D.

[00:08:33]factorial of India. That is exactly what you want. Oh, like no problem. And also your name is Punu, which is cool.

[00:08:42]I mean, you're not sure if that is your name, but okay.

[00:08:48]Use an integer I so that There is only one way. Yep, that does make sense.

[00:09:01]Here we do have to do two things. What is the problem?

[00:09:06]Let's see. Rest.

[00:09:12]Okay. So that rest never becomes equal to zero. That is why we add this here.

[00:09:26]multiply S which is whatever it is and multiply that by the factorial of C and T.

[00:09:34]Shade no not by the factorial of C and D but N minus C and D makes more sense.

[00:09:44]Cool. That is much better. 2 and 4 p.m.

[00:09:48]Why do we see four there?

[00:09:55]See out or count? You should not call it cout. Okay. Multiplied by two * 2 factorial of 2 1 into 2 3 into 2.

[00:10:35]Okay. Yes. Uh we do not have to do minus one here. Cool. Makes sense. We don't have to do minus one here.

[00:10:50]Okay, cool. That gives us the correct answer.

[00:10:54]And should we submit this? Uh, I don't want to just submit this. And then see that something went wrong. I guess we should just delete the whole solution and rewrite the solution. That would make a bit more sense. I suppose input string s. Okay. N would be and N would be the size of S.

[00:11:19]Cool. And let's say long long rest would be the final answer that we see. And now we have to go through different blocks of zeros and ones. And L= 0, R= Z while R is smaller than N and L= R after each iteration.

[00:11:40]Cool. But something is wrong. Expected something somethingwhere. Why?

[00:11:48]Oh god.

[00:11:50]Uh okay.

[00:11:53]While R is smaller than N and S L= S. R what do is R ++? Okay. And rest is multiplied by this range which is R minus L. And then we take mod what is of rest with capital M. What is capital M? M is M is the number with which we have to take modulus for the answer for calculating the answer. Okay, there.

[00:12:29]Cool. There you go. Cool. And we also need to keep count of how many things or how many blocks we see which we can do by this. Cool.

[00:12:44]Okay. And okay. Okay. Final answer would be this would be equal to rest. What we have right now that multiplied by factorial of n minus count that is n minus count is the number of digits or characters that we do have to remove from a string s. And now we have to implement the function fact int fact takes in a number x and returns the factorial for that number. Let's say long long ars equals zero or one makes more sense. Would this give error? This does not give error. Yeah. Uh there would be a automatic conversion from long long to integer which is the return type. Okay.

[00:13:33]Then for in i= 1 initially while i is smaller than x i ++. Okay. What do we have? I'll multiply rest by I and if rest is greater than M only then take modulus of rest with make sense get the modulus here and then output first of all n minus cmnt which is the number of steps then please and the actual answer which is respond this should work in first try and it does not work in first try. Something went wrong.

[00:14:23]3 * 2 factorial. Is this cool? This should be working.

[00:14:28]Take modulus of this thing.

[00:14:31]M or we could have done that right here.

[00:14:34]Modulus with M. Makes sense. factorial of n I guess factorial function is wrong.

[00:14:44]Yep. Wait, factorial function is wrong.

[00:14:47]Yep. This has to be less than equal to x. And now we should be okay.

[00:14:52]Which we are okay. Cool.

[00:14:55]Submit. I guess we have to know the problem number which is 18 97. See 1897.

[00:15:03]See cool.

[00:15:05]G++ compiler 23. Yeah, no such problem is a crazy error. 18 79 C 189 not 97.

[00:15:19]Cool. And it should work or it should work. Okay, cool. It does work. Cool. And what was the maths behind this thing? The maths behind this thing would be something like this.

[00:15:36]We have different blocks of elements and the problem is whatever the sizes of these blocks are.

[00:15:46]Let's say this one is s1 s_ub_2 that is how many elements are in these blocks.

[00:15:52]S3 elements in this block s4 elements in this block and out of these s1 elements we have to keep just one and remove all the other elements. Here also we just have to keep one and remove everything.

[00:16:06]Same here. Same here. So in total from this we have s1 number of choices of elements that we could keep. Multiply that by s_ub_2 which is the number of choices we have from this block.

[00:16:19]Multiply that by s3 which is the number of elements in this block. Multiply that by s4 which is the number of elements in this block. Cool. So that is the total number of possible ways to choose elements from these blocks.

[00:16:33]Right. Cool. And what you have to figure out is for each of this way what is the number of possible ways in which we could choose the elements which we are removing. And if we are uh let's say the total size is n and from that we are removing we will be keeping only four elements that because we are only keeping one element from a single block and there are four blocks.

[00:17:02]So in the end we will be keeping four elements. So in total we are removing n minus 4 elements. And the number of ways to order n minus 4 elements is n -4 factorial.

[00:17:18]That is how you choose n minus 4 elements for one possible selection of elements to keep. And in total there are this many possible ways. So multiply this by this number two. Copy. And that is what we are exactly doing here.

[00:17:39]We are calculating the size of blocks multiplying them at one place. This gives us this number and then we are multiplying this number that we got with the factorial here and that is giving us the right answer. And also we are taking modulus with m at every step where we could be exceeding our limits.

[00:17:59]Let's move on to the next problem or make it alternative alternating. Now strong vertices is a crazy one. [ __ ] I guess we did solve this and we did solve this. Cool. Yes, we did solve this uh on Saturday. Let's see uh let's see how we can still of course we can still solve this uh yep uh let me just read given two arrays a and b both of length n elements of both arrays uh I guess what we did was we reordered this equation not that I think that is what we did we reordered this equation and here we have something like there is a edge from the youth vertex to the V vertex. If this thing is true, V u minus B v. If this thing is true, then there is a edge from U to V. Yep.

[00:18:55]From U to V. And we have to figure out how many strong vertices are there, I guess. And which one are those? Yep. The number. And which one? Yep. Cool. Uh two things. So we what we did was we reordered this whole thing. We did a u minus b u has to be greater than equal to a v minus b v and we converted the two arrays a and array b into this form a u minus b u. Okay, let's do this again just for I guess muscle memory. Remove the old solution. Cool. We have end n I guess as the first input and then the two arrays you would make sense but uh of integers a size n and we just have a single thing diff no we'll take uh a and b for int reference x and a take in this and B and then we'll have a vector of pair of integer we'll call it diff it will have n elements pair of integers tool for an i or for auto reference x comma i in diff. That makes a lot of sense. No, that does not make a lot of sense. We have to we need the index. So, we cannot iterate like that. I is smaller than N + what we do as div or we will say here auto reference X uh V Y= I cool and X would be A I minus B I and Y would be I cool then we will sort this div dot begin and div dot end That makes sense. And now for the strong vertices we will see in uh maximum is diff dot end dot end dot second I guess. Yep. def.

