Algorithm (prob. solving) achieving fastest runtime

Question

For an algorithm competition training (not homework) we were given this question from a past year. Posted it to this site because the other site required a login.

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Answer 1

There is no need to calculate every house!!!

It's not fully developed, but I think it is worth thinking about it:

modulo N

N is the number of all houses, n shall be the "adress" (number) of some of the houses.

If you walk around the island, you will find that n is raising by 1 for each house you pass. If you reach the house where n is N, then the next house has the number 1.

Let us use a different system of numbering: increase every house-number by 1. Then n goes from 0 to N-1. this is the same way how numbers modulo N will behave.

Litres is a function of the house-number n (modulo N)

You can calculate the amount of liters for each house-Number by building the sum of all products of distance and people living there.

You can also draw a graph of that function: x is n, y is the number of litres.

the function is periodic

If you understand what modulo means, you will understand that the graph you just did draw is just one periode of a periodic function, since Litre(n) ist eaqual to Litre(n + x * N) where x is a integer (that might be negative too).

if N is big, the function is "Pseudo-continuous"

What I mean is this: If N is really big, then the amount of litres will not change very much if you move from house a to its neighbour, house a+1. So you can use methods of interpolation.

you are looking for the place of the "global" maximum of a periodic pseudo-continuous function (only really global within one periode)

This is my suggestion:

Step 1: select a distance d that is bigger than 1 and smaller than N. I can't say why, but I would use d=int(sqrt(N)) (maybee there might be a better choice, try it out).
Step 2: calculate the Litres for House 0, d, 2d, 3d, ... Step 3: you will find some values that are higher than both of their neighbours. Use this high-points and their neighbours feed them using a method of interpolation to calculate more points close to that a high points (interval-splitting).

Repeat this interpolations for other high points as long as you have time (you have 1 second, which is a long time!)

Jump from one high-point to another if you see, that the global maximum must be elsewhere.