Top K Frequent Elements - Bucket Sort - Leetcode 347 - Python

🚀 neetcode.io/ - A better way to prepare for Coding Interviews
🐦 Twitter: neetcode1
🥷 Discord: discord.gg/ddjKRXPqtk
🐮 Support the channel: www.patreon.com/NEETcode
💡 CODING SOLUTIONS: kzbin.info/aero/PLot-Xpze53leF0FeHz2X0aG3zd0mr1AW_
💡 DYNAMIC PROGRAMMING PLAYLIST: kzbin.info/www/bejne/bWTVZH6Nnqqpr80
🌲 TREE PLAYLIST: kzbin.info/www/bejne/hZ-2n2WOerZng7s
💡 GRAPH PLAYLIST: kzbin.info/www/bejne/e5isZqGLbsqnpLc
💡 BACKTRACKING PLAYLIST: kzbin.info/www/bejne/ppfMgpKGiJaabqc
💡 LINKED LIST PLAYLIST: kzbin.info/www/bejne/fWHCemCQe5WGaZo
Problem Link: neetcode.io/problems/top-k-elements-in-list
0:00 - Read the problem
2:58 - Drawing Explanation
9:42 - Coding Explanation
leetcode 347
This question was identified as an interview question from here: github.com/xizhengszhang/Leetcode_company_frequency
#sorted #array #python
Disclosure: Some of the links above may be affiliate links, from which I may earn a small commission.

[@Xeoncross]

I appreciate the time you put making and sharing all your content for free. Here is the $10 I might have spent on your udemy course.

[@NeetCode]

Thank you so much!!!

[@onlysubscriptions2152]

Does he have udemy course???

[@PeterPan-xe7qw]

@@onlysubscriptions2152 nah, just a hypothetical $10 he would’ve spent since most people pay wall this content, but he does it for free.

[@hamdi_]

As a side note, consider donating directly to the creators if they have a donation link, because KZbin takes a whopping 30% of your donation. In this case, Neetcode accepts Patreon donations, which takes a more reasonable commission of about 8%.

[@mostinho7]

@@hamdi_even 8% is too high tbh. Not for neetcode specifically, he sells his own courses and is set. But for someone else who might be in need of the money 8% is ridiculous

[@rhitamdutta1996]

I have never practiced DSA in my life, not even in college. After getting laid off, I stumbled across your videos to learn DSA. They are so crisp, informative, and to the point. I can't thank you enough.

[@e889.]

Hi you got any job?

[@rhitamdutta1996]

@@e889. not yet.

[@ozgurpeynirci4586]

Update?

[@rhitamdutta1996]

Hey guys, yes I did. Wouldn't have been possible without Neetcode.

[@hamza-chaudhry]

@@rhitamdutta1996 Nice

[@rohananjaria1009]

Best youtube channel for leetcode problems hands down.

[@justsimple6333]

i used your previous video on groupAnagrams to solve this, just hashmapped the array in to a defaultdict(int) den sorted the dictionary entirely in a descending order. Your videos have been really helpful, first time i solved a medium all by myself

[@trenvert123]

That's so cool that python has a convenient way to sort hashmap by value. I looked into it for java, and it is a nightmare. I would have to create my own comparator. I think if I'm doing that, I may as well just learn bucket sort at this point.

[@robinfelix3879]

haha, did the same, cheers

[@moveonvillain1080]

@@trenvert123 Java is sooooo verbose......

[@Albert-nc1rj]

@@trenvert123 In Go I faced the same problem, spent 20 minutes trying to sort a hash map by values (failed). So I just copied values into new array and sorted them there lol.

[@dumdum407]

@@Albert-nc1rj you can also insert the contents of your hashMap into a second hashMap, but use the values from frequency hashMap as the keys of the second hashMap. Once you have done that, take all the keys out into an array, sort the array and retrieve first k values, use first hashMap to get integers and return that.

[@maierdanefan6998]

Amazing contents! The best algorithms channel that focus on logic and thinking in a clear way. Happy to have found this channel, been writing neetcode ever since.