[00:21:46]Okay, we would have to like go to the uh if end gives us an iterator and we have to go to that iterator int max and while int i= n - 1 while i greater than equal to zero and if i dot dot second equals uh diff I dot first equals mx. What you do is I - makes so much sense and then we can just start operating.

[00:22:42]uh see out first of all the number which is n minus how many things do we have?

[00:22:46]We have n minus i things.

[00:22:50]Is that correct?

[00:22:54]That is no n minus i - one.

[00:23:06]N - i n minus i is n minus i is Oh [ __ ] No. uh n minus i is not good. N - i - and for i is smaller than n i ++ what do we do is see out if i dot second which is yeah something something and output is space or a new line depending on i= minus one that should work or that should out of bounds error that should not give out of bounds error a minus b I y okay and then we sort sorting would not give out of bounds error finding the maximum also would not give out of bounds error I this should also not give out of bounds error while I is greater than is done and first equals max minus cool makes sense and then see out let's output instead of this end of line let's see is this getting printed it is not [ __ ] uh yep this not getting printed. Hello. How to learn new topics in CP brother resource. Okay. Uh depends on how much depth you want to learn. There is CP elbows this website you can refer to this.

[00:24:56]There is there are other things too. If you want to go in like deep like literally very deep then you can refer CLS. This book intro to programming it is called CLRS.

[00:25:10]This is also very good. This is like and this is the canonical book that you would read if you want to be like one of the best in competitive programming in that topic. If you want the literal like everything you can learn about a specific topic then you would read this book and there are some YouTube channels depending on the topic and yeah cool.

[00:25:37]So this is giving out of bounce error somehow.

[00:25:44]Is it printing anything? This is printing everything.

[00:25:48]No, this is not printing anything.

[00:25:50]Standard error is only thing that it is printing.

[00:25:53]Cool. Sort we sort and then we don't see anything. So we are not getting to this line which means error is somewhere before this somehow.

[00:26:09]How could we be erroring before this?

[00:26:22]A and B. Yep. Initialize A and B. Take an input and then define a pair. Cool.

[00:26:27]Makes sense.

[00:26:29]Everything makes sense.

[00:26:33]Uh sort should also not give any error.

[00:26:38]Maximum is also fine. Get n minus one first.

[00:26:42]See out i with end line is also not getting printed somehow.

[00:27:00]Okay. Uh we got high.

[00:27:04]How did we get high like twice? That does not make Okay. uh one high we got from here.

[00:27:11]So we walk to this point and then we error at some other point depending on how many highs are printed we can deduce answer. Okay we said we got only one high this time.

[00:27:28]H there are stupid people involved.

[00:27:34]We are not taking the input. Uh yeah, we are not taking N. That is the one thing that we have to take always.

[00:27:43]Okay. Uh but still there is something wrong. Okay.

[00:28:00]H what is still wrong? Why are we still getting out of bounds? Uh I minus uh I ++ is correct. No problem with that.

[00:28:13]While I is smaller than N. Go to the second value. Cool.

[00:28:20]We are indexing I here while I is greater than. Yes. Cool. No problem.

[00:28:27]There should not be any problem with that.

[00:28:32]rerun. And why are we getting out of bounds? Does not mean since one. And why did we say two here? And also here we should say like I that makes more sense.

[00:29:04]Do not go to minus one. That makes sense. Do not go to minus one. That makes sense.

[00:29:17]That makes sense. That makes a lot of sense. 2 3 4. Okay. In case of one, why are we just printing one and nothing?

[00:29:28]else see out MSI. I I equals something and we are printing this number twice.

[00:29:47]Wrong output is crazy. Uh why do we say two here? Even though we are seeing one there 235 35 is crazy. Okay.

[00:30:06]Maximum is first makes sense.

[00:30:12]Okay. [ __ ] We can uh R end R begin and R end and let's skip this whole crazy thing and vector int response.

[00:30:26]Yep. Okay. And int first is diff zero dot first. Okay.

[00:30:41]While for i=0, i is smaller than n and def i= fs or div 0 dot first First I just plus and what do we do? Don't actually do anything. We just increment I. Nope, we do have to do something.

[00:31:19]Just press dot push back def I do second.

[00:31:28]Cool. Makes sense. See out rest dot size and then point in i= z. Uh we can here rename n equals size of this while i is smaller than n i ++. Okay. C out i and print a new space or new line depending on i=.

[00:31:55]That makes a lot of sense and hopefully it just works even though it does not.

[00:31:59]Okay, something messed up happened. And okay, we have to print them in ascending order in one thing.

[00:32:09]We can start from the end n minus one i greater than equal to 0 and - here if i = 0 then print new. Good. That should give us the correct answer and it does it does give us the correct answer.

[00:32:25]Cool. I recently started C guide for beginner. Okay. Uh if you want a guide then you need just two things. One is uh place to get problems from uh where you do not have to go and look for problems.

[00:32:41]A a dedicated place where you could just go and simply look at which problem to solve. For that you can follow this sheet CP31.

[00:32:50]It is pretty cool. go here and since you are a beginner you would go to 800 rating problem start solving from the top and go through all these problems then um go and solve 900 rated problems and so on and this is for questions and for theory you can go to CP handbook this is a pretty cool thing go there and it covers a lot of things like introduction can go through introduction what is time complexity sorting data structures in C++ and I would prefer you should use C++ as your primary language it is because C++ gives you a lot of low-level control over things but yeah and yeah I'm just biased towards C++ C++ C31 yeah that would be advice use CP31 and CP handle cool and let's move on to the next problem which would be we did make it zero wait it's not even right we were at 1300 rated problems cool uh strong vertices we did did that snowflakes uh simple version yes we did read this the last time and cool let's solve this uh let's see how much time did it take for us to solve it uh when I was solving this the last last year. Wait, these are ends. Okay, [ __ ] Uh, this one hour with editorial.

[00:34:27]Yeah, it took me editorial to solve this one year ago. Let's see how well I can do this. Now, Rudolph, this is the problem. We did read this on Saturday.

[00:34:38]This is one winter morning. Rudolph was looking thoroughly thoughtfully through the window, watching the falling snowflakes. He quickly noticed a certain symmetry in the configuration of the snowflakes. And like a true magician, Rudolph came up with a mathematical model for a snowflake. And I just remembered that I should post on X2 that I'm streaming.

[00:35:01]Yes.

[00:35:04]Cool. Uh [ __ ] Uh there were like 2,800 frames dropped while streaming. It is crazy.

[00:35:13]Cool.

