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How was your week while I went to the offline gaeaort yesterday? Offline knee sports. Nice. We're game. Yeah, kind of right. Oh, your csgo player. Yeah. Are you any good? Probably top 50%, but. Very nice. Yep. I used to play cgo. I sucked. It was really bad at the game. My cousin played valorance. I don't know if you Noah. Yeah. And he was he was top 100 worldwide. He was doing competitions and everything he said it was really fun and I would watch him play. And Hebe like this on his chair. You know, like, Yeah, that's funny. Anyway, have you done any computer science this week that you want to talk about? Okay. Let's do more databases. Is that okay with you? Yeah, all. So what do you remember about databases? I'm actually going to give you the lead here and I'm going to ask you some questions. And basically what I want you to do, what I want you to do is to literally do long answer questions. Now I know it's not fun. I know it's not that fun, but it's really important that you get to do these long answer questions so you can use just typing on your keyboard. So the question is gonna to be, I give you access. The question is going to be, what is the difference? Between one nf, two nf and three nf. And another one. And why does? Why do we want the database to follow me? Right? So. Oh, Oh my God. So please try to answer this question. And I really wanted you to try and put yourself in exam answer mode where we really, really try to figure out what the answer is and you can just write it out underneath. And we'll do two of these about databases just to get you warmed up. And then once you've warmed up, we'll go more into detail on some other stuff that we haven't covered yet. You repeat what, one and f and two and F R? Yeah, absolutely. I mean, I'd rather not, to be honest, because I'd rather see how you would answer this question in a course. So just give it a try for one or two minutes. And if really, really you're stuck, I'll give a brief refresher of what, one nf and two nf and three nfr. I know it's been a while, so I don't blame you at all for asking, but try and answer it as you would see this question just randomly on exam today. And then if we really can't do it, then I'll do an example. And this question would be four Marks. Just don't know. Give it a try. If you were in the exam, what would you do? And if you saw this kind of question right now. I would probably skip this, but Oh, okay, let's say you had loads of time and you and you you know, what do you think these four Marks represent? The explanation for one enough, two enough, and three, enough. And and why do you want a database to follow the norwritten rules? Yes, I'm pretty sure you can answer at least one of those questions. You can at least get one of those points. Which one do you think you can get? Probably one and f. No, I think you can get the normative rules. If you've never seen one, two and n, three and f four, just think logically and think like very straightforwardly. Why do we want a database to follow normative rules? Let's say you don't even know what normative rules are, right? Let's say you have never occurred of the term normative rules. You can use context clues, right? We're talking about database. We're talking about one, two, and three and f. Maybe you've heard vaguely these terms are about organizing the database. So you can literally say this, right? And this would get you a point. You can say we want to follow normative rules as it will make the database more organized and efficient. It's not very complicated and it doesn't require much thoughts, right? It's just pretty straightforward, but it's a free point. This would be a point. And you don't even need to know a one, 23 nf R. So this is just just to tell you, like you know if you see a question you don't know, look at the amount of points. If it's a high amount of points, it's almost certain you can at least say something that's going to give you a point, especially at the end of the exam, if you've done all the other questions, you really don't know the answers to this question. You might as well put something random that could get you a point onto any questions about that. No, okay, now onto one nf, three and f and two and F1 nf, is that all values? Values are atomic. What does atomic? Mean means that it's to the very basic. Yes. So for example, if you have a table. With subjects. And you had physics. Would the subjects be atomic? No, Yeah because it has physics and maths in the same place. So basically all values that are atomic. And so and this is pretty straightforward as well, but it's it's you need to say no repeating groups. So for example, by the way, can you read my handwriting? Where's my handwriting? Really bad? Yeah. Yeah can read. Okay, okay, that's good. I ask because one of my students two days ago was like, what are you? What are you writing on? The bon the board reand. I was like, so sorry, but Yeah so repeating groups, so two and f, it's easy. Must be. One and f. So if you want two nf, you need to be one and f so already. That's right. And the second thing is you should have basically all non key attributes depend on a primary key. So here and all attributes. Depend. On pk primary key, right? So basically what that means is if you have like a two a two and f sort of of grid, you're not going to have a partial dependency where you're gonna to have to say this. Let me find an example, two and f example. Yeah itbe like the student situation. Let me actually copy the image, see if I can paste it here. Yeah, nice. So be kind of like this right here. You've got a non two f table, right? But if if you see here, you've got ID students, ID Prof, right? And then you've got everything that's tied together. Like why is the idea of the student the last name and the idea of the professor and professor in the grade all the same table? And you're a smart guy. You obviously know this. You can split this up, right? Obviously, the idea of the student and the last name should be its own table, and the idea of the professor in its last name should be its own table. So you can, if you saw this kind of question on exam, you can immediately know that. And the grades, instead of having everything inside one table, right, you can just have the grades where you have the idea of the student, the grade, and that's basically idea of the student, an idea of the professor, and then the grade basically. So. That's two and f and then three and f. So already there you've got three points, even if you don't know what three nf is. Three nf is a little bit more involved, but the basic idea here is it must be. Be at least two and f. So three and f must contain everything that was in one and f and two and f and you must also