[@joshmarion8640]

Am I the only one who is a little confused as to how this solution is O(N). If you loop through the array which is the size of the array, and then in each index you might have to loop through up to N times. So how is this not o(n^2) Edit: Nevermind, I think I realize it now, I figured I would write it out for anyone who might still be confused. As we traverse through the array, we go through the whole array. So this is O(n). But we aren't doing an operation n times at each stop. We are doing N more operation throughout the entire array. So even though the for loops are nested, we are doing N more operations throughout a for loop which is N, so the total is just N+N, which simplifies to O(N)

[@abhishekdhyade7500]

Thanks a lot buddy! I was scratching my head off to find out this same doubt. Now that I saw your comment, I was able to understand it. Thanks again!!

[@ahmedmansour5032]

So essentially the inner loop is just operating on the subset of N elements?

[@quanmai5759]

I would say it's n+k rather than n+n, because the size of the res array is k. So after looping through the freq array of size n, you only need to fill the res array k times then stop, so k more operations. Still it's O(n)

[@ubermensch_1111]

@@quanmai5759 Yah I also think so it will be n+k

[@DiaaHaresYusf]

@@ahmedmansour5032 but at worest case you will face frequency = 1 for each element in nums .. and O(N) is always calcualted in worest case, I have made a commend on video please go through it , you will understand what I am saying

[@sandeshpaudel9665]

for the heap solution, it's better to use a min heap of size k rather than using a max heap and then removing max k times. Using the min heap, you would remove min and add the next frequency. by the end, you are left with k most frequent ones and removing the min gives you the answer. You can reduce this to n log k and not n log n

[@gurmukhsinghnirman4935]

and even for values like k = 1e9, logk is around 30 so the complexity is around O(30*n) which is basically O(n)

[@sandeshpaudel9665]

@@gurmukhsinghnirman4935 indeed!

[@PippyPappyPatterson]

How would you cap the size of the heap `h` at size `k`? As you're adding frequencies, `if len(h) > k: heapq.heappop(h)`?

[@sandeshpaudel9665]

@@PippyPappyPatterson so let's suppose k = 3 and you have numbers [1 , 2 , 3, 4, 5]. You can find the k-largest or in this case 3rd largest using a min-heap of size 3. As you add in numbers, your heap can grow like this: [ 1 ] [ 1 , 2 ] [ 1, 2 , 3 ] ** you're capped at 3 *** [2, 3, 4] ** add next( 4 ) and remove min (1)** [3, 4, 5] ** add 5, remove 2 ** Now the head of the heap will be the 3rd largest element.

[@MinhNguyen-lz1pg]

@@PippyPappyPatterson here class Solution: def topKFrequent(self, nums: List[int], k: int) -> List[int]: num_to_count = collections.defaultdict(int) for num in nums: num_to_count[num] += 1 min_heap = [] for num in num_to_count: if len(min_heap) < k: heapq.heappush(min_heap, (num_to_count[num], num)) else: heapq.heappushpop(min_heap, (num_to_count[num], num)) res = [] while min_heap: _, val = heapq.heappop(min_heap) res.append(val) return res

[@Thrashmetalman]

the one thing I dont like about usage of heap questions is that most of the times you havge to default to some library to do it cause I doubt any of us could code up a heap in a phone screen.

[@Number_Crunch]

The algorithm that you explained at 3:15 was counting sort and not bucket sort. What you did, however, towards the end was similar (not same as) bucket sort.

[@namoan1216]

is it different?

[@chrischika7026]

@@namoan1216 no hes wrong

[@tweefeety]

I love you man. You're an actual angel. Your explanations are always so clear. And your drawings are so easy to understand.

[@NeetCode]

Thanks, appreciate the kind words 🙂

[@fortitude242]

@@NeetCode you are n angel. :)

[@awesome_ashu]

We can optimize it more by storing the maxFrequency while creating the HashMap (which has the integer and their corresponding frequency). Then, the next iteration to get the required elements can start from this maxFrequency instead of N.

[@HimanshuPant-c1v]

this is how i did it, i basically converted the list in a dict, sorted it using values and the took out the k most frequent values but i really appreciate your videos. class Solution: from collections import Counter def topKFrequent(self, nums: List[int], k: int) -> List[int]: value = [] my_dict = dict(Counter(nums)) my_dict2 = dict(sorted(my_dict.items(), key=lambda item: item[1], reverse = True)) mlis = list(my_dict2.keys()) for i in range(k): value.append(mlis[i]) return value

[@andrepinto7895]

While counting, you can keep track of the max occurrences and then you only need to initialize freq to that max instead of len(nums)

[@amitkoushik5504]

Good one ...I spent a lot of time understanding this but finally got it..🤗🤗

[@arneishprateek6444]

Sure but it's still O(N).