[00:35:16]uh he defined a snowflake as an undirected graph constructed according to the following rule. Initially the graph only has one vertex, the one in the middle. Then more vertices are added to the graph. The initial vertex is connected. The initial vertex is connected by edges to k new vertices. K new vertices whatever k could be. Each vertex that is connected to only one other vertex is connected by k by edges to k more vertices. So we are just every k vertex. Every vertex is connected to k more vertex. This step should be done at least once. The smallest possible snowflake for k= 4 is this one. And then in the next step if we wanted we would add four ver the four new vertices for each of the this leaf vertex.

[00:36:11]Cool. And our goal is to determine whether a given number like this or this uh whether it is possible to have this many number of nodes or not.

[00:36:26]And this is some crazy problem to solve.

[00:36:30]Yes. How do we figure something like this?

[00:36:37]How do we figure out? Okay.

[00:36:40]And the smallest value for K is what is the smallest value for K? Smallest value for K is 2. Okay. for k equals to the smallest the smallest like literal graph would be with the smallest value of k and the smallest value of k is two. So that would give us something like this and these two would have two two vertices.

[00:37:08]So this is the smallest possible graph and this has three seven vertices in total. So if we have a graph which has less than seven vertices that is some that is not possible uh yeah that it won't be possible to make a graph like that and then we would add 2 3 4 5 6 7 8 more vertices. Okay, I guess given k equals something, we can figure out what would be the number of vertices at I guess uh we can divide the problem given n we have to figure out whether for some value of k. K being between two two being the smallest value of K two uh the maximum value of K could be something the maximum value of K would be 10^ 6 let's say uh it does not make sense to have K more than 10^ 6 cool so the maximum value for K is 10^ 6 okay so we have to make a function which given a number K can tell us whether It is possible to have a graph with that k that gives nodes.

[00:38:31]Okay. Given a k this uh we can apply binary search here. So for the smallest graph for a given I guess we can write a we can write a a or something like that here where the base a would be 1 + k + k² k² yes 1 + k + k square.

[00:39:08]Uh for each of these k we have k. Yeah.

[00:39:13]So that would be a and the next term should this is not a uh this is not a a this is going to be a gp because we are multiplying we are not exactly multiplying but we are doing something more crazy. The next term would be k² + 1 that is + k cube and then we would have + k power 4.

[00:39:43]The smallest is this till k². Then we add k cq k^ 4 + k power 5 + k power 6 even with two. Let's see.

[00:40:00]I guess even if we have k equal to 2 which is the smallest value for k. How many steps would it take to get to 10^ 6 which yeah which can be the highest value for n. Again I guess we can just cycle through all possible values all possible number of nodes that we could have with a given k and it would not exceed the time limit.

[00:40:23]For k equal to 2, the maximum steps would be uh log 2 10 e 6 to get to 10^ 6. Cool. So we can just brute force this and that would be easy. Yep. Let's brute force this quantity. Okay. Uh the only input that we are taking is n c n and now the string. Okay. for int and i= 0. I small or int k= 2. K 2 is the smallest value for k smaller than or equal to n k++.

[00:41:05]Okay. Now we have to figure out if this thing is possible or not. And at the end see out. No.

[00:41:16]It is possible to get to n vertices with k.

[00:41:23]That's the base.

[00:41:33]Long long c equals let's say uh sum equals zero. And we'll need one more thing. C = 1 makes sense. So sum plus equals C plus first of all add C and then C * C. Uh yes. Wait, wait, wait.

[00:42:07]Multiplied by K equal add to sum. Sum equals K or C + C multiply by C. Okay. Add multiply C 1 by K or K here. Cool. This is the smallest value. Okay. While sum is smaller then or equal to n. What you do is if sum equals n exactly equals n.

[00:42:48]Uh in the case output this with a new line and return otherwise just uh increment sum. Do we increment sum?

[00:43:02]Increment sum by C and also C. Multiply equal to C.

[00:43:07]Multiply C by C. Okay, cool. Uh, which step do we do first? That is the question. I guess we do this first.

[00:43:22]Yep, we do this first and then we add two. That makes sense. That means sense that should work. Hopefully that should work and it does not. Okay, here we are saying yes. Even though the answer is no that is wrong. That is in the fourth test case.

[00:43:50]1 2 3 4 6. Okay. Six is no. We have to say no for six. But we are somehow saying yes for six. Why do we say yes for six? Let's see for which K we are outputting this two. When k equals 2 how do if sum= n how is sum equal to n. see out sum and uh see out C.

[00:44:34]Okay. Mhm.

[00:44:38]2 64. How is K is 2. How is sum equal to two? Sum was equal to 1 here.

[00:44:47]Then we added two to it. And then we added 2 into 2 4 6 + 2 + 1 should be 7.

[00:45:01]And yet it is not because we have to do plus. Yeah, cool. Makes more sense.

[00:45:10]And still wrong answer for these two. We say no. Even though the answer should be yes. The third last and yep these are very big. This one is not very big. 15 For 15 we are saying no even though the answer is yes let's see for 15 we are saying no even though the answer is yes how for which value of k uh for let's say we can do this manually for k equals 2 we made a graph which had seven edges the next value would be uh we added four would be four into two yep 4 and 2 + 8 15. Yes, straight. Okay, for k equal to 2, we do have the right answer, but we don't output yum is smaller than equal to n. I guess in the next step, we should be giving the right answer.

[00:46:11]Mhm. Mhm.

[00:46:16]23. Why do we give 23? 23 does not make sense. Why are we after 23 7 and then 13? 13 does not make sense.

[00:46:39]13 means we are adding only six even though we should not be adding six.

[00:46:44]Multiply by eight. Ah, my bad. Okay, we have to multiply C by 2.

[00:46:53]Now still it is 139. Yes.

[00:47:04]Okay, cool. That was the problem. And now submit. Let's submit this.

[00:47:12]Yo, Bo ping. You are pinging. Hello.

[00:47:16]1846 is the problem. And E1, this one is E. Wait, this is the simple version. The more difficult version.

[00:47:26]In the more difficult version, we have N could be bigger. Let's see how big N could be.

[00:47:33]Uh, problem was 18 4 6 even 1 8 4 6 E 1 even as in like odd even.

[00:47:44]What is the more difficult version of this? We can go to a tutorial hard version. Yep. This one time limit exceeded is crazy. Time limit exceeded is crazy. Here, how big can N be? N can be 10^ 18. Okay, that is that is crazy.

[00:48:08]That is like absolutely crazy. Uh ASMR keyboard today. Wait, is it is it good ASMR keyboard? Okay, that is cool. That is cool. But time limit exceeded here. [ __ ] Time limit exceeded. It should not be exceeded.

