Do we have an editorial for Chef and Meatballs question which was a part of Dec18b Long Contest?

# Editorial for Chef and Meatballs?

**masood786**#2

Here’s the outline of my approach.

First of all, notice that in each query the problem set size can be reduced by 2 (as the answer returned by the grader can never be a part of the final answer).

So, suppose we fill up n elements in a **set** (numbered from **1** to **n**).

Now, we query first (or any 5) elements of the set and delete the 2 numbers returned by the grader (effectively reducing the set size by 2.) (Note that we do not lose any information as we stiil have the **meatiest meatball** in our set.

So, if **n** is even, we keep doing this till the set size becomes 6. ( This would take **(n-6)/2** queries).

On the other hand, if **n** is odd, we reduce the set size to 7 (costing us **(n-7)/2** queries) and then we query another block of 5 elements and delete only one of the answers.

Total queries consumed **(n-7)/2** + **1**.

So now the problem reduces to finding out the meatiest meatball among a group of 6 meatballs.

This can be done in at most 4 queries in the worst case (which comprises of replacing one meatball with the other unused one and checking as to which group they belong to , i.e, upper or lower).

So the total number of queries if **n** is even is **n/2 + 1** (more tight than the given constraints.

And for odd **n**, it is **n/2** + **2**