[@s1kebeats]

thx

[@MrHarryGaming]

count = {} maxFreq = 0 # or 1 for each in nums: count[each] = count.get(each, 0) + 1 maxFreq = max(maxFreq, count[each]) freq = [set() for i in range(maxFreq + 1)]

[@wow_donnie]

I came up with this solution originally but really appreciated the thoughtful description of the linear solution! result = defaultdict(int) for num in nums: result[num] += 1 result = dict(sorted(result.items(), key=lambda item: item[1])) return list(result.keys())[-k:]

[@VineetKrGupta]

I got my first job after following your neetcode 150, 2 years ago. now after the layoff i am here again learning the dsa.

[@pinakadhara7650]

Thanks for the video! I came up with the same solution except I assumed each element is "repeated unique number of times" from the problem statement - "It is guaranteed that the answer is unique.". So instead of looping over each lists, I just considered the first element.

[@eulier1]

That's was a very interesting way to solve a specific real life problems, by counting and working with hash and array as a way to identify K most frequent elements. This can be useful for small to big business handling inventory or when you need to pack-up your stuff to travel.

[@kwakukusi4094]

I got a similar question on my onsite interview with amazon (not the same question but same concept). I did not know bucket sort so I used the sorting method. The interviewer said there was a way of getting a linear time complexity and I did not know what to do .

[@tszyinshirleycheung4040]

I think the runtime of using heap is O(n log k), we need O(n) to construct the heap and remove an item cost O(log k) ?

[@MrACrazyHobo]

Yes, this is even what the leetcode official answers says

[@michaelchen9275]

Isn't it O(log n) to remove from a heap with n elements? And we do that k times, so that makes O(k log n).

[@theniknik0999]

@@michaelchen9275 If we restrict the heap to be of size k (since we only care about k most frequent), at worst case we'll end up popping n elements. i.e. O(n log k)

[@akarsan9121]

questions: lets assume we have array [1,2,2,3,3,4,4,4] and k = 2. Then for first most frequent we will have 4 but what about the 2nd most frequent element? cuz count of both 2,3 is 2 so how do we decide which value to insert into our resulting list??? Please help ps: "The test cases are generated such that the answer is always unique." is this statement accounting for that edge case?

[@gabrielfonseca1642]

Yes this wouldn't be included in the test cases because there are two solutions (2, 4) and (3, 4). However note that if k = 3, then (2, 3, 4) and (3, 2, 4) are both solutions, which is why the problem says to give the answer in any order

[@SHARMATUSHAR1_]

Also, you can do res += freq[i] in line 13. The problem description mentions the solution will be unique. So, we know that all the elements added if will either match k or will be lesser. So, no need to run a loop.

[@hiota45]

I used a dictionary and pair of while loops in python. Basically, while the nums list > 0, I would .pop the 0th number off the list and check to see if the number was in the dict as a key. If it was, add one to the value for that key. If not add it to dict as a key with a value of one. When the nums list was exhausted, another while loop for k > 0. Use the python max function to give a variable equal to the key with the highest value. Add that key to the results list, set the key value to 0, and -1 on k. I'm a novice though so this may be suboptimal?

[@theornament]

I did the solution with priority queue and hashmap and it seemed to have better time complexity and space efficiency than using bucketsort. I feel like this is tricky because, when we are solutions for problems, we start analyzing which data structure we are going to use, its time complexity, etc. based on how those data structures are regularly implemented. The thing is, algorithms and built in functions in languages have improved drastically that they take less time than what theoretically they should take. It's tricky but are those are things that we should consider as well?

[@randomystick]

for the return function, an alternate way is to use the extend() method in python: res = [ ] ptr = len(frequency)-1 while len(res)

[@kestiv2429]

Result size will be wrong if len(frequency[-1]) > k I think

[@vachannadupalli6133]

@@kestiv2429 The question guarantees that there is a unique solution. Hence every time we extend the result array, at some point it will(has to be) be equal to k. If it were not guaranteed, then you would be right.

