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Java Interview with an Uber engineer

Watch someone solve the k closest elements problem in an interview with an Uber engineer and see the feedback their interviewer left them. Explore this problem and others in our library of interview replays.

Interview Summary

Problem type

K closest elements

Interview question

1) Find the K closest elements 2) Find the top K most frequent elements

Interview Feedback

Feedback about Mutable Pigeon (the interviewee)

Advance this person to the next round?

Yes

How were their technical skills?

4/4

How was their problem solving ability?

4/4

What about their communication ability?

4/4

The interviewee is the first one I interviewed on interviewing.io who can finish 2 questions in 1h! 1. The interviewee asked clarification questions to eliminate the assumptions. 2. The interviewee came up with a binary search solution. 3. The interviewee proactively added the unit test. 4. The implementation is quite smooth and there it take 35 min to finish the first question. 5. The interviewee quickly came up a naive solution and an optimized solution for the 2nd solution. The interviewee also asked clarification questions. One thing to improve is to better pace the speed if the interviewer mentioned to solve 2 questions at the very beginning.

Feedback about Blue Centurion (the interviewer)

Would you want to work with this person?

Yes

How excited would you be to work with them?

3/4

How good were the questions?

4/4

How helpful was your interviewer in guiding you to the solution(s)?

4/4

he was very patient and helped me code fix compile issue.

Interview Transcript

Blue Centurion: Okay, so this is more interview for the coding. So here's whatever program language you want to use.

Mutable Pigeon: Sure, yeah, I just do Java.

Blue Centurion: Okay, so before the coding do you want to do some very brief introduction about yourself? It can be anonymous and you can also tell me what's the target so we can have a corresponding questions for that.

Mutable Pigeon: Sure. So I am a senior developer having like 1010 plus years of experience. I am actively, you know, looking out. So preparing, preparation is going on. Mainly I work in Java.

Blue Centurion: So are you looking for the front end or the back end opportunities?

Mutable Pigeon: Hello? Hello, can you hear me? Yeah, sorry. Actually I think my headphones got disconnected. I think I missed it.

Blue Centurion: Yeah, yeah. So I'm looking for the front end or back end or jobs?

Mutable Pigeon: Back end.

Blue Centurion: Okay, I see, I see, cool. So we have 1 hour so probably we can try to finish two classes. Does that sound good?

Mutable Pigeon: Yeah.

Blue Centurion: Okay, so let's start with the first question. So the first question is about an array. And you can see that there is some other integers in the array and your goal is to find the top k, closest number to the target number. Let's say target number is minus one and do you want to find the top four of them?

Mutable Pigeon: Okay. Okay.

Blue Centurion: Yeah. So is that question clear?

Mutable Pigeon: Yeah, I have a question. Is it going to be a distinct number in the audit?

Blue Centurion: It's not necessary. The distinct number.

Mutable Pigeon: Okay, so let's say if I have an array. Okay, yeah. 41236. And target is equal to, let's say five and top four. So is it going to be.

Blue Centurion: Yeah, you can, you can reach either, any, any valid solution.

Mutable Pigeon: Okay. Okay, so maybe like 1234 or 2346. Yes, 2346.

Blue Centurion: Yeah. If your target is five.

Mutable Pigeon: Yeah.

Blue Centurion: Yeah. So if there are multiple solutions you can just return any of them.

Mutable Pigeon: Okay, cool. And we. Okay, so let's say. Okay for this example, right, okay, so let's say if I, let's see if I sort the number.

Blue Centurion: Yeah, actually you can see that now. The array is already sorted. You don't need to sort it.

Mutable Pigeon: Oh, okay. Okay. So I think in my example was wrong. Okay, okay. Okay, so array sorted. That means we can use this property of sorted array and target is minus one. So let's say if I have to just find the one number, right, top one number. Then I can do binary search and I can come to this. And if I find this, then I can find like the next number would be the closest to it. If I find the first smallest number that is closest to the target, right? So let's say if I have 123468 and nine and my target is five, I would say, okay, my nearest number is. So I need, okay, so let's say I have, it's going to be three, right? Let's say. So definitely one of the number has to be, I mean the last number only then I will get six at nine. So I have to choose, I have to choose from this range 1234 and six. Let's say I, okay, let's say I have this indexes. Okay, so let's say I start with zero and four as a range. No is equal to zero and h is equal to four. My mid will be.

