Top K Frequent Elements | leetcode 347 | Hindi

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problem :- leetcode.com/p...
code :- github.com/luc...

[@RishabhVerma]

No one ever made the explanation this easy. Thanks Sir

[@codebix1096]

You are most welcome

[@sonurathod9551]

hi you have a really good voice and you explained the question really well.

[@varunsadhu9060]

Bro this was savage, better that quickselect and all sorting fuzz. I hope the interviewer would know what a bucket is 😁

[@YashNagayach]

I think in line 25 condition should be if (count < k) instead of (count == k).

[@saurabhc4278]

What an explanation code bix !!!!!!!!!!!! Never going to forget this approach

[@ramankr0022]

very helpful. thanks!

[@imajt5]

Bro you method of explaination is superb...thanks a lot

[@ridj41]

pLease also mention the time and space complexity, as only then we are able to identify the most opti.solution.

[@dev4life]

Really nice solution bro. Got to learn a new approach to these kinds of problems 😄.

[@codebix1096]

Glad it helped

[@shubhammanecr7]

💥 amazing !

[@sagardhankar1449]

Thanku bhai your explaination is at top😀

[@codebix1096]

Welcome 😊 Follow our linkedin page for regular updates www.linkedin.com/company/codebix/?viewAsMember=true

[@rishavranjan6656]

Iska brute force possible hai kya?

[@aakashsharma5901]

Very well explained sir...ur way of explation is too good

[@codebix1096]

Thank you. Follow our linkedin page for regular updates www.linkedin.com/company/codebix/?viewAsMember=true

[@iuashrafi]

great!

[@akshitajain3721]

Good expalanation !!!

[@codebix1096]

Thanks 🙂

[@saurabhkanswal7954]

thanks a lot

[@mansisharma4136]

in seraching last part it takes n2 times Then how is better you are using two loops there and doing n and k work in each loop respectively

[@soumavanag5025]

using two loops doesnt necessarily means it will be o(n^2) here , using the two loops, all the elements are being visited exactly once. hence t(n) = o(n) do a dry run and check if each element is being visited more than once or not

[@swatygupta3753]

awesome explaination, thank you.. For leetcode 692(top-k-frequent words), we have to display the list as --> If two words have the same frequency, then the word with the lower alphabetical order comes first. How to approach this one.

[@mishra1576]

Study about compareTo method in java. Hope it helps

[@saunaknandi1814]

Can u make a video on bucket sort

[@rilkedev449]

Bhai thanks

[@artone5841]

why the time complexity of o(n) , why not o(n*m)

[@shubhramishra4568]

very well explained

[@codebix1096]

Thanks :)

[@sagardhankar1449]

sir (i-button) me suggested videos/ previous videos daal diya karo please usse videos pe jyada views & like ho jate h

[@codebix1096]

Alright. Thank you. Follow our linkedin page for regular updates www.linkedin.com/company/codebix/?viewAsMember=true

[@mridulkhurana533]

Can't we just simply sort the hashMap according to values. It's Also O(nlogk) approach and quite easy

[@codebix1096]

Yes we can do that as well but I think this will be similar as our first approach so time complexity will be O(nlogn)

[@Vishalraj_1]

FOR SOURCE CODE REFER : leetcode.com/explore/featured/card/july-leetcoding-challenge/546/week-3-july-15th-july-21st/3393/discuss/740538/C++-Solutionusing-Priority-Queue-and-Unordered_map

[@ramanmanocha4800]

Can you please make a vodio on Wildcard character matching? Thank you

[@codebix1096]

Sure will make a video on it

[@gaunikasrivastava7851]

This utilises a lot of space

[@AmritAgarwal07]

public static int[] topKFrequent(int[] nums, int k) { Arrays.sort(nums); HashMapmap=new HashMap(); for (int i = 0; i < nums.length; i++) { map.put(nums[i],map.getOrDefault(nums[i],0)+1); } Listlist=new ArrayList(map.keySet()); Collections.sort(list, (a, b) -> map.get(b) - map.get(a)); int[] result=new int[k]; for(int i=0;i

[@saunaknandi1814]

Bro can u share ur linkedIn profile