[@YNA64]

Holy this is so much clearer than the quick select one.... Thank you so much

[@ChristopherElwell]

But not constant space complexity

[@netanelkaye3014]

People say you are supposed to learn enough to be able to figure out leetcode problems, as opposed to memorizing leetcode. Are we seriously supposed to have been figured this method out? This was so specific...

[@edd4851]

When you encounter a similar problem next time, you will think , well, i already saw it somewhere. I can do this.

[@netanelkaye3014]

@@edd4851 Really? How many leetcode problems are solved this way?

[@_carrbgamingjr]

@@netanelkaye3014 1😅

[@wotizit]

its best not to think about it lol

[@beaglesnlove580]

@@netanelkaye3014none of

[@mohamedeltawab]

Your explanation is like art! Thank you!

[@saifmohamed1776]

i didn't see any body came up with this solution in the discussion on leetcode , all of the solutions were use heap or may be some of them use quick select ; so i was afraid that i analysis my algorithm wrong but after watching you i know that i was right about my solution .

[@CarlJohnson-iv7sn]

Infact the top solution in the discuss is using bucket sort itself.

[@yarnehermann]

I'm thinking you can just invert the frequencyMap to {frequency: list of values with that frequency} and then sort the keys in that inverted map. This sorting would be O(sqrt(n) log(sqrt(n))) (which is < O(n)) because there cannot be more than O(sqrt(n)) different frequencies (if each value has a different frequency, then n = O(c * (c+1) /2), with c being the number of distinct frequencies). Then it's just a matter of iterating over the reverse sorted keys and adding values to a resultArray until that array reaches length k lookupDict = defaultdict(int) for n in nums: lookupDict[n] += 1 inverseDict = defaultdict(list) for key, v in lookupDict.items(): inverseDict[v].append(key) sortedKeys = sorted(inverseDict.keys(), reverse = True) sortedKeysIndex = 0 res = [] while k > 0: values = inverseDict[sortedKeys[sortedKeysIndex]] if len(values) > k: res.extend(values[:k]) else: res.extend(values) k -= len(values) sortedKeysIndex += 1 return res

[@pragnyatata491]

really love the approach taken, thank you

[@gabrielfonseca1642]

This is still O(n) because of the for loop on line 2, but yeah it's slightly more efficient

[@johnzheng849]

Your video got a me a job as an SDE at AWS!!

[@NeetCode]

Congratulations 🎉🎉

[@jwastken8814]

Hey John, which neetcode 150 questions did they ask? I have a phone interview coming up

[@symbol767]

This is perfect, thank you bro, took me a while to understand this problem, gonna need to redo it without looking at the solution in a couple days. Thank you, liked and commented again to support

[@denshaSai]

Got this question for google, what to do then if input is streaming (like a log)? guess we keep updating the count (histogram), and rebuild the freq array everytime?

[@prithvirajchavan9186]

So basically the map of frequency to values can actually be of size O(n*n)? The first 'n' is due to worst case frequency being 1 to n and the second 'n' is because at most 'n' numbers can have the same frequency of 1. I kinda understand how its not really O(n*n) and a bit less than that as the duplicates are being consumed into a single key, but its still more than O(n) right? Another observation: The only possible values of frequency that have a value attached to them are the subsets that add up to N. In the given example, 100 occurs once, 2 occurs twice and 1 occurs thrice, so sum of frequencies is actually 1+2+3=6 which is N in this case. So the worst case space complexity is actually O(n*length of longest subset of frequencies that sum to N). Dont really know how I ended up with a Subset Sum problem just because i couldnt justify the O(n) space complexity lol. Any clarifications are appreciated :D

[@SHARMATUSHAR1_]

You can also use count = Counter(list) This will count in itself. No need for a loop. Another day thanking Guido van Rossum for making my life easier.

[@md-ayaz]

One thing to note in python. Please do not intialize like this ( if you are ) bucket = [[]] * (len(nums) + 1) this references to the same list and your output will be wrong. Use the bucket = [[] for _ in range(len(nums)+ 1)] , as mentioned in the video.

[@ronakpatel7911]

What I don't quite understand is that when I implemented this solution, the speed and memory is not as efficient compared to my initial solution in ``` class Solution: def topKFrequent(self, nums: List[int], k: int) -> List[int]: count = {} for num in nums: count[num] = 1 + count.get(num, 0) res = sorted(count, key = count.get, reverse = True) return res[:k] ``` Even though using sorted() here causing it to be n log n... Can someone explain why this one is appearing to be much quicker than the solution in the video?