[00:48:26]Time limit exceeded. Time limit exceeded.

[00:48:29]Time limit should not be exceeding.

[00:48:33]That is Oh [ __ ] That is like the one thing which should not be happening.

[00:48:39]Time limit exceeded.

[00:48:44]H ASMR keyboard. Time limit exceeded with the ASMR keyboard. Not possible.

[00:49:06]Time limit exceeded.

[00:49:09]10^ 6. Our solution is taking n login and 10^ 6 is totally fine with n login.

[00:49:19]Yep.

[00:49:22][ __ ] This was written by AI. Uh this is m dash. Crazy.

[00:49:28]Time limit exceeded. Should not be brute forceful implementation math. Why are tags being shown here?

[00:49:36]tags should not be shown.

[00:49:46]Time limited. Uh okay, let's rewrite the solution. int n that is like the only way fix the thing and rewrite the whole thing again. Uh int n take n. Okay, and then we will iterate through all the pos possible values of k. K the smallest value for K is 2 while K is smaller than equal to N K + but what is smaller than equal to N that is K. Okay. And then we will have two values sum which would be equal to 1 initially and long long C which is current would be equal to K. All right.

[00:50:26]And the new value for C is plus equals and the original value plus C + C into K. And C is also multiplied by K. Now we will have an internal loop. This loop runs n times. And this internal loop will will not will run log n times. So while sum smaller than equal to n. Here we'll check if sum equal to n then z out. Yes, we found what we are looking for and return otherwise here sum plus equals or before plus equals C * K and some plusals C makes sense and cool.

[00:51:23]That makes sense. That makes sense. Um but if we exit uh if we are getting to this point then we are not able to find anything as the answer.

[00:51:34]That makes sense. That makes sense. That makes a lot of sense.

[00:51:38]That should give the right answer. That does give the right answer. But it is giving time limit exceeded somehow.

[00:51:47]How?

[00:51:51]Compare 1846 1846 E1. Let's submit 1846 E1.

[00:52:05]Submit. Submit. Yep. Show everyday is the main thing. Running. Running actually running. That is cool.

[00:52:14]Running on test case 2. And uh I hope it does not give time limit exceeded. But I guess it will give time limit exceeded.

[00:52:25]Uh-huh. Yep. It will like definitely give time limit exceeded. Uh time limit should not be exit exceeding.

[00:52:33]What is the problem with seven?

[00:52:40]Seven maximum number of test cases. And this maybe we can do some pre-processing for this. That would make sense. We can do Can we do some pre-processing for N?

[00:53:09]We should do some pre-processing. [ __ ] We We should do some pre-processing, I guess. And with some pre-processing, we would even be able to solve the hard version of this problem, which would be dope.

[00:53:33]Announcement here.

[00:53:41]around 883 is crazy. Uh this one with some pre-processing and by pre-processing I mean we can just iterate through the smallest value of K which is two to the biggest value of K which we'll have to figure out if it is not too big in the hard version. uh this one the value for n can be very big 10^ 18 we have to see which what is the minimum maximum value of k that would give a number that big which will definitely not be in uh in the range 1 e 9. So yeah still this would not work in the hard version but we would be able to do something in the small in the like easier version. So what we are going to do is we are going to iterate through all the values of k which is from 2 to 1 e6 we will iterate through all these values. Okay. And yep that is O of N time complexity. And inside within each iteration we will go through uh what are the possible number of nodes that a graph could have with this K and that would take login time to go through. So that total time complexity would be n log n which is well within our range and is something that we can do void pre-processing pool and we can have a bit set which is pretty cool bit set and the size of the bit set comes here inside this thing I guess ps that makes sense that does not make sense uh constructor or destructor how Do we initialize a bit set? This size comes here.

[00:55:42]Where is the size for the bit set coming from? Uh bit 10 C++ ASMR keyboard is cool. Okay, let's see.

[00:55:54]Yep. Size comes inside this thing. Cool.

[00:55:56]Makes sense, right?

[00:55:59]From integer bits set all bits initialized to zero. uh in this case but in this decimal 42 converts to by yeah cool good cool. You have to give the size inside this thing.

[00:56:16]Con maximum equals 186.

[00:56:27]Yep. Maximum let's make this 186 + 5 just in case. Okay. We will go through all the possible values of K. K equals Z while K is smaller then maximum K ++ and yep we will have some internal stuff for each value of K which will be long long current sum initially let's say equals 1 and long long C initially K and the first value of sum is sum ++ equals C + uh C into k. Okay. And y sum is smaller than smaller than mx. Makes sense. We will do ps sum equals to or we can write this in a more fancy way. uh take zo or base equals take the uh okay so bs some or take the or of current value of ps with one which simply is like one uh yeah no problem okay and then multiply C by K and sum is incremented by C would make sense that is the pre-processing that we will have and our main answer has now changed see out depending on BSN BSN is true then output yes Otherwise output no with a new line. That makes sense. That should give the right answer. And that does not give the right answer for these two.

[00:58:35]H that does not give the right answer for these two C * K sum plus.

[00:58:47]And sorry this is not the thing which is giving the wrong answer.

[00:58:54]Yep, that is not the one. Not the culprit. Your keyboard sound is kill.

[00:58:58]Wait. Uh, is it like bad or good?

[00:59:03]Uh, [ __ ] Please tell me. Is it bad or good? If it is bad, then I would have to do something very noisy. Okay. Um, okay. I will have to do something about this in the next streaming. It is very noisy. Okay. Uh, noise suppression is on. Uh, how is this one?

[00:59:36]This is like the better noise suppression. If this is good, then I'll keep this one.

[00:59:46]Okay. Uh, this is not giving the right answer. This should be giving the right answer.

[00:59:53]See out depending on BS and if VSN is yes then yes. If no then no return line. Cool. This all has to go inside a bracket.

[01:00:14]Cool. Very good. Okay. [ __ ] That is cool. That is cool. There were two options here. two noise suppression.

[01:00:20]There is RNN which is something and then there's the older version which is SP X some noise suppression thing and my internet was down. Okay, it is up and it is down. Why? Wait, what?

[01:00:40]Why is the net going down?

[01:00:43]Wait, what is downloading? What the hell is downloading? It is downloading something.

[01:00:52]I check my stream is going on here also.

[01:00:54]That is pretty cool. I can watch my stream.

[01:00:59]It is not giving the right answer.

[01:01:02]Un cool.

[01:01:11]Okay.

[01:01:15]We are getting wrong answer for 1 2 3 4 6 for the sixth one. This one 15. Why are we getting the wrong answer?

[01:01:30]15. Let's print something for I equals Z. R equals Z. I is smaller than N or there is no N here. I ++ see how PS I with some space uh I comma BSI with a new line.