Blue Centurion: How did you get it four in the first place?

Mutable Pigeon: So I need top three numbers, okay, and if my four is the index, only then I can get the three number right. I can't, like this has to be the one of the highest index, only then I can get the top three right. Like 689. You see what I'm saying? Like you can't. If you choose five, I, if I get the lowest number, eight, I will only left with two numbers. I won't get the, I won't have. So, so my target is to find the lowest number that is closest to the target. I see, yeah. So that's why like I took this zero and four. And let's say if I made is two, right, I would say, okay, this is three, number three. And I have to see, okay, five minus two. Let's say, let's say this is the closest number, like the number three on target, on index two, the top three will be three, four, six if I consider this. Else if I. Now I have to see whether to go left or to go right. Okay, now how can I decide that? Like I have to compare it with something to decide whether I have to go left or I have to go right. So let's say if I. This is the smallest, this is the first window. Okay, let's say I have mid to and from mid to from middle two, I have three and six, top three. But the answer is. Okay, let's. Okay, so let me just write one thing here.

Blue Centurion: So every time you want to find a lower bound of the sliding window, um, sorry. So is that case that every time you try to find the lower bound of a sliding window with size k uh uh, lower bound because you started from six. So that lower bound for now for the first round is six. Yeah. Then you use binary search to try to find a new lower bound.

Mutable Pigeon: Yeah, yeah, yeah. Like the number which is closest to five. Let's say this is the, so let's say this is the number five here. Like this is the number line. Okay. The closest is like this. Four and six, both are the same distance. Now three is actually more closer than seven like the eight. Right. Because five minus three is two and six. Five times five minus eight is three. Like absolute value. So I'll have to take three and two is same as eight. So instead I will go with 234 as the window. Right. Instead of going, no, actually three, four, six should be the answer here. Yeah. Because four and six is this. So yeah. Three, four and six should be the answer of this guy. So, so answer is three, four and six. Okay. For top three. So let's say I got this number three and I compare it with the, what is the next window? Next window going to be three and 1 second. Yeah. So this is the two and two plus three is going to be eight and if, okay, 1 second. So this is one window. And if I compare it with three, four, six. Yeah. Three, four, six. The next, I now have to decide whether to go left or to go right. So to, let's say dope size four and then it will be three, I get one. And I'm just taking example, if say I have to find top is equal to four, what, how would, what would happen then? So I'll have a window from zero to three. In case of top four, I'll get one.

Blue Centurion: I see, okay, so, so remember in a previous example you start with six, eight and nine, right? That's the suffix of the array. And now you start with a prefix of the array. So what, what makes the change?

Mutable Pigeon: No, no, no, I'm not changing. I'm saying, uh, I'm saying for index four is going to be my last index. It has to be the value. The answer should be between zero and four like the starting point.

Blue Centurion: I see.

Mutable Pigeon: So answer cannot be starting from the five because we won't be having the three. Yeah, so, so answer live in this range in case I need three. And if I need four, answer lies in the range three. So in this case I get two here. Okay. And let's say two is going to two is one window and other window is going to be 2346. And let's say other smallest window is going to be six. Then if this is if this, this is the closest number. Then if. Yeah. From here I have to decide whether I need to go. So let's say. Okay, right now my low is zero and four. Now I have to decide and it is two. Now I have to decide whether I need to stay between zero to two or I have to go between three and four. So with three and four, right, I have zero to two. Is this guy zero, two, two. Mid three is this. Now my next number is three, four, six. And my move to the, let's say if I, okay, let's say if I compare the next starting of the window. Okay, so let's say I have, this is my mid two for top three. And then I chose. I do. Okay, what is the next window after this? Like this is, this is the next three numbers are going to be this. Like this is. Okay, so let's say this is the potential answer starting from three, four, six. Now from three, four, six. I see. Okay, what is this number? Is this number, is that like after the window is this number is closer to this. Okay, right now it is. No, because eight minus five is three and three minus five is two. So I will, I will have to move to the, I don't need to go to the right because I know, like I, then I'll make my search zero to two. Like zero to mid. Right. And now next time my mid will be one and I will see, okay, from two, three, four it is. And six and five. Then six and five are closest. That, that means I have to move to third again. So yeah, third will be the answer. Should I explain? Okay, let me try to explain here. You get from mid to. Then I'll get three, four, six. Now you compare nums at mid. Okay, whatever. Is the array at mid to. Array at mid plus k. Okay, if array of middle. And if. Okay, now if array of mid, if. Okay, let's say I have this guy. Okay, map, dot absolute of array of middle minus target is greater than math. Dot absolute of array of mid plus k minus target. Now, okay, so yeah, so let's say in our case we have this equation. Actually, if you're, if the right hand side is greater difference, then you have to make your high is equal to mid. Else you have to. Then, you know. Okay, if right side is bigger, that means you are sure that it is not going to be the mid. It's going to be the mid plus one at least. Right? So we have to keep narrowing down our search using this thing. And then I think by the end you will left with one element that is going to be the lower bound, closest to the target. And from there you can just take, okay, numbers.

