After checking the leaderboard of Codeagon 2020 I would Like to send NCB to the Codenation Office.
I Would like to know the Algorithm that was used to rank the person who solved only 2 question with 100 penalties upper than the person who solved 4 question
Contestants might have been given partial marks on the all the problems that they attempted but could not solve completely. But if that is the case, They should have shown the partial marks during the contest itself as at that time it would have encouraged everyone to go for partial solutions as well if not fully correct ones.
If thats the case then check the person at 82 rank ..... I am not blaming anyone just asking if partial marks of question 3rd can be more than full marks of question 5... if this is the case then i dont think its logical from any sense
What is the cut-off rank for interviews?
Recently, In InterviewBit's live interaction on YT, Codenation HR said that they will interview the top 100 or so.
I heard from a senior that his friend whose rank was around 120 got an interview call in past.
I tried to calculate points for all problems...
p1=9.621pts
p2=6.888pts
p3=27.586pts
p4=3.491pts
p5=17.931pts
p6=34.483pts
The distribution sure seems bit odd, don't you think so? XD
p5 has most number pf points?? I dont think so and its clear from the scoreboard that p5 does not have most points
Yeah, sorry. Fixed.
I guess they are decided by the percentage of people solving each problem.
Problems had different scores. They also had partial scores. At one point in time, I got 250/500 points in that XOR problem. Also, no one knows how are ties broken due to time and penalties.
But providing such massive points to visible test cases, such that entire rank list could be affected is wrong I guess.
Wait one minute. I never said I got 250/500 points by just solving visible test cases. How did you come to this conclusion?
It was a soln which was intended to be TLE under given constraints. But still decent enough to deserve some points on that problem.
Did the sample test cases themselves have a score ?
Still you mananged to get such a great rank! :D Orz
Later on, I did change it to a solution which passed all test cases instead of just a few. I had AC on all except necklace one. Now we know P3 is necklace one.
Since rank 4,5 and rank 10-51 have a green tick on same problems. We also know that rank 4,5 had partial scores on the last problem which interviewbit just shows with a cross.
Those problems which shows a "cross" is a complete wrong answer. Those which display a "cross" with "x tries" means you got some partial marks in that question
Ashishgup, FastestFinger, AbhishekVanjani
Maybe, these people know it.
Is 57 rank good enough to land an interview ?
No, only rank 1, 2, 3 and 4 are gonna be called for interview. Rest all are not considered. XD
Honestly, I don't have an idea about how much their intake is so I don't know how many students they consider. That's why I asked.
Yeah, pretty sure you'll get a call.
In problem discussion session HR said they will call more than 100 people for interviews not just top 20.
This is not the first time interviewbit did this, in scaler hiring challenge first they mentioned they will shortlist whose rank is in top 200, there was no announcement of partial marks anywhere, then after receiving no reply after days(obviously I was not applying,just curious what happened, I mailed them), they said they changed the criteria to somewhat points.
And now this,after solving 4 questions, where each problem had 100 points, people who solved 3 problems are ranked higher, it's very frustrating to see, and guess what you might be thinking that they ohh they might have submitted the partial points solution but no most of the guys just hard coded the given test cases, where many test cases were visible, and I don't why visible test cases have points rewarded. I have seen the case where person solving 0 questions is ranked higher than solving 2 or 3 questions completely. It is just unfair.
How do I search any user?
What is the total number of participants?
Use binary search on page number. In some steps you will get total participants