[@pruthvipegasus]

sorted() uses quick sort which has O(n log n) time complexity in the python interpreter.

[@1000marcelo1000]

It should be `for i in range(len(freq)-1, -1, -1):` instead of `for i in range(len(nums)-1, 0, -1):` right?

[@vdyb745]

I was wondering where you were going with the bucket. This is so clever !!! Brilliant !!

[@rakshitkumar1141]

I think your solution is O(n2). because in last part you performed n operations inside n. for i in range(len(freq) -1, 0, -1): ==== O(n) for n in freq[i]: ==== O(n)

[@naveennvnkumar4615]

I am bit confused, at 12:25 and lines 12,13 wouldn't it become a N^2 solution if all the elements are distinct

[@ParryHotter73]

yeah, the example he showed is actually the worst case and it may be O(n²) but we can't say n² cause we don't find n elements at every index, this is an exceptional and not like general nested loop

[@naveennvnkumar4615]

@@ParryHotter73 ok got it. thanks buddy

[@sivasangaranv4746]

Sir, good explanation; but could you please show the heap based solution? O(Klog(n)) is better than O(n) in some situations, particularly when K

[@propropropropuser]

p sure he said it wrong. it's supposed to be nlogk

[@xrdevelopment1390]

is it just me or after struggling on a particular part for a while.... then it hits!!!!! best feeling ever. NeetCode, I am following along every problem and have the confidence that I'll get my dream tech position. thank you, its the same feeling I had when I found out about khan academy in high school

[@Milan-vi1bq]

we're both gonna get the dream job homie!

[@arungowda]

We can do a little space optimization by having max(counts) size for bucket instead of nums.length

[@LeetCodeSimplified]

Good point!

[@shreyaskaup]

i had actually thought of the 2nd array implementation you said with N array size.. but i didnt think on how I would extract the top K as you did by going backwards! you're a genius!

[@chesea9790]

Awesome video! One small thing, shouldn't it be O(n log k) instead of O(k log n) for the heap solution since there's n elements, heap of size k means log k time to heapify and n calls so n * log k?

[@AustinCS]

Yes, it isn’t O(n) like he said

[@indoor-adventurer]

Does this become linear time because, we iterate the list of nums (n), then the map of counts (n), and eventually the frequency list (n). Which ends up as O(n + n +n) = O(3n) = O(n) ???

[@mach1ontwowheels632]

where do you get that the max input array size is 6? The problem says the length of the array can be up to 100k

[@shuvbhowmickbestin]

Can we not use a heap/priority queue instead of using an arrya? Wouldn't that automatically keep the most frequent elements at the top if the sorting was done according to the count?

[@ARJUN-op2dh]

More simpler way from collections import Counter def dx(nums: list, k: int): x = Counter(nums).most_common()[:k] ls = [] for i,j in x: ls.append(i) return ls

[@acecool1715]

Thank you for your time teaching. Can i ask what software you use for the black board in the background, or anyone know? Thank you all.

[@ekimsr123]

Do you think just using a heap for an interview is sufficient?

[@bundayyolayinka3352]

My current status: Understands the Questions, Have an Idea of how to start (not how to finish), Watches just your drawing explanation, Realize what is missing, Writes the whole solution.

[@anonymoustv8604]

that's great man. Keep going

[@pacomarmolejo3492]

keep grinding. You will get there

[@Ervaibhav75]

Should we consider the input will be always sorted?

[@seanbarel2]

Is this solution also works when theres no limit on the values of the initial array? Or it assumes that the values in the array are bounded?