have no transitive dependencies, no transsitive. So what does that mean? Let me go back to this website actually because they've got a three and f table as well, and I think it's really good. No dependency between primary keys. Yes. Yes, there must be. No, I'm trying to paste it. Yeah. No dependencies between primary keys, right? So you've got that right. I mean, here the idea is that here you've got the ID of the vendor, right? And then here you've got the bank code number. Okay, wait, a me vendor and the bank code number and the bank you obviously the bank code number and the bank. This is its own kind of thing, right? So it's going to be unique and so you can split this up into its own table. It's a bit harder to spot for three and f. Let me try and find you a good rule of thumb here. I saw this three. Nice cool rule of thumb. Yes. So basically what you want to do is that every source of non key attributes, right? So bank right is a non key attribute or name is a non key attribute, should depend only on the primary key, right? So name should depend only on the primary key, which is ibank should depend only on the bank code number, which is the primary key ID like here, name should not depend on ID, which is unique, and bank code number, right? It should have only one primary key that is dependent on unless it's unless it's a tubecause. You have keys, primary keys where you have two variables, but that might make sense for certain other values. So here, because bank and vendor idea are different things, you don't need them. A bank will only ever depend on this. Bank will only ever depend on this. And name will only ever depend on this, right? Name will depend on ID and bank code number. That will never happen. If it did, then this would be three and f. If they depend it, but if they don't, then it's not. Does that make sense? Yeah. Yeah, perfect. So with that in mind, I'm going to ask you one more time to now that we've done one, 23 and f, I'm going to ask you one more time. I know it sucks, but I think it's really, really useful. Just try and write an answer question to my question over from over here. I'm going to copy it down here. And just try to answer this in exam style question and and then we'll do one more also about databases, about the database management system. So maybe try to warm up your head there and we'll see what we get or maybe I'll do about something else. Only one value in its record, atomic. Yes, you want to use the word atomic. But I don't want to give you too many hints. I really want you to try your best, and if you don't know, just don't write it down, or write down what you think can work. I'm actually not sure how to write it, yes, but I really want you to to do it because it's the whole point of the practice. If you if you don't know how to write it, don't write it and then we'll go over at the end how to how to write it properly. But I really recommend giving your best shot, even if you're just blabbing and seeing where it goes. Allows linkly primary and foreign key. We're not. Looks pretty good. Let me know when you've finished and we can go over. All right. So I would award this one point here. Perfect. And I would reward this one point here. So this would be two points. I think one key detail here that you could have added is one f only allows for atomic values for and you don't need to add in each record because it convolutes so only allows for atomic values, and you'll be good there. And one of the properties as well from one I every, if you look here, all values are atomic and no repeating groups. So you can just quickly add no repeating groups. But it should give you the points here. One thing you missed is two nf allows linking a primary and a four and key. Three and f allows three keto link together. So these are not correct definitions of two and nf and three and f. But one thing you could have added here, which I know that you know, is two nf needs one nf, right? That could get you three points. And three and f needs two and f, right? So that's three points right there. And two and f, let's go back to definition. Two and f means it must be one and f, and all the attributes end depend on one primary key. So it's not really about linking a primary and a foreign key. It's more about every single ID here must depend, all attributes depend on one primary key, right? Bernaconi depends on ID Prof, right? Schmidt depends on their ID here. So basically it's removing redundant entries inside of a large table. So you don't have so you've got separate tables every time. And three and f is not three keys. It's more about if you have, for example, this situation here and name links well to ID, but let's say bank code number also or bank also links to ID by coincidence, right? Let's say that every vendor ID has a different bet. Sorry, go for it. No, no, I just okay. Then every bank here is going to depend on the ID, but not because it should just out of coincidence, what it really depends on is a bank code number, right? And so even though it does fit two and f because in two and f all the attributes depend on one primary key. All of these depend on one primary key. They don't necessarily have they've also got transitive dependencies, right? Bank. What that means is bank depends on bank code number, and bank code number is implicitly linked to ID, but bank should only ever be linked to bancode number. Does that make sense? So we want to get rid of that. And so even though this is two and f, right? It's it fits because everything depends on ID, right? We don't want everything to depend on ID because bank doesn't actually depend just on ID. It depends on bancode number and bancode number depends on ID. So we need to separate these two out into its own table. And that's three enough. So if you see jumps like this where bank depends on bank code number and bank code number depends on ID, or maybe student address depends on student address ID, which depends on student ID, then you should split these out and it's not three enough. Does that make sense? Yeah. All right, let's move on to something else. It's going to be another long answer question. I know you hate them, but I have to do this to you. I'm very sorry. We'll do something that we did a lot of work on, though. Please explain how ip works. I'll. Pit for Internet it. Right. So I'm just going to ask you to just answer this question, long answer form. And I know it's not related to what we were talking about today, but I think you will know the answer to this. And I'm going to give you I think this will be with let me know, think about it. Probably four Marks this thing or six. Yeah, I was thinking four Marks. I think four Marks would be would be doable. Six, we could also do six. But let's stick to four. Make it shorter. Does