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
2 жыл бұрын
Thank you so much!!!
@onlysubscriptions2152
Жыл бұрын
Does he have udemy course???
@PeterPan-xe7qw
Жыл бұрын
@@onlysubscriptions2152 nah, just a hypothetical $10 he would’ve spent since most people pay wall this content, but he does it for free.
@hamdi_
Жыл бұрын
As a side note, consider donating directly to the creators if they have a donation link, because KZitem takes a whopping 30% of your donation. In this case, Neetcode accepts Patreon donations, which takes a more reasonable commission of about 8%.
@mostinho7
11 ай бұрын
@@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
5 ай бұрын
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.
4 ай бұрын
Hi you got any job?
@rhitamdutta1996
4 ай бұрын
@@e889. not yet.
@ozgurpeynirci4586
Ай бұрын
Update?
@rhitamdutta1996
Ай бұрын
Hey guys, yes I did. Wouldn't have been possible without Neetcode.
@hamza-chaudhry
2 күн бұрын
@@rhitamdutta1996 Nice
@tweefeety
3 жыл бұрын
I love you man. You're an actual angel. Your explanations are always so clear. And your drawings are so easy to understand.
@NeetCode
3 жыл бұрын
Thanks, appreciate the kind words 🙂
@fortitude242
2 жыл бұрын
@@NeetCode you are n angel. :)
@rohananjaria1009
3 жыл бұрын
Best youtube channel for leetcode problems hands down.
@justsimple6333
2 жыл бұрын
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
Жыл бұрын
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
Жыл бұрын
haha, did the same, cheers
@moveonvillain1080
Жыл бұрын
@@trenvert123 Java is sooooo verbose......
@Albert-nc1rj
6 ай бұрын
@@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
4 ай бұрын
@@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.
@joshmarion8640
Жыл бұрын
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
Жыл бұрын
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
Жыл бұрын
So essentially the inner loop is just operating on the subset of N elements?
@quanmai5759
Жыл бұрын
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
Жыл бұрын
@@quanmai5759 Yah I also think so it will be n+k
@mohamedhassan-ub4kj
Жыл бұрын
@@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
2 жыл бұрын
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
2 жыл бұрын
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
2 жыл бұрын
@@gurmukhsinghnirman4935 indeed!
@PippyPappyPatterson
2 жыл бұрын
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
2 жыл бұрын
@@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
2 жыл бұрын
@@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
@maierdanefan6998
2 жыл бұрын
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.
@leofastov9205
Ай бұрын
I'd like to suggest a minor improvement: for n in nums: count[n] = 1 + count.get(n, 0) if count.get(n, 0) > max_val: max_val = count.get(n, 0) freq = [[] for i in range(max_val + 1)] The max_val variable is used to track the maximum frequency instead of using the length of nums. This can potentially save some space if the maximum frequency is significantly less than the length of nums. Addition of max_val adds a small const time overhead
@Number_Crunch
2 жыл бұрын
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
7 ай бұрын
is it different?
@chrischika7026
3 ай бұрын
@@namoan1216 no hes wrong
@Thrashmetalman
3 жыл бұрын
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.
@andrepinto7895
2 жыл бұрын
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
Жыл бұрын
Good one ...I spent a lot of time understanding this but finally got it..🤗🤗
@arneishprateek6444
Жыл бұрын
Sure but it's still O(N).
@s1kebeats
Жыл бұрын
thx
@MrHarryGaming
Жыл бұрын
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)]
@netanelkaye3014
7 ай бұрын
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
3 ай бұрын
When you encounter a similar problem next time, you will think , well, i already saw it somewhere. I can do this.
@netanelkaye3014
3 ай бұрын
@@edd4851 Really? How many leetcode problems are solved this way?
@_carrbgamingjr
2 ай бұрын
@@netanelkaye3014 1😅
@wotizit
2 ай бұрын
its best not to think about it lol
@beaglesnlove580
25 күн бұрын
@@netanelkaye3014none of
@awesome_ashu
Жыл бұрын
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.
@SHARMATUSHAR1_
2 ай бұрын
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.
@SHARMATUSHAR1_
2 ай бұрын
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.
@donnieellstrom7643
2 жыл бұрын
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:]
@md-ayaz
21 күн бұрын
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.
@samihk1
2 жыл бұрын
For some reasons you have solved all blind-75 related Heap problems in some other way :D Best one is this for better runtime.
@mohamedeltawab
3 жыл бұрын
Your explanation is like art! Thank you!
@pinakadhara7650
Жыл бұрын
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.
@johnzheng849
2 жыл бұрын
Your video got a me a job as an SDE at AWS!!