[01:01:59]Let's see. Okay. Uh it is not possible to have all these seven. Seven is possible which is pretty cool. 11 is also possible in some way or form. But then 15 is never possible which is messed up. 15 is possible.

[01:02:16]While sum is smaller than MX increment summ okay because c has multiplied by a here and now we are okay probably we can let's see we are okay we can even submit this that is how much okay we 1846 I suppose 1846 that is correct 1846 E1 and submit and let's see hopefully we don't get time limit and we won't get time limit exceeded like because we are not yeah yeah we are like barely doing anything cool But is there a way by which we would be able to solve the more difficult version of this problem which would be pretty cool if we can solve that problem.

[01:03:26]And also before moving on let's delete the solution for this one. And we rewrite it again. We have to do some pre-processing. And constant MX = E 6 + somehow. And we need a bit set with size MX. We call it P as like possible or not. and some pre-processing for int= to while k is smaller than mx and k + k makes sense.

[01:04:11]Now internally we need a few things.

[01:04:13]Long long sum initially equals 1 and long long C= K initially sum plus equals C + C into K and also C is multiplied by K. Then while sum is smaller than the maximum value possible we will do BSI equals 1. It is full but there is no BS.

[01:04:39]There is only cool stuff and no BS uh sum is possible cool and after each step here C * Y= K multiply C by K and then sum plus equals C makes sense okay we'll do the pre-processing at some point you go take the first input pool and then here N take N And and when depending on and depending on whether n is possible or not we will output yes or we might output no we never know.

[01:05:29]Okay I does not exist but I exist.

[01:05:33]That is okay and that should like work here without no problem.

[01:05:38]1846 that is 1846 E1 1846 E1 here paste submit.

[01:05:54]Now uh it would be very cool uh to see about the more difficult version of this problem.

[01:06:04]this one.

[01:06:09]But I guess before that we should read the editorial for the easier version that we just completed. Uh this one hints.

[01:06:18]Yes, that is how yeah that is like the formula that we used. What maximum value of k can be reached for the current constraint on n 10 1 power e 10 power e 10^ 6 can we pre-calculate for each n if there is a appropriate k yes that is like exactly what we are doing for the current current given constraint you can pre-calculate whether it is possible to obtain each n for some k to do this we can iterate through all possible values of k from two to 10^ 6 and for each of them calculate the values all of these values where P is such that this thing is always small.

[01:07:01]Yeah, we will only multiply K till the point while it is smaller than 10^ 6 for this version of the problem it is enough to calculate values for P smaller than 20. Yep. Not that the maximum number of snowflake layers is three. Wait. Uh, for this, this only works for this version.

[01:07:24]But for 10 power 18, how many times do we have to go through this? 10^ 18. If we take log of 10^ 18, can we take log two?

[01:07:45]Two.

[01:07:47]How do we log out? uh our calculation uh before this before this let's uh do it on our side. Wait, I guess it should just be uh 10 with 2 power 2^ 64 is 1 e 9. And if you just square on both the sides and 2^ 128 gives us 1 e 18 and therefore instead of p here where they are seeing that p is equal to uh less than equal to 20 here p equals 128 h but the maximum value of k is very big even So uh yep but the maximum value of k is very big. Maximum value for k would be 1 e8 which is yeah we cannot iterate through all the values and this solution would not work for the hard version.

[01:08:51]So yeah [ __ ] this solution would not work for the hard version. We will have to do something crazy. Let's see. Uh we can read the editorial. It is fine. No problem with that and easy. Uh it would just teach us something. What maximum value of k can be reached for the current constraint on n constraint on n is that the maximum value for n is this this n maximum value is 1 e 18 10^ 18 okay and the smallest value The smallest number of nodes when given a k is 1 + k + k² k².

[01:09:46]So if you make this equal to the maximum value of n uh which is 10^ 18 h these two things don't count when we have this. So we can write k² nearly equals 10^ 18. And if we do under root on both the sides we have k here which is nearly equal to 10^ 9. So we might just be inside but then okay so we will have to iterate k times which is 10^ 9 but in each iteration we will do 128 steps that is what we saw above 128. Yep. which exceeds our limit which means the same solution would not work should not work technically speaking.

[01:10:48]See more hints. Can we pre-calculate for some n snowflake with four or more levels if there is appropriate K with four. What we can do is maybe stop at the point.

[01:11:10]Yeah, stop at some point. Let's see.

[01:11:13]Let's see. Let's see if we can use it like same solution here also.

[01:11:18]That would be cool to see.

[01:11:23]Okay. uh n taken n word but before n the main thing comes con long long math is 1 e 18 + 1 let's say or plus 7 let's say uh we need a bit set of size max which is messed up. [ __ ] Uh, yep.

[01:11:55]Uh, [ __ ] We cannot use a bit set of size max. It is just too big.

[01:12:02]Uh, that is too big. So, we will not be using a bit set. What we will be using is a set set of integers or set of long long numbers.

[01:12:16]We will call it false.

[01:12:21]It is not initialized here.

[01:12:32]I guess it is initialized.

[01:12:35]uh we will see uh if there is some error then for long long a= 2 while k is smaller than max what you max uh 1 e 9 + 1 k ++ Let's long wrong long smallest smallest is 1 + k + k into k which would exceed our limit.

[01:13:24]I guess it won't. I guess it won't not. Okay, good. And the other number would be long g which is equal to K and K while SM is smaller than or equal to max. What we will do is post insert SM saying meaning SM is possible to reach and then C * K makes sense and SM plus equals C makes even more sense.

[01:14:10]That should be enough pre-calculation.

[01:14:12]And here if pause dot s contains and see out password contains and question mark if it contains then new line otherwise nope that should work.

[01:14:42]Okay. Wrong answer is much better than no answer. So wrong answer is much better here. We are getting wrong answer here that is some internal problem.

[01:14:57]Okay. Uh wrong answer means we can go through all this precalculation which is good. So we should be able to do this difficult version also.

[01:15:14]C multiplied by K and SM plus equals C.

[01:15:18]That is giving us a wrong answer somehow.

[01:15:23]I guess the problem is with boss.

[01:15:30]We are outputting no for everything.

[01:15:32]Okay. The problem is then with boss.

[01:15:37]It is not with pause. We are not calling three here. Okay, that was the problem.

[01:15:42]And if this gives TLE then uh yep, it will it will give T. Yes. [ __ ] That is taking too much time. It will give time limit exceeded.

[01:15:57]Yep. Uh the limits are just too big.

[01:16:06]Why is the Wi-Fi messing up?