protocol have two lor? One protocol, one l. I would get a point for saying that it stands for Internet protocol or would I got for a four point question? I would be careful relying on that since four points is not much in a six point question, definitely in a four point question. I would not rely on that too hard. But it would be good. So then I would mention ipv four and v six to get a mark. So here I would say, please explain how ip works and how packets are routed over the Internet. I would really focus on the how packets are routed over the Internet part of the question. I think you will not lose points and I think you will gain respect by the person marking your paper if you talk about ipv four and ipv six. So that's actually a good point. I think if you have space when you're writing, I would definitely, definitely mention ipv four and ipv six, but I would also not rely on a point from that. For packets that are routed, should I only focus on the routing part? I would focus on what's inside a packet and how they're routed. Knowledgement code was something I forgot. What's the name for that? Well, it's just acknowledgement. Yeah, it seems good to me. All packets have a self correction code break. I don't want to give you too much. I really just want you to I want to help you, but I think you'll get the most if you write it yourself and then we we go at the end and we talk about it. So I'll just I won't say anything from now on and just tell me when you're done. All right, ip stands for Internet protocol. It is how different Internet interconnected devices use to look at each other using different addressing messages. Ipv four and ipv six. Yes, the ip, ipv four, the pack of a sender, the address of a receiver, metadata, exactly. And self correction data. Nice. These packets are added through servers that are interconnected through the designated address. Very nice. So honestly, very, very good. You basically explained everything that you need. I would give this between 34 points. Now let me explain why for someone who is strict, right? They might give you three points. And the reasoning for this is the following. I'm gonna to be very, very pedantic, right? So pedantic by that, I mean I'm gonna really be aggrenot aggressive, but like I'm gonna really nitpick because some some markers is really nitpick a lot. So here what I would knitpick nknitpick and again, not a personal tabut Internet protocol here it's spelt with an o. So just to know it might put some markers off. So it might be worth looking at some grammar and stuff like that, making sure that you've got it done very well using different addressing messages ges such as ipv four and ipv six. This is good. This is really good, because even though it's not a point inside of the marscheme necessarily, if it was a six points question, you would have gotten an extra point here. But really, what this tells the person marking you is okay. This guy knows what he's talking about, right? Like he's knows ipv four. Ipv six typical address says the address center address, the receiver metadata, and that's perfect. And tcp would include acknowledgement and udp would contain self pressure data. This is all correct. The thing is, it doesn't pertain to the question because they don't ask about tcp udp. But again, if the marker who reading this was like, wow, this guy knows about tcp udp knows what right? Hebe like or theybe like, Oh okay, I'm gonna to be more lenient when marking this guy, but if they're less lenient, they might say, well, we asked about ip, not tcp. And a lot of the markers, you know they don't actually know much about computer science, they just follow the mark scheme. So it's a bit of a gamble. I think ipv four and ipv six is less of a gamble. I think udp, tcp is a bit more of a gamble. So maybe I would avoid that. Now this is the meat of the section. These packets are routed through servers that are interconnected to reach their designated address. This is actually worth two points. The reason why is packets are routed through servers. This is a whole point, right? That are interconnected to reach the designated address is another point. So it looks ridiculous. Like one sentence gave me two points, but it really is that you really want to say Internet protocol, right? This is how I would have answered the question. Well, okay, so that's the three to four points, right? So if I was leaning into I would have given you an extra points because you mentioned ipv six, ipv four and I like, okay, he knows what he's talking about, but the key word that's missing here are routing tables. Do you remember what routing tables are? Yeah, there are tables that the exactly that's really the only thing that's quite that's missing just routing tables somewhere in there you could have said these packets are routed through servers using routing tables and are in counterconnected to reach the designated address. Boom, four points. So it sucks at is that way. But I would say very good job and you're definitely improving on that front. So good job. I would say as well here, acknowledgement, it's a very complicated word, but here you don't need the H, so it's acknowledgement and you got the rest correct. Very, very nice. Okay. Any questions? Okay okay, show you some leacode now, all right. Receive. Best time to buy and sell all we did already keep. Oh. Maybe majority element number 169. I'm just thinking of getting a counter for every for every number. Pretty good, pretty good strategy, I would agree. But I think that would be beautiful. It would it would I think it might work here. How about ts, for example? How many times would you need to see a number before it's definitely majority and divide it by two times? So that would cut the processing time a little bit there. Yeah, but there's could be. A variety of them. That is. There's actually a trick to this one, but it's a mathematical trick. I wonder if this is the best question to be doing. I don't think the brute force option is necessarily bad. I think you could try implementing that. You'll learn a few things and then Yeah and then we can kind of go from there without I use a mat for this or Yeah, I would use that. If you don't know how to use maps. I started doing some c sharp, and I realized how bad my c is terrible. They're different languages. Yeah, I know I thought they would be quite similar, but they're very different stuff. Meaning, see what four plus is. It didn't throw narit, so it's still good. I think it's all right. I think you're pretty close. Yeah. I'm just thinking of how to actually get the. Things from the map. I think there's two ways you can do it. I think the one you're thinking of or the most, the one that immediately comes to mind is the you go through all the elements inside of