Blue Centurion: Uh, so for this example, what is the initial value for low and high?

Mutable Pigeon: So initial value for this example, target five and top three numbers is going to be zero and four. Okay, now from zero and four, you see, uh, the array of mid is. Array of mid is. Mid is two. Now, array of mid is three and array of mid plus k, which is k is three and mid three is eight. Okay, now we see, okay, this guy, array of mid minus target is five and no, the array of mid two. So this guy is, this the one which I highlighted is going to be value two. Right. Two is less than eight, minus five is three. Three. That means this guy is closer. Like the index three is closer than index five. Index two is closer than index five. Right. So what we'll do is we'll make h as two. So our range will be now our range will be now zero to two. So we'll have low is equal to zero and high is equal to two. Right. And now mid will be. Mid will be one. So again, if I run the same logic at this time, the mid will be one and one is two. Absolute difference will be three. And three is going to be less than. If this condition, like, so we have at index mid, we have two and five minus two is three and mid plus k is six. And five minus six is one. So which is on the right side, we have the smaller number. That means we have to move to right side. Right. So we will make our. L is equal to three and, uh oh. L is equal to two. Sorry. L is equal to two because mid is one and mid plus one is two. Yeah. And our high is. Yeah, we'll move low to the right side. Right. Like we will increment the low so that the window will move to the right and high will be two also. Right. So we'll get the index two. That is going to be the lower element that is closer. So three, four and six we can reduce which is the closer from that window.

Blue Centurion: I see. Yeah. So what's the termination condition for this binary search?

Mutable Pigeon: So we don't have any specific element. We don't know. Right. What we are searching. So I think we have to shorten the window. And we will be having. When your low and high will. Yeah, when you, when you're low and high will be at same point. Right, I see, yeah. I mean, when, after this search, when you increment the low, I mean, as long as your low is less than high. You can do search. But when they meet, I think that time it should be terminating condition. I see.

Blue Centurion: Yeah, that's great. So I think you can try to implement it.

Mutable Pigeon: Okay. Okay, I think I have to right click class solution. Okay, so we have to search. Okay, let me just copy this. Okay, let me here. So we have to return the, we have to return. All right. Like the closest number get. Yeah. Okay, closest. Closest. We will be giving the array and we will be giving the target and we will be giving the cake. Right. So, yeah, and let's say we have this result. We know that it's going to be of length k. Okay. Return result. Okay. Yeah. So let's say int low is going to be zero. And yeah, we have high element, let's say. Okay, enclose this guy and height is going to be int n is equal to dot length. So it's going to be n minus. Okay, let's. In this case we have six. Seven is the length, seven minus kk is three. Right. So yeah, so this is going to be the high. Right. Because if we see we have length of seven and seven minus three is going to be four and fourth is what we need. Our target is. So let's say while as is as long as we don't have the answer. Let's say we have mid is l plus this guy and mid. Okay. Before this, I just want, maybe I can have a quick test here. Okay, let's say I have, okay, one. Let's say I got this number. Okay, so int array and we expect result from new solution. Dot get k closest array and our target is five and k is three. Yeah, I just want to be, you know, result, print r. Okay, so, yeah. Okay. Yeah, so this is the, now if I made is this. I just want to see if it's still compilable or any issue. I just ran it. I should expect empty everything. Okay. I think it is looking okay. It's not.

Blue Centurion: Because.