[@Hrzybs]

Instead of using a second loop while adding to the result I just checked if the values in the array exist like that for i in range(len(tables)-1,0,-1): if (tables[i]): result.extend(tables[i]) if (len(result)>=k): return result for me it's easier to understand this way

[@chamikaonyt]

Excellent video as usual.... Can someone clarify whether this is an efficient solution. This is what I came up on my own initially and it passes all test cases as well. I just want to know if I wrote this in an interview will this be accepted as an efficient solution class Solution: def topKFrequent(self, nums: List[int], k: int) -> List[int]: # let n = len of nums answer = [] my_counter = dict(Counter(nums)) # {1: 3, 2: 2, 3: 1} -> O(n) priority_queue = [(-value, key) for key, value in my_counter.items()] # O(n) heapq.heapify(priority_queue) # O(n log n) for i in range(k): # O(k) answer.append(heapq.heappop(priority_queue)[1]) return answer # total time ==> O(n log n)

[@venkatshiva8275]

First of all thank you very much @neetCode for this amazing explanation. Anyone has suggestions or improvements on the below solution?? class Solution { topKFrequent(nums, k) { let countObj = {}; const resArr = []; let ctr = nums.length; for(const i of nums){ countObj[i] = (countObj[i] || 0) + 1; } while(ctr > 0){ for(const key in countObj){ if(countObj[key] == ctr && resArr.length < k){ resSet.push(key); } } if(resArr.length == k){ break; } ctr--; } return resArr; } }

[@FieryDragon883]

Why would we initialise freq with (length of array + 1) empty lists instead of just len(array)

[@yxngboypolo]

11:42 you could also use `for i in reversed(freq):` and modify the code as needed

[@John-z7m]

I guess this is what makes the question medium? I just did a quick hash count of numbers and then Object.entries()'d it out and sorted by second index, then pulled the first k values out of that. So O(n log m). I wonder if this is weak in an interview. Offering bucket sort solution (or sometimes other niche solutions) just feels like I musta done the same question or a nearly identical question to pull it quickly out of an interview though... I'm surprised Neet code starts with two mediums whose solution feel a bit hyper specific. Is it intentional?

[@angelinazhou667]

If we were to use the max heap approach, how would we apply heapfiy if heapify only takes in a list and not a dictionary? Thanks for the amazing video!

[@VarunMittal-viralmutant]

The final array that you create may have a lot of holes. Say there are 1000 elements, but consisting of only 1's and 2's, then the list will be full of holes. It can be further optimized by keeping a 'max_freq' variable which will provide an upper bound on the size of the array. This max_freq can be updated while creating the hash-map.

[@mearaftadewos8508]

or may be this may not be the most efficient way to do this for all kinds of inputs like the one you said and unsorted numbers. It doesn't give the top frequent one's. Dictionary with maxheap is the one that can handle all possible inputs.

[@nishanttripathy8275]

For the heap the time complexity should be n.Log(k) since the max size of the heap can only be k and n is the number of elements

[@firasyousfi2269]

Nope, if you are using a maxHeap it will be k log n. Because for that you need to heapify the whole thing first so n elements would be in the heap. Then you would pop 'k' times from the heap of size 'n'. So O(k log n). This is Valid if you use a maxHeap!!! If you are using a minHeap then you would be correct, then the heap would have a max size of k as you said. And you would loop n times and push then poll when size reaches k.

[@karthik_jain]

Hey! Thank you for sharing this. It is very helpful. I was confused with the space complexity of this solution. Can you please answer this?

[@Not_TVs_Frank]

Is using a defaultdict frowned upon for interviews? I can see how a Counter would be a bad call for an interview since it basically creates the multiset and counts up the number frequency for free. A defaultdict is a little different and saves you the bother of creating the zero values in the dict used for tallying up the frequency of values. It's a small convenience, but one I like to use sometimes. What do you think, is using a defaultdict a bad look in a coding interview?

[@mark1A100]

explain it and why your using it. expect to be asked to do without

[@bhargav1811]

Does coding in python for doing Competitive coding has any side-effects? I know python and am doing coding in python and do i need to learn c++ ? Open for suggestion !!!

[@TomerBenDavid]

Is the bucket sort always about having the number of buckets as the size of the highest frequency number or which kind of bucketing we use in the standard bucket sort?

[@nagadeepakkaza3433]

I have a doubt. Time complexity of building a hashmap to get frequencies is O(nlogn) right ? How can it be O(n), the overall time complexity should be O(nlogn) right

[@danny65769]

What is the space complexity of the solution? Is it O(n + k + n) == O(n)? n for hashmap of counts, then array of k size, but an array item can contain a list of n items in the worst case being all numbers being distinct? How does space complexity work with list of lists? Thanks.