the table and then you return the maximum one. But think about it else like in a different way, right? Every time I look at something in the table, where every time I add something to the table, I'm immediately accessing that element, right? If that makes sense. So you can keep a tracker for what's the maximum you've seen and just return that. But it's that Yeah. And then once once you do that, you can check is this n higher or lower than what we've seen before? Don't understand. So every time I check n right, so let's say I've got, let's say, let's take example, one, three or let's take example two, two, two, one, one, one, two, two, right? First iteration two. I add two to unordered map. And so I've seen two once, then two again. I've seen two twice, then one. I see it once. I see it twice. I see it three times. Now I've seen 13 times, which is more than I ever saw 22 times. So maybe I have a variable somewhere which is like, okay, the maximum I've seen so far is two, but I've just seen 13 times. So now I replaced my maximum with one. And then I go, I keep going. And then I see another two. And I'm like, Oh, I've seen it three times, but it's not more than one. So let's just keep going. And I see two again, and then I say, wow, I've seen four. I've seen two, four times. So now that should be the maximum one I found. Did that make sense? I don't know how to implement it. Like you're so close to logic. Yes, you're quite close. I would say think about indexes, right? Think about indexes store. I don't want to say it outright, but when you increment, check if this is the biggest you've seen and compare it with your stored index of the biggest one you've seen so far. So instead of storing values, store indices and think about it like that or not, indices maybe keys or the hash table or the map. So the cmathe current max needs to be padigm, right? Yep, so you could say, well, you could always put something like minus one, right? You could always just say, Oh, it's I doesn't need to be a pair. I see your question. Yeah because I would need to store both the current max and the count, but you've got the count ts in the map right inside of your map. You've got the count already. So if you just keep store the index or the key of the n, you actually just need once. But how will I initialize little? My personal recommendation is to give any index which is not initialized a value of minus one. And in your if statement, I would and in your if statements, I would check is Cmax minus one. If it is, then automatically the first number we see should be equal to Cmax. Well, it says here, it's wait. Okay, here it's cliitself for some music numbs okay, then I would use something like null, we can't say null. Then I would just make it equal to the first element in the array. For example, my mouse is sort of moving itself now. Oh, my God, my, Oh, it's me. You can move my mouse. That's hilarious. That's not very good. I shouldn't be able to move your mouse. I don't know why that's happening. I'm not sure why that's happening. I'm going to stop moving my mouse until the end of the of the class. I'm just gonna to Yeah leave you to it. But I would say Cmax should be equal to the first element of the array then as long as because numdot length is always at least one, so you can always guarantee that it's going to be equal to numum zero. So you can all just equal to numum zero. It's moving again. What, sir? No, my mouse was moving. Oh, sorry. I was scrolling onto your screen to see the answers, to see the the results of the run. You're very, very contromy mouse though. Yeah, I don't know why that's happening. I will say I can't see the other run because I've zoomed in up to your screen. But I will say you're very, very close. And I would think about if you need to return the map of cmap or if you just need to return Cmax, think about you know because map of Cmax, it's just the wait that's Oh Yeah that's returning the count. Yeah Yeah exactly. Oh, we don't not end Cmax almost. Did that work? I can't see. I can't see. Okay. Great. Yeah, there we go. Three milliseconds. Very good. So to give you a rundown here, this is perfectly acceptable. There are algorithms that use Yeah you should not try to. I can't read this. Yeah, I wouldn't try too hard. People do programming competitions on on this kind of thing. I'll all for this. Yeah, literally, I'm going to move your mouse again. I need to scroll down. Yeah, there you go. Yeah, no, it's they do programming competitions. Actually some of the hedge funds and Google and Amazon and Facebook, they take easy questions like this and they're like, okay, lowest runtime gets the job and then you just have to, Yeah, you got to do it. It's actually great. Es ist die. Was was. Did that work? Yeah. The way this algorithm works here is a bit different here. You're not going to have a you're not going to have an array or a dictionary. The way they do it here is they use something called, I forgot the name of the algorithm, but they use an algorithm where the idea is that every time you see a new number, you start counting from zero. So basically, it's quite interesting. So I think it's worth talking about basically, let's say you've got an array here. I see you've got two, two, one, one, one, two, two. What you do is you have accounts and a nun. If you see this one, you set it to two and counts equals one. You see this one count equals two. Then you see a one here, right? And so you decrease the counts. So it's one, and you see another one, you decrease the counts. And if it's zero, the num is equal to the new numbers, it's one. So it becomes zero. And so what you end up with is you end up with being in this position, and nuum is equal to one, and then account equals one, but then you end up on two. So you decrease the counts, it equals zero. And then you set this to two. And then here you increase the count, it's one, and then you return two. So you're only using two variables. And the and does that make intuitive sense for why this would work? Does make sense. Nice. And you're a very smart man because a lot of people don't understand this. And it's a very cool trick. Ger, the first time I saw it, I had to be told about it because I didn't really understand what was going on. But the idea is that you're almost battling out the numbers. And so the majority elements will always win. So there's actually a medium question, which is the exact same thing, but with less memory allowed. Okay, we have five minutes. We can start an easy if you want. I have, I have to. I owe you ten minutes anyway, are you free for 15 minutes? We can do more, more. Okay, let's do that. Scroll down quite far. Let's let's get some more. This is going to sql. Oh, let's see here, number of one, bits 191. What's this one again