@NeetCode
2 жыл бұрын
Congratulations 🎉🎉
@jwastken8814
2 жыл бұрын
Hey John, which neetcode 150 questions did they ask? I have a phone interview coming up
@YNA64
2 жыл бұрын
Holy this is so much clearer than the quick select one.... Thank you so much
@ChristopherElwell
5 ай бұрын
But not constant space complexity
@yarnehermann
2 жыл бұрын
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
9 ай бұрын
really love the approach taken, thank you
@tszyinshirleycheung4040
3 жыл бұрын
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
3 жыл бұрын
Yes, this is even what the leetcode official answers says
@michaelchen9275
3 жыл бұрын
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
2 жыл бұрын
@@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)
@bryanleow5024
Жыл бұрын
for the return function, an alternate way is to use the extend() method in python: res = [ ] ptr = len(frequency)-1 while len(res)
@kestiv2429
Жыл бұрын
Result size will be wrong if len(frequency[-1]) > k I think
@vachannadupalli6133
Жыл бұрын
@@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.
@symbol767
2 жыл бұрын
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
@kwakukusi4094
2 жыл бұрын
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 .
@VineetKrGupta
3 ай бұрын
I got my first job after following your neetcode 150, 2 years ago. now after the layoff i am here again learning the dsa.
@eulier1
20 күн бұрын
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.
@chesea9790
6 ай бұрын
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
3 ай бұрын
Yes, it isn’t O(n) like he said
@naveennvnkumar4615
2 жыл бұрын
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
Жыл бұрын
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
Жыл бұрын
@@ParryHotter73 ok got it. thanks buddy
@xrdevelopment1390
Жыл бұрын
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
Жыл бұрын
we're both gonna get the dream job homie!
@vdyb745
2 жыл бұрын
I was wondering where you were going with the bucket. This is so clever !!! Brilliant !!
@bundayyolayinka3352
2 жыл бұрын
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
2 жыл бұрын
that's great man. Keep going
@pacomarmolejo3492
7 ай бұрын
keep grinding. You will get there
@arungowda
2 жыл бұрын
We can do a little space optimization by having max(counts) size for bucket instead of nums.length
@LeetCodeSimplified
Жыл бұрын
Good point!
@yxngboypolo
5 күн бұрын
11:42 you could also use `for i in reversed(freq):` and modify the code as needed
@saifmohamed1776
3 жыл бұрын
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
2 жыл бұрын
Infact the top solution in the discuss is using bucket sort itself.
@VarunMittal-viralmutant
2 жыл бұрын
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
2 жыл бұрын
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.
@ronakpatel7911
10 ай бұрын
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
4 ай бұрын
sorted() uses quick sort which has O(n log n) time complexity in the python interpreter.
@ARJUN-op2dh
2 жыл бұрын
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
@hiota45
8 күн бұрын
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?
@venkatshiva8275
Ай бұрын
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; } }
@jugsma6676
4 ай бұрын
The simplest method is: def topKFrequent(self, nums: List[int], k: int) -> List[int]: cache = {} for c in nums: if c not in cache: cache[c] = 1 else: cache[c] = cache[c] + 1 cache = dict(sorted(cache.items(), key=lambda i:i[1], reverse=True)) return list(cache.keys())[:k] O(nlogn)
@tarunkathuria2705
5 ай бұрын
You don't need bucket sort for linear time. You can just store the number, freq. pairs and run quickselect.
@sangramshinde2211
2 жыл бұрын
nick white, kevin, neetcode best guys to get the perfect explanation...
@prithvirajchavan9186
2 жыл бұрын
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
@javidabderezaei3632
Жыл бұрын
Heap solution (faster than 95% of solutions on leetcode): dic = {} for i in nums: if i not in dic.keys(): dic[i] = 1 else: dic[i] += 1 heaped = [] for key in dic: heaped.append([-dic[key], key]) heapq.heapify(heaped) res = [] for i in range(k): freq, x = heapq.heappop(heaped) res.append(x) return res
@nishanttripathy8275
2 жыл бұрын
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
Жыл бұрын
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.
@denshaSai
2 жыл бұрын
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?
@geekydanish5990
2 жыл бұрын
class Solution: def topKFrequent(self, nums: List[int], k: int) -> List[int]: _map = {} res = [] for num in nums: _map[num] = 1 + _map.get(num,0) heap = [(val,key) for key,val in _map.items()] for val,key in (sorted(heap,reverse=True)[0:k]): res.append(key) return res
@devashishubale1565
9 ай бұрын
I was solving 692, and I got AC, I remembered this video came back to this. Thanks for such detailed videos.
@rakshitkumar1141
Жыл бұрын
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)
@sammyj29
2 жыл бұрын
Amazing content as usual!! But I still don't understand, why is it O(n) even if there are 2 nested for loops? Thanks for creating these helpful videos!