[@mastermax7777]

what about c++ videos and explanations? i looked at the website solution and some of the answers in c++ have pretty BAD RUN TIME on leetcode. (bottom 10% compared to different solutions.) The question i am talking about is Group Anagrams

[@devashishubale1565]

I was solving 692, and I got AC, I remembered this video came back to this. Thanks for such detailed videos.

[@rajivshah3661]

Just a question, What if after building the hashmap, we sort the dictionary by values using sorted() and return k keys?

[@hype-r3076]

The sorted() is done in O(N log(N)). Although ur solution is correct it’s not the most efficient

[@hype-r3076]

And in worst case N^2

[@atulkumar-bb7vi]

Trust me bro, you are amazing explaining things. Thanks a lot for such content. Pls keep posting..

[@akashp4863]

is log(n) better solution than nlog(n) and log(n)?.... can you create a video on how to find big o and which one is better than which? i know there are many videos in youtube, but yours will be the best

[@amrojjeh]

In this case, a heap sort might be more optimal. Even though your code is O(n), it's making a ton of allocations, which may make it practically slower in some cases.

[@34535fff]

Man I could not do it myself because I didn't understand problem clearly, after 50 seconds of the video I understand and did it with ease, thanks a lot. Now I am going to watch rest of the video to learn optimal solution)

[@fakruu]

how does it work for negative numbers ? For eg : arr [ -1, -1] and k = 1

[@leetcodemermaid9804]

For the first heap solution, you can do a total complexity of n Log(k). Instead of heapify the whole dictionary ,you can heap push and heap pop when you go through the dictionary. When len(heap) > k, heap pop.

[@MehmetDemir-xi3yy]

nlogk generally bigger than klogn k=1200 and n=1500 nlogk ≈1500×3.0792≈4618.8 klogn=1200×log1500 ≈ 3811.2

[@lolvinnieboylol]

my solution involves the sorted buildin function, after seeing your solution I felt like I cheated... should I prevent using these buildin algorithms?

[@srinadhp]

Got the same question in one of the phone screens. I came up with map and heap approach. The interviewer asked for an optimization for memory. And, I could not think of any other solution. It costed me next round unfortunately. Only if I notice your solution!!!! :-( Again! Great explanation and "neat" solution!!

[@harigovind11]

This solution is also not memory optimised. Using "min" heap instead of max heap with size k would be memory optimised.

[@kobeissi721]

@@harigovind11 Wouldn't it still be O(N) since you still need to add them to a map to store the frequencies?

[@sampatkalyan3103]

@@harigovind11 "Using "min" heap instead of max heap with size k would be memory optimised. " can you explain how ?

[@numberonep5404]

I love how your videos are always so damn clear :) ! Thanks alot Btw, here is a weird(?) quicksort version that beats 93%: def topKFrequent(self, nums: List[int], k: int) -> List[int]: count=list(Counter(nums).items()) def quick(l,r): pivot,p = count[r][1],l for i in range(l,r): if pivot>=count[i][1]: count[i], count[p] = count[p], count[i] p+=1 count[r], count[p] = count[p], count[r] if p>len(count)-k: return quick(l,p-1) if p=ind]

[@NeetCode]

Nice!

[@heathergray4880]

Mine beats 98.62 time and 90 on space and is two lines long :)

[@lucaslau8379]

@@heathergray4880 would you share your code for learning please?

[@sangramshinde2211]

nick white, kevin, neetcode best guys to get the perfect explanation...

[@WinnerSingh]

Well what mistake I did is - I watched other python tutorials What good thing happen is - I am watching your videos But you are geniuses you explain well but I have to watch two times every video to understand correctly. Thank you

[@ADC3131]

i used a heap to solve the problem, doesn't the below solution have an O(n) time complexity? i only loop through the input array and use heapify, which has a linear time complexity (i push the frequencies by first negating them, as the heapify method returns a minheap not a maxheap) import heapq class Solution: def topKFrequent(self, nums: List[int], k: int) -> List[int]: counts = {} for num in nums: if num in counts: counts[num] = counts[num] + 1 else: counts[num] = 1 counts_tuples = [(-counts[el], el) for el in counts] heapq.heapify(counts_tuples) res = [] for i in range(k): res.append(heapq.heappop(counts_tuples)[1]) return res

[@parthshah1563]