here? Yeah, let's. Let's see this one. I already have wait, you've already done it. No, no, I have a method of right, shifting it. There's a method. You know I was thinking about rice shifting it until it's zero, but this doesn't I mean, I almost think that I looked it up quickly in c, in C++. And there's just functions to do this already. So I'm not sure this is the best problem. To give you an idea, I'll just tell you, I think it's it's easier. In C++, you have a function called underscore underscore built in underscore pop counts x and this will give you the bcount. Outit's standard, okay. Yeah, exactly. So it it almost feels it almost feels redundant to do this Yeah. Then Yeah. Let's roll. Roll down. Let's see. Let's see. Let's see. Isomorphic strings 205. Let's see this one. Yes, this is a good one. Okay, so I make a map again. Yeah, I would make hash. I just hash each value and then see if they are equal. Yeah, good plan. Good plan. You might encounter some difficulties, but you've got the right idea. You just have to think about edge cases and how things would work, but you've got the right idea. Could you give me an example of an edcase? Example two is pretty good here. Oh, I'm moving your mouse. Okay, I'll quickly explain. And no, no, it's okay. That's actually not moving it something. Okay, cool. Okay, perfect. Okay. That's that's that's good. Think about this foo and bar, right? Yeah. F maps to be no problem because there's no problem later on. But here, o maps to a and o maps to R, right? O maps to A, O maps to R. So it's false because you can't have that kind of problem. But think about if you have, for example, let's say you've got fo, let's say here fo, and then let's say you've got B A, and I'm trying to come up with a Yeah and let's say you've got c and then this will map to d, right? Cd. And then here maybe you've got let's see. D that maps chain a, right? This maps to a and this maps to a, but a maps to o and a maps to d. So it's it's not possible, right? So you got to make sure that it is a two way correlation between each of these letters. If you have a breakage, start to add function. That would work. Absolutely. I think I've given you some clue and you already solved this. Basically, you've had the right idea, so I'll just let you try and solve it and let me know if you get stuck. If you need some help, I'm going to quickly look at there. Yeah, this function would do it. If it contains the primary key, I can't quite see Yeah, Yeah, absolutely. I think the best way is just to try it out and see if it works. And I think you'll find it will work out well for you. But just make sure it's a two way street, that the map works in both directions. Make that makes sense. Just give it compiler video. What c equals already. I'm not sure you might just have to do it for inying or maybe not. If I had a good editor, I could just hover over it and see the type. Maybe it's because you need the c one, c two, since you're looping over s one is two. This creates a unnamed struct, and then it assigns c to that struct. So c zero would be s zero. S one is two, is three, then c one would be okay. I C. It just loops over like c dot s and c dot t. Well, I think that's how it would. Although that's for the hash map rethat's for the the hash map rights you need to because right now you've only got the raw the raw strings, you've only got s and t. So Yeah, I think it's more more effective to first to just loop through the what's it called to loop through the the eyes like because they're gonna Yeah, I just indedexed because they should be the same length I believe. Yeah just want to see if it's but then I need A T. Then also the way I would approach this is, well, first we should check the isomorphic strings description and see if s and t are the same length. And if they're the same length, then we should just do a four I in range loop. They are the same length. Then I would just I would stick to a four I and range loop so you can access the integer and si and ti. Tell me what the considerator is. I'm not sure what that is, but what I would say is here, I think the approach should be more in terms of, okay, we start to index zero of s and t. We know they have the same blank. So wehave to worry about that. Let's say it's egg and add, right? Then what I would do is I would say, okay, we want e to match to a and we want so e from from s should match to a of t, string t, then we should have string T, A should link to e, right? So we should try and think about that and then we should go to the next one. And then we say, okay, g of s links to d of t, right? And d of t leads to g of s and then keep on going. And every time we do that, right, every time we check, you know, every time we add these connections, then what we should do is check, okay, are one of these connections already defined, right? So if I, for example, take example two, f links to b and b links to f easy, o links to a, and a links to o easy. There's no link already, but then o links to R boom, error, right? Because ocean, not links to R, it's already linked to something else. So that's how you can know, Oh, wow, there's already a link. So false. But if you go through the whole if you go through everything and you don't have a break in these links, then you return. So I would kind of think about that sort of approach there. I'll give you one more minute, but then I'm going to have to wrap up the lesson. Just take your time and let me know once you're done. Plus I need to contain name. I'll let you solve this one in your free time as we'll have to end the lesson here. But as usual, Henry, great job. Sorry. Then we have like four minutes of, Oh, wait, it's ten minutes. Sorry, sorry. We can keep going for a minute or two. I'm not sure how much I mean, I can give you the answer very quickly for this one, but I feel like it's more fun. Maybe if you try to solve it yourself, since you're so close. So I think giving you the answer would be almost a bit of a shame. So perhaps, maybe as homework where I just try to finish this one, I can ask you about it next session and see if you can solve that. I think you're very, very close. However, I'll give you a little bit of a hint. You need two maps for this word for this to work. So think about that. Don't think you need to maps. So like I think I'm very close now. Yeah, Yeah you'll see I mean don't I don't want to say you won't I believe you probably be able to but just think about using two maps if you get stuck on on that note, I'll let you Oh no, maybe you're almost done here. Let's see if this works. Now I just need to fix this part and that's all. I'll leave you to it, but if it doesn't work, think about two maps, just to let you know. All right. Well, great job as usual, Henry. Have a great day. Bye bye.