Mutable Pigeon: Yeah, yeah, I just noticed, I mean, I'm not doing anything here. So, yeah, so, okay, let's say if I copy this guy here and if I see if map of abs is equal to. Basically I wanted to check my compilation, but I see. Okay. I haven't done anything there. So. Okay, so let's say if I have my map, abs of array of mid minus target is less than. Okay, something is math. Abs of this is less than a target and I have high equal to mid. Else I am doing okay now. Okay, so n minus k result. And we know that for int I is equal to lithe and I is going to be l plus k I. So just do int. J is equal to zero. Okay, so this is, this result of my result of j is going to be air of I. I is equal to l. Okay, so yeah, so math else is. Okay, so I. I have target and. Okay, I think I just want to check my compilation issue. If I have compilation issue. Okay, I think I don't have any. Air of I. So this is going to run for l. Then I plus this would be six, three. Okay. Okay, so let me just put this guy made. Okay, so we have l plus this one, four. I expect that mid should be two. Then I will see. Okay. Error of mid is this and out of method is this. If left side. If right side has bigger distance with the target, that means we have to move, move left. Else we know that. Move. Right. And now we know once we have the target. Okay, now I have air I. So, okay, so the input is 123-4689 and nine. Okay, let's say this is 0123 and target is three number. Right. So this is going to be window 1st. 012345. And I'll get the mid two. And in two we will get three, which is distance is two. Difference between five and three is two. And next is eight. Eight is eight minus five is three. Then I expect my high will become 20 to two. And. Okay, shall I just try running it?

Blue Centurion: Yeah, yeah.

Mutable Pigeon: Okay. Okay. Mid to what is credit outside? Oh, mid is two, 1 second. I can make it more clear. Mid is two and then mid is one, which is expected. Mid is two. Mid is one and mid is one. This one. So. And target is five. So okay, so you get low now. Okay.

Blue Centurion: For this example, the readout will be three, four, six.

Mutable Pigeon: Yeah, yeah. But it is coming as 60 zero because I think, oops, I think, let's say I forgot to do j here.

Blue Centurion: Okay.

Mutable Pigeon: Yeah, I think three, four, six. Okay, so mid, two, one. And we have three, four, six. Okay, I made one. But next time when you do eight plus one, your l will be two and this condition will break. So we have the shortest window of l and it will then. Yeah, I think I can try maybe some more example.

Blue Centurion: Yeah, yeah. Maybe you can try some edge cases.

Mutable Pigeon: Yeah. So what do you think? Yeah, I think if your if array dot length is equal to zero, get it basic one. Okay, so. And also other thing possible is just return empty array. Okay, so other thing possible is if your k is actually, you know, higher than the array of lengthen you know, so you won't get it. Or k is greater than. Because at that. I mean, this is not possible, I think because if your length is seven and you're asking for. Give me nine. Unless until we take the min of alidot length. Okay, so let's say we have duplicates here. Okay. So let's say I want closest to eight. Then I should get answer 899. Right. Because this is six. Six is definitely more than nine in terms of difference. So. Okay. Eight, nine, nine. Let's say about.

Blue Centurion: Try a very big number.

Mutable Pigeon: Negative number, you said.

Blue Centurion: I mean a very big number. Or you can also try negative number. Actually, that's basically you have return the prefix or suffix of the array. Right?

Mutable Pigeon: Yeah, yeah, yeah, sure. I think let's target for nearest number to minus one. Let's say prefix will have, let's say any five numbers. So we should get 1234 and six. Just before I run, I just want to go through over it. One, two, three. We need five numbers. Right. So this will be. This will be our window. Yeah. And I would say. Okay, we'll get mid as two. Now, two minus absolute value. Okay. I think. Not sure about the absolute value. How will it work? Two minus four? I think either it should work because we are targeting two and then the five. This. This guy eight, which I think we'll see. This is. This is greater. I will move my. This will be my next zero. And high and one and. Okay. I think should work ideally. Yeah. 12346. Okay. And let's say if I do ten and I need nearest, then I should get 46899. Okay, cool. Yeah. So seems like, I mean, any other test case is. I think it should work if.

Blue Centurion: Yeah, I think it's good enough. Yeah, we can.

Mutable Pigeon: Okay.

Blue Centurion: We can. We can jump to the second one if you want.

Mutable Pigeon: Yeah, sure. Yeah, let's do it.

Blue Centurion: Yeah. So, um.

Mutable Pigeon: I mean, we did not discuss this.

Blue Centurion: I'm sorry.

Mutable Pigeon: We did not discuss the time complexity. So.

Blue Centurion: Yeah, I think for this one, you. You already mentioned it's binary search, so it's very clear.

Mutable Pigeon: Okay. Okay, sure.