[01:16:23]I guess it is the hard version for some point for some. Okay. Uh let's see the next snowflakes with three. Wait, what do we what do they mean by can we pre-calculate for some n and snowflakes with four or more levels if there is appropriate k given an n can we pre-calculate for some n and snowflakes with four or more levels if there is appropriate k if there is appropriate k if there is appropriate k how can check with three levels separately. What does that even mean? With three levels separately tutorial, let's see tutorial. Uh just gazing from above.

[01:17:16]We will do some precalculation but some crazy pre-calculation on these constraints. It is also possible to pre-calculate whether it is possible to obtain certain values of n for some k.

[01:17:26]To do this we can iterate through all possible values of k 10^ 6 uh 10^ 6 10^ 6 10^ 6 for each of them calculate the values this p where p such that all of this is wait [ __ ] if that is where have they given the limit for k they have not given given a upper bound for K. That is why I went for 10^ 9 but we are seeing that there is an upper bound for K. It's messed up.

[01:18:04]Uh huh.

[01:18:07]That is true. I guess we can just say 106 here and that should work.

[01:18:19]That also does not work. [ __ ] That also does not work.

[01:18:26]Okay, that should have worked.

[01:18:35]What if we make this five?

[01:18:46]That should make the solution like 10 times faster. [ __ ] 10^ 5 is also too big.

[01:18:54]Now 10^ 4 [ __ ] that is also taking too much time.

[01:19:01]Uh 10^ 3 which is just like 1,000.

[01:19:09]Okay so that is also taking a bit too much time.

[01:19:15]How about 100 100?

[01:19:21]How is it even taking this much time? Uh yes. How is it even taking? It should take 10 power log to 10^ 18 which is equal to 128.

[01:19:46]Okay. [ __ ] Even that is taking too much time. There might be some problem with our implementation in that case. Some we are yes we are doing everything okay is maximum just like way too weak or maybe yes we also have this part where we are inserting into a set that insertion also takes some time.

[01:20:17]That session also takes some time.

[01:20:24]Compilement test case.

[01:20:27]Let's see how much time does it take to run.

[01:20:33]Okay, that is taking way too long. That is taking way too long.

[01:20:44]Let's see the editorial.

[01:20:47]for this tutorial.

[01:20:54]Cool.

[01:20:56]Cool. Okay. Yes, that is true. However, in order to obtain all possible values of n, we have to iterate through 10^ 9.

[01:21:05]Yes, that is what I said that yes, which exceeds the time limit. Uh therefore, it is pre-calculated all possible.

[01:21:14]Therefore, let's pre-calculate all possible values of N that can be obtained for P within this range. We will store all the obtained values. For example, in a set, we cannot use an array called used here due to the constraints on N. Yes, N is just way too big. It is better to perform this precalation before iterating through the test cases of the current test case. Next, for each test, we only need to read the value of n and check if obtained value is in the e4 calculation. If you obtain it, then we have yes, otherwise you have to know. If we did not obtain it, uh we need to separately check if we could obtain that value for t = 2. Okay. To do this, we solve the quadratic equation. This is uh if it has integer roots that satisfy the problem. Why do we have to solve quadratic information? What part are we cutting out?

[01:22:14]We are cutting out P = 2 and going directly to Q.

[01:22:23]Okay, we going directly to cube. Does that help?

[01:22:28]That does help. We can iterate to this.

[01:22:35]We can have a upper limit for k which would be equal to 10^ 6. It's cool. It's cool.

[01:22:43]Holy [ __ ] That makes sense. That makes uh like uh literally a lot of sense. But we messed up somewhere else. We messed up somewhere else. For k= 0 and k. Okay, there's no problem with this. We can change this to see. Okay, sum is initially equal to this. Yep, that is also K.

[01:23:04]We should say K long long and here also multiplied by one long long that might have been a problem state that literally was the problem I guess.

[01:23:21]Multiply by long long multiplied by long long pause.insert Insert SM.

[01:23:27]Are we able to pre-calculate for this much?

[01:23:47]We should be No, we are not. We are still getting time limit exceeded. So we change this to do.

[01:24:01]We should get a wrong answer but not time limit exceeded. And we are going to get time limit exceeded.

[01:24:08]Okay, that is messed up. That is literally uh quite a bit messed up.

[01:24:20]K is long long which is fine.

[01:24:25]Still K is long long. So uh there was no point in doing uh this part or even this long long hair. But now K isn't. So it makes sense to have one long long here and here also why are we like literally not able to do this thing. Uh we are also like literally we are incrementing SM uh and inserting into pause.

[01:25:03]It might be pause is the problem.

[01:25:09]Let reference pause. Okay. And here sit pause should be given here also.

[01:25:26]And here set of long long post initialize call it here boss.

[01:25:40]What is needed is long long path here and this long makes sense. Makes sense. But I guess that was not the problem.

[01:26:09]Yep. That wasn't the problem.

[01:26:14]Go back. Excitement. U here. But why is it taking this much time?

[01:26:38]Uh okay.

[01:26:43]Let's are we able to do this for one value of K? Okay. Okay. Good.

[01:26:53]We are able to put something. Okay. Our five. Can we do this for five?

[01:26:59]We can.

[01:27:02]Cool. Our 10. 10 is also fine. How about red? That should take 10 times more.

[01:27:10]And that is taking 10 times more [ __ ] Why is it taking so long?

[01:27:25]I mean the bigger numbers should be taking a smaller time than smaller numbers.

[01:27:53]Okay, we are not even getting time limit executed. We are getting Yes, that was the problem. Negative, negative, negative. Our number is getting a bit too big for us to go. Golang plus back end.

[01:28:06]Uh, good for remote job as a fresher. Is it possible to land one? Your opinion?

[01:28:11]Golang plus. Yeah. Uh it is good for remote job. There have been some openings. I gave a interview for ago but I was not able to pass it. But yeah there are openings. It is good for fresher but no no uh as a fresher it is good to have them but uh it is not decided that you will get the job.

[01:28:29]The place that I wanted that I gave the interview at they wanted a experienced go developer who has three or four years of experience. But you might find fresher job. Uh it is good to have but don't make it your main thing because there are in quantity speaking a lot more openings for TypeScript jobs as a beginner and not a lot of openings for go jobs as a beginner. So yeah, but you can get a go job as a fresher. But yeah, the problem is just the quantity of the jobs.

[01:29:14]There's a lot more competition but also yeah uh a lot of people are not preparing. Go play chess now. D uh just let me solve this problem. There's something wrong with this. Then we can play chess. We can play all night if you want to. We can even play uh we can even watch a movie if you want. I want to watch this movie.

[01:29:37]Gul you can watch this and uh this movie is on YouTube. You can watch the like whole literal movie on YouTube.