处理时间: 29385 秒 | 字符数: 34,093
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{
    "header_icon": "fas fa-crown",
    "course_title_en": "Language Course Summary",
    "course_title_cn": "语言课程总结",
    "course_subtitle_en": "1v1 CS Lesson - Database & Algorithms",
    "course_subtitle_cn": "1v1 计算机科学课程 - 数据库与算法",
    "course_name_en": "CS Henry",
    "course_name_cn": "CS 亨利课",
    "course_topic_en": "Database Normal Forms (1NF, 2NF, 3NF) and IP Routing\/LeetCode Problems",
    "course_topic_cn": "数据库范式 (1NF, 2NF, 3NF) 与 IP 路由\/LeetCode 问题",
    "course_date_en": "Date not specified (Based on context '0125')",
    "course_date_cn": "日期未指定 (基于 '0125')",
    "student_name": "Henry",
    "teaching_focus_en": "Reviewing database normal forms via long-answer exam simulation and practicing LeetCode style algorithm problem-solving (Majority Element, Isomorphic Strings).",
    "teaching_focus_cn": "通过长篇问答考试模拟复习数据库范式,并练习 LeetCode 风格的算法问题解决(多数元素,同构字符串)。",
    "teaching_objectives": [
        {
            "en": "Student can recall and articulate the definitions and purpose of 1NF, 2NF, and 3NF.",
            "cn": "学生能够回忆并阐述 1NF、2NF 和 3NF 的定义和目的。"
        },
        {
            "en": "Student can attempt to answer a structured, high-value exam question under simulated conditions.",
            "cn": "学生能够在模拟条件下尝试回答结构化的、高分值的考试问题。"
        },
        {
            "en": "Student can apply data structure knowledge (maps\/hash tables) to solve algorithmic problems (Majority Element, Isomorphic Strings).",
            "cn": "学生能够应用数据结构知识(映射\/哈希表)来解决算法问题(多数元素,同构字符串)。"
        }
    ],
    "timeline_activities": [
        {
            "time": "Start",
            "title_en": "Casual Check-in & Game Discussion",
            "title_cn": "闲聊与游戏讨论",
            "description_en": "Discussed student's week and brief discussion about CS:GO\/Valorant.",
            "description_cn": "讨论了学生的一周情况并简短讨论了 CS:GO\/Valorant。"
        },
        {
            "time": "Core 1",
            "title_en": "Database Normal Forms (1NF, 2NF, 3NF) Review via Exam Simulation",
            "title_cn": "通过考试模拟复习数据库范式 (1NF, 2NF, 3NF)",
            "description_en": "Student attempted a long-answer question on the differences between 1NF, 2NF, 3NF and the reason for normalization. Teacher provided detailed feedback on structure and content.",
            "description_cn": "学生尝试回答一个关于 1NF、2NF、3NF 之间区别以及规范化原因的长篇问答题。教师对结构和内容提供了详细反馈。"
        },
        {
            "time": "Core 2",
            "title_en": "IP Protocol Explanation Review",
            "title_cn": "IP 协议解释复习",
            "description_en": "Student explained IP addressing and routing. Teacher provided detailed feedback on completeness, pointing out missing keywords like 'routing tables'.",
            "description_cn": "学生解释了 IP 寻址和路由。教师对完整性提供了详细反馈,指出了缺少“路由表”等关键词。"
        },
        {
            "time": "Core 3",
            "title_en": "LeetCode Practice: Majority Element (Map approach)",
            "title_cn": "LeetCode 练习:多数元素(Map 方法)",
            "description_en": "Student proposed a map-based counting solution for Majority Element and refined the logic for tracking the maximum count seen so far.",
            "description_cn": "学生提出了一个基于 Map 的计数解法来解决多数元素问题,并完善了跟踪迄今为止最大计数的逻辑。"
        },
        {
            "time": "Core 4",
            "title_en": "LeetCode Practice: Isomorphic Strings (Two Map approach explained)",
            "title_cn": "LeetCode 练习:同构字符串(解释双向映射)",
            "description_en": "Discussed the problem, focusing on the need for a two-way correlation (two maps) to ensure 'isomorphic' mapping.",
            "description_cn": "讨论了问题,重点强调需要双向关联(两个 Map)以确保“同构”映射。"
        },
        {
            "time": "End",
            "title_en": "Wrap-up and Homework Assignment",
            "title_cn": "总结和布置作业",
            "description_en": "Extended time slightly. Assigned the Isomorphic Strings problem as homework.",
            "description_cn": "略微延长了时间。将同构字符串问题作为家庭作业。"
        }
    ],
    "vocabulary_en": "Atomic, Repeating Groups, Partial Dependency, Transitive Dependency, Primary Key (PK), Vendor, Bank Code Number, Internet Protocol (IP), Acknowledgement, Isomorphic, Edge Cases.",
    "vocabulary_cn": "原子性,重复组,部分依赖,传递依赖,主键 (PK),供应商,银行代码号,互联网协议 (IP),确认,同构,边缘案例。",
    "concepts_en": "Database Normalization (1NF, 2NF, 3NF), IP Packet Routing, Hash Map implementation for frequency counting, Bi-directional mapping for isomorphism.",
    "concepts_cn": "数据库规范化 (1NF, 2NF, 3NF),IP 数据包路由,用于频率计数的哈希映射实现,同构性的双向映射。",
    "skills_practiced_en": "Long-form written answer construction, technical explanation, algorithmic problem decomposition, application of data structures (Hash Map).",
    "skills_practiced_cn": "长篇书面答案构建,技术解释,算法问题分解,数据结构(哈希映射)的应用。",
    "teaching_resources": [
        {