Blue Centurion: Yeah. So for the second question, it's also a sim. So we're providing an array and just some number.

Mutable Pigeon: Okay.

Blue Centurion: And the question is to find still the top k, but the top k, the most frequent elements.

Mutable Pigeon: Okay.

Blue Centurion: Let's say two. In that case, you will return one or two.

Mutable Pigeon: Okay.

Blue Centurion: One appears five times.

Mutable Pigeon: Five times and.

Blue Centurion: Yeah.

Mutable Pigeon: To appear four times. Okay, cool. So I think.

Blue Centurion: And the array is not sorted. And this, for this example, is sorted by chance.

Mutable Pigeon: Okay, sure. Yeah. So, yeah, I mean, let's say we have, I'm not, I'm not coding. I just want to, okay, let me do it there only. Yeah. So one way we definitely, for this top K problems, right? One thing is we can take count of all, count of all the elements, right? And put it in a map. Say one occurs five times and two occurs four times and three occurs three times. We put it in a map and put these map entries in heap. Should be max heap, right? Should be min heap actually, because, yeah, let's say if you have smallest element three.

Blue Centurion: So we are returning the top k, most frequent item.

Mutable Pigeon: Yeah, yeah. So let's say, okay, so let's say you have, okay, let's say you have this data, right? Okay, most frequent, yeah, most frequent items. Yeah. So that means, and then let's say, let's say we have one number here that is like four occurs, you know, six times and then this is number seven that games five times. Or, you know, let's say it games eight times. So in that case, if k is equal to two, then my answer should be four and seven, correct?

Blue Centurion: Yeah.

Mutable Pigeon: Yeah. So let's say we put, take a heap of size two and we put first two elements, okay? With the map entry of one, five and two four. And it's a min heap. So when I get the third, when I get the third entry, that is three times three and my minimum is at least four. So I won't, I won't insert it, I will just discard it next time. If I get four that occurs six times and my min is this guy, I will replace it with four and six. And next when I get seven times eight. Now this is the min heap as I mentioned. So we will poll this guy and I will put seven times eight. So by the end min heap will have the two most frequent elements. So, but the complexity of this solution is going to be ok because we are going to insert the heap rhythm size k, right? So we will be having log k time of insert and we will be doing worst case and insertion. So I would say it would be ok with this solution. But there is one way, which if we want to improve is let's say we take the count array and we have like five and four and nine and 312. Let's say we take count of array dot length plus one, because that is the only possible counts we can have, right? 1212, one in case we have all different elements, if we have all different elements, and, and twelve if we have all same, okay, so what I'm doing here is we have a count of array and in this count we will actually store list. So this will be something like list of integer. Yeah. So something like this. And I will be an array of it.

Blue Centurion: An array of list, is that right?

Mutable Pigeon: Yeah, yeah. So what I will be doing is like this is the array of, so these are buckets of array, which is, you know, 0123 till twelve. Right, because that's what the length is. And for each guy will have.

Blue Centurion: Each.

Mutable Pigeon: Guy will have list or set, whatever. So let's say in case we have number. So each bucket index will represent the count here. And for this I will put two here in the list. And here I will put the five. I know the count is one because element one has this and this guy will have three. And if I continue my example, I'll have also six here, right. And six will have four. And now seven, list of index. Seven will have. Not seven, actually seven will be empty, but I will have eight. Eight will have seven because this is what, so now we know we want k, biggest number. And we can start from the back of the array. And twelve will be empty. We will let it go. And when we come to eight, eight, count eight.

Blue Centurion: Sorry. So the count is a list of arrays. So each element will be at least here, right?

Mutable Pigeon: Yeah. No, each element. So each element will be, each element will be a list. That's correct. Yeah, that's correct.

Blue Centurion: And that is example, it's an integer.

Mutable Pigeon: You mean in line 95?

Blue Centurion: Yeah. So here the count is a variable of array of list. Is that right? It's an array of, yeah. And so here each entry will be least instead of an integer. Right?

Mutable Pigeon: Yeah, I just showed it like an integer, but it will be a list. Like say there are two elements of count eight. Then it will come here.

Blue Centurion: I see, I see.

Mutable Pigeon: Yeah. Then if I take top k elements, then I will in top case and case two. Then I will just take these two elements like seven and nine.

Blue Centurion: I see, so the index here, this is the count. Yeah.