@nathanwailes
2 жыл бұрын
The inner "for" loop on line 13 generally won't be iterating through an O(n) list, but instead just the number of input numbers that occurred a particular number of times.
@sravankumar4195
2 жыл бұрын
I think, in worst case, it will be O(n^2). Let's consider the case, nums = [1,2,3,4,5,6] and k =6, then on line 13 the inner loop freq[1] = [1,2,3,4,5,6] will iterate "n" times. But in Avg. case it is O(n).
@chanpreetsingh007
2 жыл бұрын
@@sravankumar4195 nope its o(n) only.
@Logan-mj3wx
2 жыл бұрын
@@sravankumar4195 I believe you are correct, in the worst case this would be O(n^2) for the situation you described. You have to iterate over the whole frequency list(size n) and then when you get to the solution set(in your case index 1) you have to iterate over n elements once more. I had the same thought
@SlakOffs
2 жыл бұрын
@@Logan-mj3wx What about, if instead of a for loop, you simply did res += freq[i]. Is that operation faster O(1)?
@amrojjeh
4 ай бұрын
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.
@KardboardCode
8 ай бұрын
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)
@ShreksSpliff
2 ай бұрын
Watching this after attempting on NeetCode is soooo good!
@numberonep5404
2 жыл бұрын
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
2 жыл бұрын
Nice!
@heathergray4880
2 жыл бұрын
Mine beats 98.62 time and 90 on space and is two lines long :)
@lucaslau8379
2 жыл бұрын
@@heathergray4880 would you share your code for learning please?
@xinyiwu2837
2 жыл бұрын
Man you light up my leetcode
@aleksey8230
Жыл бұрын
sorting works faster and les memory in result it almost the same 98ms Beats 99.28%of users with Python3 class Solution: def topKFrequent(self, nums: List[int], k: int) -> List[int]: ans = collections.defaultdict(int) for n in nums: ans[n] = 1 + ans.get(n,0) reorderd=sorted(ans.items(), key=lambda item: item[1], reverse=True) return [item[0] for item in reorderd][:k] #return [item[0] for item in reorderd[:k]]
@AsafVertman53
4 ай бұрын
Note that for the last loop using reversed(freq) is faster and more readable
@34535fff
8 ай бұрын
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)
@leetcodemermaid9804
2 жыл бұрын
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
5 ай бұрын
nlogk generally bigger than klogn k=1200 and n=1500 nlogk ≈1500×3.0792≈4618.8 klogn=1200×log1500 ≈ 3811.2
@mgst4699003
Жыл бұрын
Thank you man! Blessed to have you!
@theornament
6 ай бұрын
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?
@racerx4684
5 ай бұрын
Thanks Neet! I burned my brain cells trying to come up on my own. This helped.
@sharvitomar3809
Жыл бұрын
Firstly, thank you for being so amazing with your videos! Actually I was wondering in the for loop on line 12, since inside that loop we keep checking for the length of res, isn't that increases the time complexity from the linear expectation?
@jambajuice07
Жыл бұрын
nooo actually the inner runs only for n times . soo thats n(outer loop) + n(inner loop) = 2n which is O(n).
@syedzami-ul-haquenavid9392
Жыл бұрын
The explanation was so amazing that I understood how to solve half way through the video!
@BECEGtirth
3 ай бұрын
2:20 time complexity of heapify would be log n and creating heap would take n log n time
@tanayshah275
3 жыл бұрын
Made it simple but efficient as always!
@atulkumar-bb7vi
Жыл бұрын
Trust me bro, you are amazing explaining things. Thanks a lot for such content. Pls keep posting..
@CodeWithRVR
5 ай бұрын
man this was the video , great great explanation
@normanfung7124
Жыл бұрын
Simpler approach: def topKFrequent_simplest(nums: List[int], k: int) -> List[int]: count = {} # key: n, value: count of n for n in nums: count[n] = 1 + count.get(n, 0) count = [(k, v) for k, v in count.items()] count.sort(key=lambda entry : entry[1], reverse=True ) return [ entry[0] for entry in count[:k]]
@il5083
Жыл бұрын
The solution and thought process is genius! Can't come up with this optimal solution by myself, thanks a lot.
@vasumahalingam5162
10 ай бұрын
Nice algorithm but it looks like it will fail when the nums list is [-1,-1] which makes c as -1 throwing index out of range exception.
@Nohope__
Жыл бұрын
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
10 ай бұрын
ur solution is much faster for me, thank u
@1000marcelo1000
2 ай бұрын
It should be `for i in range(len(freq)-1, -1, -1):` instead of `for i in range(len(nums)-1, 0, -1):` right?