            "en": "Visual examples\/diagrams used for explaining 2NF (student\/professor\/grade table) and 3NF (vendor\/bank dependency).",
            "cn": "用于解释 2NF(学生\/教授\/成绩表)和 3NF(供应商\/银行依赖)的视觉示例\/图表。"
        },
        {
            "en": "LeetCode problem descriptions (Majority Element, Isomorphic Strings).",
            "cn": "LeetCode 问题描述(多数元素,同构字符串)。"
        }
    ],
    "participation_assessment": [
        {
            "en": "High engagement, especially when discussing technical solutions like IP routing and LeetCode algorithms.",
            "cn": "参与度高,尤其是在讨论 IP 路由和 LeetCode 算法等技术解决方案时。"
        },
        {
            "en": "Willingness to attempt challenging long-answer questions despite initial reluctance, showing commitment to exam practice.",
            "cn": "尽管最初不情愿,但仍愿意尝试有难度的长篇问答题,表现出对考试练习的投入。"
        }
    ],
    "comprehension_assessment": [
        {
            "en": "Good recall of 1NF definition (atomic values, no repeating groups).",
            "cn": "对 1NF 定义(原子值,无重复组)的回忆良好。"
        },
        {
            "en": "Correctly identified the core logic for IP routing (servers, designated address) but missed the key technical term 'routing tables'.",
            "cn": "正确识别了 IP 路由的核心逻辑(服务器、指定地址),但遗漏了关键技术术语“路由表”。"
        },
        {
            "en": "Strong conceptual understanding of 2NF (partial dependency removal) and 3NF (transitive dependency removal) after initial review.",
            "cn": "在初步回顾后,对 2NF(消除部分依赖)和 3NF(消除传递依赖)有很强的概念理解。"
        }
    ],
    "oral_assessment": [
        {
            "en": "Clarity improved significantly during the technical explanations, especially in the second half of the session.",
            "cn": "在技术解释过程中,清晰度显著提高,尤其是在课程后半段。"
        },
        {
            "en": "Able to articulate complex algorithmic ideas (Majority Element tracking) when prompted with structural guidance (using the map's count).",
            "cn": "在获得结构性指导(使用 Map 的计数)时,能够清晰地阐述复杂的算法思想(多数元素跟踪)。"
        }
    ],
    "written_assessment_en": "Student's written attempts in the chat showed good faith but lacked the precise, structured phrasing required for full marks in an exam setting, although key concepts were present.",
    "written_assessment_cn": "学生在聊天中的书面尝试表现出诚意,但缺乏考试环境中获得满分所需的精确、结构化的措辞,尽管关键概念是存在的。",
    "student_strengths": [
        {
            "en": "Strong foundation in data structures, readily suggesting hash maps for counting\/mapping tasks.",
            "cn": "数据结构基础扎实,能迅速为计数\/映射任务提出哈希映射的建议。"
        },
        {
            "en": "Ability to grasp abstract algorithmic concepts quickly (e.g., the 'battle out' mechanism for Majority Element).",
            "cn": "能够快速掌握抽象的算法概念(例如,多数元素问题的‘互相抵消’机制)。"
        },
        {
            "en": "Good engagement with potentially 'boring' review material (Normalization) when framed as exam preparation.",
            "cn": "当规范化复习被定性为考试准备时,表现出很好的投入度。"
        }
    ],
    "improvement_areas": [
        {
            "en": "Structuring long-form answers to directly address all parts of the question and use required technical terminology precisely (e.g., 'routing tables').",
            "cn": "需要改进长篇问答的结构,以直接解决问题的各个部分,并精确使用所需的技术术语(例如,“路由表”)。"
        },
        {
            "en": "Distinguishing between required knowledge (IP routing) and related but non-requested knowledge (TCP\/UDP details) in exam answers to maximize relevance.",
            "cn": "在考试答案中需要区分必需的知识(IP 路由)和相关但不被要求的知识(TCP\/UDP 细节),以最大化相关性。"
        },
        {
            "en": "Ensuring full bidirectional mapping logic for Isomorphic Strings (requiring two maps for the most robust solution).",
            "cn": "需要确保同构字符串的双向映射逻辑的完整性(最稳健的解决方案需要两个 Map)。"
        }
    ],
    "teaching_effectiveness": [
        {
            "en": "Highly effective in using the exam simulation format to motivate the student to recall complex database theory.",
            "cn": "使用考试模拟格式来激励学生回忆复杂的数据库理论非常有效。"
        },
        {
            "en": "Effective in guiding the student through the algorithmic thought process, providing hints instead of direct answers.",
            "cn": "在引导学生完成算法思维过程中非常有效,提供提示而不是直接答案。"
        }
    ],
    "pace_management": [
        {
            "en": "The initial database review was deliberately slow to ensure foundational recall, which worked well.",
            "cn": "最初的数据库复习是故意放慢速度的,以确保基础回忆,这很有效。"
        },
        {
            "en": "The session successfully transitioned between database theory and coding practice, utilizing the extra time effectively.",
            "cn": "课程成功地在数据库理论和编码实践之间进行了转换,有效利用了额外的时间。"
        }
    ],
    "classroom_atmosphere_en": "Interactive, supportive, and task-focused. The teacher successfully balanced drilling for exam precision with encouraging algorithmic creativity.",
