## Editorial for CCC '15 S3 - Gates

Remember to use this editorial

**only**when stuck, and**not to copy-paste code from it**. Please be respectful to the problem author and editorialist.**Submitting an official solution before solving the problem yourself is a bannable offence.**Author:

A simple greedy algorithm goes like this:

For each plane, assign it to the gate with the largest number that isn't occupied.

It's not hard to see why this is correct — indeed, intuitively, we're using the "worst" gate we can for each plane (since a gate with a lower number can serve at least every plane a gate with a higher number can).

Implementing this naïvely will result in a complexity of , which is enough for the weak data used on the actual CCC. To speed this up, we can use a balanced binary search tree to store free gates. In C++, this is the `set`

data structure, with a final time complexity of .

## Comments

Why do we need to use set?

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O(P(2logG)) = O(PlogG) since we disregard the constant factor