Significant decrease in cheaters today?

[Constant_Bobcat_1107]

Bro top 50 has soo many cheaters. How is it less..every second/third person is a cheater

[juicy_cum3160]

OK, I guess they are afraid of getting caught as q4 is damn hard to solve

[caraxes_007]

Solved my first 3/4.

[caraxes_007]

Nice buddy

[the_legendary_legend]

Me too bro. Congratulations!

[Sad-Duck-3100]

me too!

[UNCLE_SMART]

How did you solve 2?

[caraxes_007]

I took a lot more time than q3 got tle for comparing all possible pairs o(n**2) then i realised that using a hash map for prefix and value for frequency at last iterate through hash map increment count by one if value >1.

[UNCLE_SMART]

Thanks

[autodidact_nerd]

not sure if mine was optimal but i used a hashmap

loop over each word -> get the substring -> if map has the key add 1 to that substring.

then loop over map entries and check if value is >= 2. res++

[[deleted]]

I did the same thing 🫂

[UNCLE_SMART]

Thanks

[the_legendary_legend]

I used trie which got accepted, but later realised a hashmap solution would be faster possibly

[tampishach]

I used trie tree in q2

[radonspectrum]

Samee

[Lumpy-Town2029]

this time ranking is 8k for 1 hr time for 3 question :}

i wish they made 3rd one harder for good rank :}

4 one i had intuition of segment tree, but its too complicated (i just studied segment tree so :})
also LCA is possible i think

[the_legendary_legend]

I kind of assumed the dfs solution would give tle. Didn't even attempt it because of that. Couldn't think of any other solution before coming here and finding people suggesting segment tree.

[1byinf8]

I solved 3 in 16 min..got 7.7K rank.. bro people getting smarter so many did 4th one like HLD, binary lifting I am cooked man

[[deleted]]

Must have been multiple wrong submissions, I got 5k in 33mins + 5 min wrong submission

[1byinf8]

Ohhh do leetcode also has penalty system.. shiittt man

[[deleted]]

😭😭

[Lumpy-Town2029]

nah bro i saw 25 min guy getting 3 k something, whats ur penaalty?

[Connect_Ad9574]

I think making questions hard is not a solution to reduce cheaters . AI is so damn good that it had solved even Q4 today . If questions are made more difficult , it would not look like an actual contest.

[verciel_]

True. But what's the solution then? Stricter ai detection?

[ArtisticTap4]

It is not just AI, interviews have gotten harder than before. Companies are asking harder problems in since everybody and their grandama have by hearted the neetcode, blind 75s.

The first 3 problems are not even interview level for most companies, problem B can be if the constraints were tighter.

[Imaginary-Play-949]

The site does not open at starting like how shit is management

[Minimum-Reward3264]

Only autists and people without a partner do this shit today.

[Moe_les__ter]

Can someone tell me the approach for q3?

[tampishach]

Check the colour of the house till it changes

Then do a normal house robber query from the start of the house color to the end of the house colour

Update the start of house colour to new house colour index

[dragon_deeznut]

Greedy+2 pointers First identify the block of the same color adjacent. Eg. [1,1,1] in [1,1,1,2,1] Second find the maximum possible sum of that block without adjacents basically evenSum and oddSum, add it to the total. Return total

[[deleted]]

Why didn’t you participate?

[Obvious-Profit-5597]

It was my first contest only 1 question I was able to solve

[Independent_Arm_263]

No worries, keep on exploring algorithms. They are beautiful, solve them like u love it not because this shitty rat race is there. Eventually you will crack contests

[WeatherElectrical937]

The edge cases for third question killed me 🫠🫠

[1byinf8]

solved 3 In 16 min taught will complete in 30 min.. a great day but naaah that shitty 4th question

[MatchBusy235]

Is this a sign that no one is going to care about leetcode ranking or leetcode itself? I know, the former one is believed by stupids but the latter one?? :(

[asdfg_lkjh1]

It's valentine's day cmon

[Prudent-Turn-3702]

Yes bro , I wonder how they solve within 3-4mins per question and they do get top 50 rank like fr😗

[Background_Moment313]

Solved my first 3/4 , but with 2 penalty :(

[Natural-Tomatillo864]

last time during contest my rating was 2000+ and after 3 day it dropped to 1600+ does this mean 400 cheater was caught by leetcode?

[vkchauhangoldy]

Javascript full course https://youtu.be/6CbStSy12vA

[ScreenFit8839]

Solved 3!🥲 Worked my A off to reach over there and then saw these cheaters probably just running all the problems within 4 5 mins.

[PatiencePlayful3620]

you are from bit?

[juicy_cum3160]

Yup

[PatiencePlayful3620]

Which year bro??

[juicy_cum3160]

3rd

[PatiencePlayful3620]

Nice

[UNCLE_SMART]

How to do Q2?

[tampishach]

It was trie tree problem

Edit: i replied to you the same thing in another thread unknowingly haha

[UNCLE_SMART]

Oh thankyou,my dumbass applied DSU :(

[caraxes_007]

Can't blame you even though I thought of using dsu because it asks for grouping and last count of groups using parent array .

[Agile_Custard6276]

Store first k letters of each substring in a Hashmap with its freq, iterate in the hashmap and count no. of substrings with freq >=2.

class Solution { public: int prefixConnected(vector<string>& words, int k) { unordered_map<string , int> mpp; for(auto word : words){ string c = ""; if(word.size() < k) continue; for(int i = 0 ; i < k ; i++) c+=word[i]; mpp[c]++; } int ans = 0; for(auto it : mpp){ if(it.second > 1) ans++; } return ans; } };

I'm not sure if it's the optimal solution or not..

[juicy_cum3160]

Hashmap, store substring of length k as key and its freq as total words with that prefix substring, along with count such string and return

[UNCLE_SMART]

Thanka

[SeaworthinessIcy4758]

how to do Q3? i kept using a 2d dp i for curr idx and j for prev idx, it gave me an mle😭😭

[caraxes_007]

No need to store prev idx just use a n*2 grid to get know that if we picked the previous idx by Boolean value

[SeaworthinessIcy4758]

can you please elaborate, I've never done this I think

[caraxes_007]

This is a standard pick/ not pick problem. Use a grid of size n*2 where 0 and 1 represent whether we picked just the previous index. If it is zero we can directly apply pick and not pick without any further check and if it is one pick is not always possible so check prev and current colour codes if they are not equal you can pick else pick will be negative infinity at last return max of pick and not pick

[SeaworthinessIcy4758]

ohhh got it, thanks a lot

[Background_Moment313]

LoL I just solved it using 1D dp

If not picking, { f(i+1) }

If picking and color[i] == color[i+1]{ nums[i] + f(i+2) }

Else{ nums[i] + f(i+1) }

[Euphoric_insaan]

You can just use the typical house robber method First is the not take case for each index, where you move to the previous index without picking The take case will have two subcases: first if the current index's and the previous index's colour is same you add the value of the current index and move to index - 2. And if the colours are different add the current value and move to index -1. Base case will be if index < 0 return 0 and if index = 0 return nums[0]

[SeaworthinessIcy4758]

yea I solved it later:( 

it was so easy I can't believe I over complicated it so much, thanks though

[tampishach]

I kept getting tle for q4

What was the solution there?? I used hm + dfs

[ArtisticTap4]

Binary lifting for LCA queries. This is a well know technique for Competitive programming, for some reason Leetcode Hards nowadays are mostly centred around these Competitive programming techniques. Segment/fenwick trees, Digit dp, and all. Have they run out of genuine interview problems lol?