    "classroom_atmosphere_cn": "互动、支持性和以任务为中心。教师成功地在追求考试精确度和鼓励算法创造力之间取得了平衡。",
    "objective_achievement": [
        {
            "en": "Objective 1 (Normalization Recall): Largely achieved, with clear articulation of 1NF and guided understanding of 2NF\/3NF.",
            "cn": "目标 1(规范化回忆):基本实现,能清晰阐述 1NF,并对 2NF\/3NF 有指导下的理解。"
        },
        {
            "en": "Objective 2 (Exam Practice): Achieved through direct simulation and detailed post-mortem feedback.",
            "cn": "目标 2(考试练习):通过直接模拟和详细的事后反馈实现。"
        },
        {
            "en": "Objective 3 (Data Structures in Algorithms): Demonstrated strong ability, especially with the Majority Element problem.",
            "cn": "目标 3(算法中的数据结构):展现了很强的能力,尤其是在多数元素问题上。"
        }
    ],
    "teaching_strengths": {
        "identified_strengths": [
            {
                "en": "Skillful use of exam simulation ('exam answer mode') to drive high-quality recall.",
                "cn": "熟练运用考试模拟(“考试作答模式”)来驱动高质量的回忆。"
            },
            {
                "en": "Providing precise, actionable feedback on written answers, linking structure to mark allocation.",
                "cn": "对书面答案提供精确、可操作的反馈,将结构与分数分配联系起来。"
            },
            {
                "en": "Excellent transition into and scaffolding of complex LeetCode problems.",
                "cn": "在引入和脚手架复杂的 LeetCode 问题方面表现出色。"
            }
        ],
        "effective_methods": [
            {
                "en": "Breaking down the value of points in an exam question to encourage students to write *something* relevant.",
                "cn": "分解考试问题中分数的价值,鼓励学生写出*任何*相关的内容。"
            },
            {
                "en": "Guiding the student to the 'two map' solution for Isomorphic Strings instead of giving the answer immediately.",
                "cn": "指导学生得出同构字符串的“双 Map”解决方案,而不是立即给出答案。"
            }
        ],
        "positive_feedback": [
            {
                "en": "Student's IP explanation was 'very, very good' and showed high understanding beyond the basic scope.",
                "cn": "学生的 IP 解释“非常好”,显示了超越基本范围的高水平理解。"
            },
            {
                "en": "The student is 'definitely improving' on technical explanation tasks.",
                "cn": "学生在技术解释任务上“绝对在进步”。"
            }
        ]
    },
    "specific_suggestions": [
        {
            "icon": "fas fa-database",
            "category_en": "Database Theory & Exam Technique",
            "category_cn": "数据库理论与考试技巧",
            "suggestions": [
                {
                    "en": "When answering normalization questions, ensure you explicitly state the prerequisites (e.g., 'Must be 2NF to be 3NF') to secure all potential points.",
                    "cn": "回答规范化问题时,请确保明确说明先决条件(例如,“要成为 3NF 必须是 2NF”),以确保获得所有潜在分数。"
                },
                {
                    "en": "For terminology like 'Internet Protocol', double-check spelling ('protocol' vs 'protocal') to avoid small deductions in formal exams.",
                    "cn": "对于像“Internet Protocol”这样的术语,请仔细检查拼写('protocol' 而非 'protocal'),以避免在正式考试中被扣小分。"
                }
            ]
        },
        {
            "icon": "fas fa-code",
            "category_en": "Algorithm Implementation (LeetCode)",
            "category_cn": "算法实现 (LeetCode)",
            "suggestions": [
                {
                    "en": "For the Isomorphic Strings problem, focus on implementing the two-map (s->t and t->s) structure to strictly enforce the bidirectional link requirement.",
                    "cn": "对于同构字符串问题,重点实现双 Map(s->t 和 t->s)结构,以严格强制执行双向链接要求。"
                },
                {
                    "en": "In the Majority Element problem, practice initializing the max tracking variable (Cmax) using the first element of the array (num[0]) when its length is guaranteed to be >= 1.",
                    "cn": "在多数元素问题中,练习使用数组的第一个元素(num[0])初始化最大值跟踪变量 (Cmax),因为其长度保证 >= 1。"
                }
            ]
        }
    ],
    "next_focus": [
        {
            "en": "Completing and reviewing the Isomorphic Strings problem implementation.",
            "cn": "完成并复习同构字符串问题的实现。"
        },
        {
            "en": "Further exploration of advanced database concepts or moving onto a new algorithmic concept if time allows.",
            "cn": "如果时间允许,进一步探索高级数据库概念或转向新的算法概念。"
        }
    ],
    "homework_resources": [
        {
            "en": "Complete the LeetCode solution for Isomorphic Strings (205), focusing on the two-map approach.",
            "cn": "完成 LeetCode 问题同构字符串 (205) 的解决方案,重点关注双 Map 方法。"
        },
        {
            "en": "Review notes on 3NF to solidify understanding of transitive dependencies.",
            "cn": "复习 3NF 的笔记,以巩固对传递依赖的理解。"
        }
    ]
}
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