ZCO 2019 Discussion

I’ve attached a spreadsheet for ZCO 2019 Scores. Please update this Spreadsheet after you finish the ZCO 2019 Contest. Solutions can be discussed here after the contest. :slight_smile:

Click the below link to access the spreadsheet.

ZCO 2019 Scores Spreadsheet Link

Important Notice: This is not an OFFICIAL Ranklist. It is for discussing and estimating cutoffs for ZCO 2019 amongst the participants before the Official Results are announced by the Organizers of ZCO 2019.

Is there an ICO WhatsApp group?! I didn’t know about this, could you please upload the group link.

How did you do the first problem(singing competition)?
I did it with brute force, and it was partially accepted.

@jagoshom I got a 40 point solution.
What I did was visualise the ranges on a straight line.Since there can’t be any draws the pair with the lowest lower range should have the highest upper range as well.So I stored the ranges in a vector of pairs and sorted the vector. The ans for the ith is 2*(n-i-1). Then I undid the sorting.
This clears the second subtask.
Someone help me with the 100 point solution please.

I did a brute force for the 15pt one, for the 40pt one I did the sorting thing but I did not get the right answer because I was not able to output the answer in the original order.How did you overcome this problem?

What was the method for the first one?

I had hell of a time figuring out the way to undo sorting. I finally kept 1 ordered map where keys were elements and values were all 0. And seperately stored the original order in a vector. I then printed 2*(N-i-1) for each i.

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@kayak your code is difficult to understand since it is uncommented. Just post your logic

@bdyutish How did you undo the sorting?

@jagoshom I kept a map of pairs as keys and the original indexes as values.You calculate the ans for the sorted vector and assign the answers to an array in the original location using the map.

I just hope the cutoff is 15.

Can anyone tell me the solution for the first problem? Some where saying using mergesort…

Please post your solution for the first problem if anyone solved it fully. code not needed. thanks.

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Here’s my code for problem 1.

The logic is pretty simple.
Construct two arrays of pair of integers, L and R.
In the first position of each element l of L we put in the value l we receive and in the second position, we put in the index. Create a similar array for R.

Now sort the array L in decreasing order, and R in increasing order.

The score of the i-th person would be the sum of the positions of its corresponding l in L and its corresponding r in R.

For example, if n = 3, and the l and r values are

(10, 20), (13, 15), (14, 16)

The L and R would be, after sorting,

L: (14, 2), (13, 1), (10, 0)
R: (15, 1), (16, 2), (20, 0)

Now, \text{score}[0] = 2 + 2 = 4, \text{score}[1] = 0 + 1, \text{score}[2] = 1 + 0.

Though sadly I was getting WA in the test itself, I have no idea why. My logic is sound.

So, I just wrote a quick brute-force solution and got 15. :confused:


Can grp admin pls upload the link to the ICO WhatsApp grp? Also @kristopher , people are uploading their marks in both the spread sheets. Pls copy the data till now, and delete one, else analysis is going to get pretty messed up.

Is there any chance the cutoff goes down to 15?

When are the results released?Like, any idea?

Please share the logic for the second one.

Can someone who has got 100pts in the first question share their logic.

@mg21 it is most probably not going to be 15 since almost anyone could get 15

For the second DP problem there are few basic observations.

  1. We always put an integer because it will always increase the answer.

  2. The answer can either be the maximum length of UpDown subsegment already present+1 ( put any integer at last in the subsegment) or it can be combination of 2 subsegments which are just separated by 1 element which doesn’t obey the property. We can make the “bad” element “good” by inserting an integer beside it.

Let dp1[i] = maximum length of UpDown subsegment ending at index i

Let dp2[i] = maximum length of UpDown subsegment starting at index i

Now all that is left is to go through each index and find maximum value of dp1[i]+dp2[i] .
We can calculate DP and do this in O(n) with very small constant factor.

P.S. I may have missed some details and it is left as an exercise for the reader :stuck_out_tongue:


Is there any 11th student here who is interested in making a WhatsApp group to prepare for next year’s ZCO?