[01:29:51]If you beat me, I'll download a new and delete. Okay. Uh the stakes are high. Uh send me a link. We'll play.

[01:30:02]We'll play. We'll play. Send me a U invite. We'll play a match.

[01:30:07]You will have to delete new vim if you do win and you will have to delete V code and download new V.

[01:30:23]Let's go speaking. Why do they have put scazard logo here?

[01:30:32]And in the meantime, uh in the meantime, we can switch to the main topic of the stream which is competitive programming.

[01:30:45]Okay. Something is getting way bigger than it should be getting.

[01:30:49]That's what she said.

[01:31:01]Did you watch Odyssey? Yes, I did watch Odyssey. It was very good. It was very good. Uh, this one was uh much more psychological than the past Nolan films. And this one was much more straightforward than his past movies compared to something like Tennet or Open.

[01:31:33]This thing, this word can even be read as P.

[01:31:37]That would not be a mistake.

[01:31:51]pause entering. Okay. Uh then file SM is smaller than this.

[01:31:59]Makes sense. We need a bigger number or we need to make this thing smaller.

[01:32:28]smaller than max.

[01:32:31]We should not uh overflow and sm is greater than zero.

[01:32:46]Okay. And my laptop just got stuck or at least the new part got stuck. So should I start TypeScript? I already started any language actually and want Blender remote job is possible in one year or one and a half year. Would you recommend? Yeah. Uh yes, you can start with Typescript because Typescript has a lot more opportunities than Go and then you can increase your range or domain you can do some domain expansion stuff and learn and go afterwards. But yep uh if you want to get a job then TypeScript is your best option. There are a lot more TypeScript jobs. Then after you have some exposure to programming and stuff, you can choose more languages and start targeting niche things like Go or Rust.

[01:33:40]If uh even you delete any of them, okay?

[01:33:43]No. No. Uh no. Uh I will not play chess with you. I will not talk to you if this is the deal. No. Nope. I will not take this. It would be crazy if you accept.

[01:33:51]No, I will not accept this challenge. It is crazy. Yep.

[01:33:55]I don't know how to use VS Code at this point. I forgot it is easier for me to use new web.

[01:34:04]The game is on. Cool.

[01:34:15]Uh probably will win just by time.

[01:35:38]Oh [ __ ] Makes sense.

[01:38:01]Wait. [ __ ] Is this checkmate? [ __ ] Is this checkmate? Wait.

[01:38:06]>> I can see. Uh, I guess this is checkmate. Uh, I guess you are losing.

[01:38:11]You are going to lose this. How can you come out of this?

[01:38:18][ __ ] [ __ ] That this is checkmate. This is like literally quite literally. Uh, you will have to sacrifice your queen.

[01:38:26]Oh yeah, you will have to sacrifice your queen.

[01:38:32]Yo, cool stuff.

[01:39:37]Attack key. Attack key.

[01:40:36]Oh [ __ ] Uh I don't lose this on time.

[01:40:45]Wait, why was that approve? Okay. [ __ ] It is your choice.

[01:42:17]I really do not know how to convert this into a victory. Uh, yep. Okay, fine.

[01:44:15]Okay. Flag. Uh, GG. That was good. Uh, that was good. Bit crazy. Okay. Uh, okay. We got it. By the way, uh, what they typically look for when they are hiring for remote roles like uh making good projects, opensource, building in public and posting about do they hire based on things. Yes, they do hire based on thing.

[01:44:37]Yes, I did get you there. They do hire based on above things but mostly when you have joined they would be looking for responsibility. That is one thing that I have come to know after actually being hired. So they will look for responsibility but yes if they know that you are a new like fresher then they will let pass different things and these would be the parameters on which they would be hiring you. If you do get an interview, there are two things. One is getting an interview and then the second thing is passing the interview. So getting an interview is a different thing. Uh for that you need a lot of things on paper and after you have gotten an interview.

[01:45:22]It depends on your knowledge on about different things like uh database, networking, there is a lot of studying to do how databases work.

[01:45:35]networking, what is SSH, public private key pair and stuff. What is TCP? There's also something they might ask or they might ask you about DSA stuff, competitive programming as you are planning to do. Yep, there's a a lot of stuff there. And do you see it bro? Uh this is the first time that I defeated you, but this was not a real victory because first of all you blundered and second of all I didn't checkmate you and that was just lagging. So yeah and it was not like there's no cause for celebration will be killed still. Maybe I could have won by time. Yes, you could have one late night. Uh this is like my morning stream.

[01:46:25]Okay. So I can do you send me a link whatever. Uh this is my username Jan Raga 50. Send me a friend request and then we can do it. Let's see. And that would be the end of spring that [ __ ] Why is it getting stuck? Somehow new is getting stuck.

[01:46:51]This is not working.

[01:46:59]Yep. Start.

[01:47:03]Nothing is moving.

[01:47:06]Okay, cool. Uh, no. Uh, second time, but uh, it was my second game. Oh, [ __ ] Uh, this is me defeating you second time.

[01:47:13]Cool. Uh, bro, actually, you know, I had a chance at last get your queen and one move on move 27. How? On move 27.

[01:47:28]How I made sure that your knight was not doing anything crazy. How you could have done this two-way attack on my queen. I could have moved and yep, your knight was not going to give me.

[01:47:47]Wait, what? Uh, I'm not muted. I'm speaking.

[01:47:52]Rook check.

[01:47:54]Okay.

[01:47:57]Rook check. And then I move this.

[01:48:02]Or you mean like after? You cannot.

[01:48:04]There is something wrong with your headphones. Then king to H2. Yes. Then king to H2. Rook check. King to H2.

[01:48:13]Oh [ __ ] Okay. Makes sense. Makes sense.

[01:48:17]Uh makes sense. Uh makes a lot of sense.

[01:48:21]Makes sense.

[01:48:24]There is probably something wrong with your headphone. Uh voice is going from my side or there might be something from wrong from my side also.

[01:48:33]There is a possibility.

[01:48:36][ __ ] That was cool. But that is also like hard to spot. It is a uh yeah this this this and yo let's go start from today. There is no tomorrow.

[01:49:01]Uh F1 is better.

[01:49:08]there. Check.

[01:49:11]Could go there.

[01:49:13]And then take and then take.

[01:49:19]Okay. Yo. Uh, you now have ears. That is cool.

[01:49:25]How is F1 better?

[01:49:27]You mean Louis Hamilton or Maxwell Sto?

[01:49:35]In this case, my king would be here. But if we went for F1, my king would be on F1 at the end. That would be the only difference.

[01:49:48]That is why F1 is better because my king would be at the back side and here my king would be kind of in the center of the board and would be controlling more things. So that's why F1 is better. Too many ads.