@rohanaurangabadkar951
3 ай бұрын
Best explanation this helped me in solving Top K Frequent Elements and Sort Characters By Frequency
@freddy5638
2 жыл бұрын
My man! I've spent HOURS watching you. FYI you can do the counter with collections, and save few lines of implementation
@yossarian2909
2 жыл бұрын
How is that?? Can you pls share
@hamzasayyid8152
2 жыл бұрын
@@yossarian2909 from collections import Counter, then Counter(nums) gives the frequency dict
@balavikashkandukuri6139
Жыл бұрын
@@hamzasayyid8152 continue the code
@haarismirza2044
2 жыл бұрын
How come the solution code doesn't work when we instantiate our freq array like this: freq = [[]] * (len(nums) + 1)
@dharmatejabandaru3344
2 жыл бұрын
Such an awesome explanation and solution. Thanks, Man! Love it.
@tomskrovan6622
Жыл бұрын
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
7 ай бұрын
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)]
@WinnerSingh
Жыл бұрын
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
@MistaT44
2 жыл бұрын
This solution blew my mind! excellent video as always :)
@avadheshsingh4255
8 ай бұрын
without knowing what is bucket sort I was only able to come with the hashmap sorting solution thanks mate for the o(n) soln.. great explanation
@EmergencyCringe
4 ай бұрын
My solution in Javascript, if anyone needs it :) It's a little different from NeetCode's solution, which i also emulated later in JS but it showed worse memory and runtime on Leetcode. var topKFrequent = function (nums, k) { const count = {}; for (let i = 0; i < nums.length; i++) { if (count.hasOwnProperty(nums[i])) { count[nums[i]] += 1; } else count[nums[i]] = 1; } // you cant sort objects, so move values into an array of two element arrays // where 0-idx element is the number and 1-idx element is the count of its occurrences const sortedArr = []; for (let num in count) { sortedArr.push([num, count[num]]); } //sort in descending order sortedArr.sort(function (a, b) { return b[1] - a[1]; }); const res = []; for (let i = 0; i < k; i++) { res.push(Number(sortedArr[i][0])); } return res; };
@ashadahmad651
10 ай бұрын
Cannot believe I figured this on my own, definitely took some time but I was able to figure it out on my own.
@shreyaskaup
2 жыл бұрын
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!
@akarsan9121
9 күн бұрын
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?
@srinadhp
2 жыл бұрын
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
2 жыл бұрын
This solution is also not memory optimised. Using "min" heap instead of max heap with size k would be memory optimised.
@kobeissi721
2 жыл бұрын
@@harigovind11 Wouldn't it still be O(N) since you still need to add them to a map to store the frequencies?
@sampatkalyan3103
2 жыл бұрын
@@harigovind11 "Using "min" heap instead of max heap with size k would be memory optimised. " can you explain how ?
@tedtran7855
2 жыл бұрын
Clever solution! I came up with the nlogn solution immediately and thought the problem was over since the Leetcode page only wanted that. Then I watched your video and I was shook when you said there was an O(N) solution haha.
@StfuSiriusly
Жыл бұрын
leetcode page says to find a solution that is BETTER than n log n
@melihsafacelik1166
4 ай бұрын
I came up with a short solution but the complexity seems like O(nlogk) from collections import Counter class Solution: def topKFrequent(self, nums: List[int], k: int) -> List[int]: count = Counter(nums) if k > len(count): return [] return [word for word,cnt in count.most_common(k)] I know it is not a optimal solution since Navi (as always) finds a way to solve in the most optimized way! But I want to hear what do you guys think?
@actiondelucas4959
Жыл бұрын
Mentioned maxheap solution: def topKFrequent(self, nums: List[int], k: int) -> List[int]: # count the frequency frequency = Counter(nums) # convert frequency to a maxheap of tuples maxHeap = [(-count, num) for num, count in frequency.items()] # return top k elements return [heapq.heappop(maxHeap)[1] for _ in range(k)]
@Ankitkumar-kk5rm
Жыл бұрын
Bucket Sort !! Name was actually scary than what it is.
@AJ-ju7tl
Ай бұрын
this is a mistake at the end, heap solution is O(N logK ) not O(K logN)!!! otherwise it is a very good explanation, thanks!
@mehmetnadi8930
Жыл бұрын
I'm speachless! thank you, NeatCode!
@kushagrakulshrestha2966
6 күн бұрын
that's descending , unless you're talking about input values, not the count
@akashp4863
2 жыл бұрын
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
@haardpatel4242
7 ай бұрын
using the two loops in the last actually increases the time complexity.
@BEEFnCHEESE44
2 жыл бұрын
Thank you so much, your explanations are so easy to understand, I would be lost without you
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