Hi Neetcode, I got following solution. Can please tell me it is optimal or not? dic = Counter(nums) dec = dic.most_common() res = [] for i in dec: if len(res) != k: res.append(i[0]) return res

[@crnpowerimmortal]

Amazing solution, I have refered refered your solution and created a similar version. Will the time complexity of this solution be same ? def top_k_frequent(self, nums: list[int], k: int) -> list[int]: from collections import Counter counter = Counter(nums) res = [] for ele in counter.most_common(k): res.append(ele[0]) if len(res) == k: return res

[@hamoodhabibi7026]

wwow .most_common(), and I believe so

[@hithambasheir3283]

Nice explanation and iteration from lowest performance to the best one, but I'm confused about something, where we're sorting the words lexicographically?

[@MohitKumar-er1qu]

How does the heap solution cost us (Klogn)? Should it not be (n + k logk)? as building a heap costs us O(n) and getting the max value log(k) k times.

[@sami9323]

Why is O(n) better than O(k log n)? Would that be the case for very large n?

[@leandrormor]

Does the "for i in range(len(freq) - 1, 0, -1):" goes only until 1? (not zero right? which is fine for our case)

[@YawarMurtaza-z9k]

Trying to understand why we need to sort the map.. why is it not acceptable to linearly go through the map, take the items has have value (occurrence) greater or equal to k? It will still be a linear time solution right? What am i missing? Thx

[@YawarMurtaza-z9k]

got it! It has to be in order of occurrence

[@Nohope__]

or alternatively sort the count table by values, and then return first k elements : # create hash_table where num in nums is ke # increment value for each key repeatation table = dict() for n in nums: if n in table: table[n] += 1 else: table[n] = 1 # sort by values table = sorted(table.items(), key=lambda item: item[1], reverse=True) # create a list answer = list() # store to answers for item in table[:k]: answer.append(item[0]) return answer

[@basedasuka]

ur solution is much faster for me, thank u

[@pratikmhatre4815]

I am not sure about Python but for Java, adding item to a dynamic list takes O(n) time. So above solution becomes O(n^2) ?

[@danobot12]

I used a HashMap and treated it like a "count occurances of characters in a string". At the end, I simply ignore map entries where count is less than required K. This made the entire problem very easy and I did not have to think about Heaps and bucket sort and all that. Is this an acceptable solution? What am I missing? ``` class Solution: def topKFrequent(self, nums: List[int], k: int) -> List[int]: countMap = defaultdict(int) for n in nums: countMap[n] += 1 return [key for key, count in countMap.items() if count >= k] ```

[@KardboardCode]

Hey everyone, I have one question. Why is this solution considered O(n) where n is the size of the nums array? Why are we not looking at the worst case complexity for lines 11-16 at 12:52 If all items are unique Line 12: i will iterate over the range (0 , n) //Size of freq is n+1 Line 13: if all items inside nums are unique, every item will be present at the first index This will lead to looping through the entire length of the array n => this gives us O(n^2)

[@tomskrovan6622]

I believe the length of the frequency/bucket array could be lower than len(nums) + 1 because of the rule that the # of distinct elements >= k. This tells us that there will be at least k distinct elems and since each one must have at least one frequency, no single number could occupy all len(nums) spots (unless k is 1). Therefore I think it could be further optimized (albeit minimally lol) to: freq = [ [ ] for i in range( (len(nums) + 1) - (k - 1) ) ]

[@ichigokurosaki7762]

Could you explain how the for loop works within the array, I didn't understand that part? Freq=[[] for i in range(Len(nums)+1)]

[@Grace-gk9og]

Can someone explain why in line 12 of the code you have to -1 from len(freq)? since with the example of nums = [1, 1, 1, 2, 2, 3], the length of freq is 6, if you decrement from len(freq) - 1, which is 5, wouldn't you go 4, 3, 2, 1 and completely miss the largest number? I'm so confused, please help

[@AJ-ju7tl]

there is a mistake at the end, heap solution is O(N logK ) not O(K logN)!!! otherwise this is a very good explanation, thanks!

[@arnavgoel1022]

This won't work for negative values. The question you are referring to from Leetcode also has negative values as inputs in the array

[@rohanaurangabadkar951]

Best explanation this helped me in solving Top K Frequent Elements and Sort Characters By Frequency