[01:50:02]Uh And that is just so many money. So much money. Hello Archra.

[01:50:08]Uh it has been a long time. How are you?

[01:50:10]How is everything going on? You are from Borakur, right? I know that.

[01:50:17]Yeah, we both equal. And where is your chess? Uh D. Uh [ __ ] You have not sent any request. Lavender, bro. Yes, yes, yes. Send me the request. I saw his two moves. Uh move later.

[01:50:30]Oh [ __ ] You were able to discover this on your own. That is crazy, man. I lost all motivation doing nothing nowadays.

[01:50:38]Wait. Uh, this sword. I thought Kilawa was your alt account. Okay, I guess I was wrong.

[01:50:51]You saw this two moves later. [ __ ] That is crazy.

[01:50:55]What happened to your chess project? Uh this project is there kind of there are some bugs in deployment you can go play rapid blitz or bullet or whatever you want and also not local host codes if AWS has not yep it is also there but uh there are bugs in deployment some simple le bugs which is if you go to networks the requests log. Yep. The requests are going to local host instead of the actual back end which is messed up that is the current state of things in hello past lender.

[01:51:54]Now send me a uh request for game and yeah I switched to Zen browser. It is pretty cool. Zen is pretty cool. Uh it is lagging a bit I guess because of stream or something.

[01:52:07]Why is 100% of the GPU being used somehow it is being used.

[01:52:14]Laptop is hot. Yep.

[01:52:16]Docker is uh running. Uh docker is running. I can turn off docker. How do I turn on Docker container from the terminal?

[01:52:41]Uh, pseudo system C docker. Yeah, but I guess this won't work because I'm on Mac. Yep. System CDL. I am on Mac.

[01:52:55]Docker desktop start.

[01:53:00]Wait. There are other commands for desktop. Yes, I can start Docker from here.

[01:53:08]It is already running. No problem. Then stop.

[01:53:18]Stop. That will stop.

[01:53:21]My internet is so bad. [ __ ] Then how are you able to see the stream ad is messed up. Uh maybe I'll fix them uh when I get them. Oh, okay. That would be cool. Uh I will merge your free pull requests.

[01:53:41]I can uh play send link works in deployment. I am a professional Microsoft developer without uh even working in Microsoft. Wait, no.

[01:53:54]I made a project recently and order matching engine. Okay, I was able to reach 49 to 51 million operations per second, which is pretty crazy.

[01:54:06]I don't understand the maths of operations per second.

[01:54:12]8 49 to 50 million operations per second.

[01:54:17]Yes, that is very good because in competitive programming it's entering in competitive programming if a loop is running uh like 1 billion times that we take it as it would take 1 second to run and 50 million is like 20 times less than 1 million. This is pretty cool because those things are written in C++.

[01:54:50]Okay, this is pretty cool.

[01:54:53]Because uh if it were more than this uh you would be hitting theoretical limits.

[01:55:13]Scandinavian is what it is called.

[01:55:17]I don't know the variation.

[01:55:51]There was no point still early That's the one.

[01:56:15]Yo.

[01:56:16]Uh.

[01:56:18]Yo.

[01:56:23]No [ __ ] Uh, you have that.

[01:56:26]Let's move. Let's move. Let's move.

[01:56:30]That's I thought I got your queen.

[01:57:18]Goodbye.

[01:57:45]My company people are messaging me.

[01:57:48]Boss, the co-founder.

[01:58:22]That move made like perfect sense.

[01:58:26]Uh, you open a company. No, I joined a company. I am employed now.

[01:58:34]Why are chats coming late?

[01:58:38]Uh cash misses were as low as this much.

[01:58:44]Let's have some theoretical studying to do.

[01:58:49]Let's open the screenshot.

[01:58:52]It is deployed. But there are some problems with the deployment. Is there somewhere I can play without for? Yes, you can. No, you cannot play. Yes, thank you.

[01:59:03]I joined a company. I'm employed now.

[01:59:12]That is why I'm not in my home. I am somewhere else.

[01:59:32]That would be a very bad No, please.

[01:59:56]Yep. After learning TypeScript, should I learn Go or do learn something between?

[02:00:04]It's uh that would be you would be thinking about that after you have learned TypeScript, right? And also not just learning a language, you need to learn the whole ecosystem. a bit bit by bit like you would have to know what happens when you press enter on a URL in a browser. What actually happens?

[02:00:25]What is DNS? Just basic networking stuff. It would help a lot.

[02:00:38]I don't know what to do. [ __ ] [ __ ] Okay, I didn't see this. My bad.

[02:05:20]also lavender. [ __ ] you are from America. Uh, okay. I didn't know I I [ __ ] up. I actually up here.

[02:05:46]Yeah, there are no moves.

[02:05:52]Literally speaking, I got a piece. I messed up, bro.

[02:06:03]chat.

[02:06:05]There is chat here also.

[02:06:08]Dramatic. I'm from Pakistan. My country is set to America for some reason. Oh, hello.

[02:06:15]F people are watching.

[02:06:29]That didn't even do anything.

[02:06:58][ __ ] And now I do the last call.

[02:08:10][ __ ] Mhm.

[02:08:39]Uh, I guess you better. Uh, yes.

[02:09:08]Okay. Uh yeah, you will pass.

[02:09:43]Okay.

[02:09:45]Yeah. Segmate, segment, segment, segment, segment. That was good. That was good. Uh, GG. Where do I write? GG.

[02:09:51]Jeez. Let's go. That was good. That was good. Uh, one win and one loss.

[02:09:58]Cool.

[02:10:00]I'm from Pakistan. My country reset to Amelia. Yes, that is something crazy.

[02:10:04]From almost winning to losing lol. Yes.

[02:10:06]Why do we make sound? Uh, easy, bro. Yes, that was good. Uh, no, that was not good. It was very bad. It was very bad. I need to work on my chess skills once I once I'm better in programming, of course. Like that is the main thing, the main task. And yeah, cool. That I guess would be it for today.

[02:10:30]What did we do? We uh did some problems from here. It this was like very inefficient.

[02:10:39]Uh yeah, we did this this and [ __ ] strong vertices and make it alternating but not new problems. We did redid them on Saturday and yeah today we just revised them and this is the only fit.

[02:10:56]Even this one is something that we did on Saturday. We didn't do any problems today. That is messed up. [ __ ] That has to change. That will change starting today. Starting from tomorrow, from the next stream, uh we will have to focus more on problem solving. And internet is bad.

[02:11:15]Internet is very bad. Someone is uploading something.

[02:11:19]Yep, that would be it for today. We didn't do any problems. We only messed around and that has to that is not good, that is bad. And thank you for being here.