Mutable Pigeon: Yeah, that's what I said. Like one if we have all different elements and twelve if we have all same, then the count will be twelve. Right. The method. Yeah, I see. Yeah. You want me to code it?

Blue Centurion: Yeah, sure, sure.

Mutable Pigeon: Okay, so I'm using the same class if you, if you're okay. Okay, cool. So let's say we need to return top k elements. I would say, let's say, uh, maybe I would take list of integer. And I will go with the second approach because this guy is o of n. We are not doing like log k operations here. So I will go with this one. So yeah, public. Okay, let's say get top k and we would get integer of array and int case here. And let's say we'll have, we'll have a count of, okay, so let's say we have a, listen, you have a count. Here. Okay, before, because before I did this, I have to actually go through the map of integer, integer I am expect, okay so you said it's an integer only, so.

Blue Centurion: Yes, integer only.

Mutable Pigeon: Yeah. So let's say I go on for int a of array and let's say I have maps dot sput a map, dot get or default a of zero plus one. Okay, so I put the entries in the map, the count, once you have the map then you have to have a list of integer. Array is equal to, let's call it count and new, or a list of, let's say we have int n is equal to length and I have two have n plus one because we need one and twelve both. So yeah, ten plus one, I have discount. And let's quickly do four of four integer over map dot key set. Right. Let's say I'm going over map dot key set. And so before I do this, okay, I have to do if count if, okay, so let's say map dot get map, put a a will have one and map get key. Okay yeah, map get key. And this will be the, I would say occurrence of that key. Right? Now get your count. Okay, if count of that occurrence, the count of that occurrence is not equal to null then actually it has to be equal to, yeah. Then create a new arraylist of this and arraylist is going to be type integer. And let's say you get the new arraylist and say you added count occurrence dot add key. Right? So once you have this guy ready, you have to start iterating from let's say for and I is equal to array dot not array. Let's say we use count dot length minus one and I is greater than equal to one and we have our, we have k greater than zero and then I minus minus. Okay, seems like this is the one. Okay, so this is the result array. I'll add base cases here. Now if I have here and if count of I is equal to null. Right. If I just, if we don't get this count and just continue else get the list of integer, let's say I'll call it.

Blue Centurion: Just call it whatever.

Mutable Pigeon: Yeah, let's say I call it, our list is going to be part of result so let's say I get the count of I here and. Yeah, so I have, let's say if this is having top three elements, right? Top k. Okay, so let's say if this guy has top k, one, if what? If the counts are same, they will still be counted. Counted as like if both of these guys are, let's say if I have this one, one and one and then I have two two, right? And I want top, top two elements, top k, frequent elements, then I should be getting one two or I should be getting like nine also because two and one have the same count.

Blue Centurion: So, so you, if you get top two, right. So that would be one or two because, okay, both have frequency of four, but nine have only, only have a frequency of two. So that's the third most frequent one.

Mutable Pigeon: Okay, cool. So yeah, I think then I have to take care of that. For int, s is equal to zero. S is less than our list size. S plus, plus. Okay. Yeah, so this is my result. Result dot add list, dot get s. And here I have to actually mention that a is greater than zero. R list. And I'll do, okay, I think this guy, I'll do k minus, minus here. So, yeah, let me quickly see if I have any compile time issues. Okay. Cannot find symbol. Where is the line number? Is 70. Okay, public list and teacher get up. Okay. New array list. Okay, this guy. It is key. Okay, now I have what? New hash map at line number 73. What is this saying? List, integer result.

Blue Centurion: What is this?

Mutable Pigeon: New array list. New list of results. Cannot find symbol list. Do I need to import?

Blue Centurion: Yeah, probably you need to import. I'm not sure whether this editor or, or something.

Mutable Pigeon: Yeah, that seems like I need to import. Okay, yeah, let's see if I have any compile time issues. Okay, I have only one now, which is going to be at line 82.

Blue Centurion: You didn't return the result.

Mutable Pigeon: Oh, okay, that's one. Okay, that's one. And extends object declared in. It seems like I have issue here because I don't have angular bracket list of. It should not have any issue list of. No, it's fine, actually. Why this showing list of integer count is equal to new arraylist and is not this one.

Blue Centurion: You don't have a type to cover.

Mutable Pigeon: I don't think I need type as such. Yeah, I don't think I need type. I think it's going to be what is it saying? Cannot create list of count. I don't need it actually. Yeah, I don't need this guy for sure. And yeah, I think it should be fine.