Here is a less mathematically inclined solution that works in O(n).
Let us partition the houses (indexing starts at 0) into two disjoints sets:
F, "front", where people walk CCW to the houseB, "back", where people walk CW to the house
and a single house p that marks the current position where the plant would be built.
I have based my illustration on the example given in the image.
By convention, lets assign half the houses to F, and exactly one less to B.
F contains 6 housesB contains 5 houses
With simple modular arithmetic, we can easily access the houses by (p + offset) % 12 thanks to Python's sane implementation of the modulo operator, quite unlike some other popular languages.
If we arbitrarily choose a position for p, we can determine the consumption of water in O(L) trivially.
We could do this all over again for a different position of p to arrive at a runtime of O(L^2).
However, if we only shift p by one position, we can determine the new consumption in O(1) if we make a somewhat clever observation: The amount of people living in F (or B respectively) determine how much the consumption of F changes when we set p' = p+1. (and some corrections because F itself will change). I have tried to depict this here to the best of my abilities.
We end up with a total running time of O(L).
The program for this algorithm is at the end of the post.
But we can do better. As long as no houses change between the sets, the cs and ws that are added will be zero. We can calculate how many of these steps there are and do them in one step.
Houses change sets when:
- When p is on a house
- When p is opposite of a house
In the following diagram, I have visualized stops the algorithm now makes to update the Cs and Ws.
Highlighted is the house, that causes the algorithm to stop.
The algorithm begins at a house (or the opposite of one, we'll see why later), in this case that happens to be a house.
Again, we have both a consumption C(B) = 3*1 and C(F) = 2 * 1. If we shift p to the right by one, we add 4 to C(B) and subtract 1from C(F). If we shift p once again, the exact same thing happens.
As long as the same two sets of houses move closer and further away respectively, the changes to the Cs are constant.
We now change the definition of B slightly: It will now also contain p! (This does not change the above paragraphs regarding this optimized version of the algorithm).
This is done because when we move to the next step, we will add the weight of the houses that are moving away repeatedly. The house at the current position is moving away when p moves to the right, thus W(B) is the correct summand.
The other case is when a house stops moving away and comes closer again. In that case the Cs change drastically because 6*weight goes from one C to the other. That is the other case when we need to stop.
I hope it is clear how and why this works, so I'll just leave the working algorithm here. Please ask if something is not clear.
O(n):
import itertools
def hippo_island(houses, L):
return PlantBuilder(houses, L).solution
class PlantBuilder:
def __init__(self, houses, L):
self.L = L
self.houses = sorted(houses)
self.changes = sorted(
[((pos + L /2) % L, -transfer) for pos, transfer in self.houses] +
self.houses)
self.starting_position = min(self.changes)[0]
def is_front(pos_population):
pos = pos_population[0]
pos += L if pos < self.starting_position else 0
return self.starting_position < pos <= self.starting_position + L // 2
front_houses = filter(is_front, self.houses)
back_houses = list(itertools.ifilterfalse(is_front, self.houses))
self.front_count = len(houses) // 2
self.back_count = len(houses) - self.front_count - 1
(self.back_weight, self.back_consumption) = self._initialize_back(back_houses)
(self.front_weight, self.front_consumption) = self._initialize_front(front_houses)
self.solution = (0, self.back_weight + self.front_weight)
self.run()
def distance(self, i, j):
return min((i - j) % self.L, self.L - (i - j) % self.L)
def run(self):
for (position, weight) in self.consumptions():
self.update_solution(position, weight)
def consumptions(self):
last_position = self.starting_position
for position, transfer in self.changes[1:]:
distance = position - last_position
self.front_consumption -= distance * self.front_weight
self.front_consumption += distance * self.back_weight
self.back_weight += transfer
self.front_weight -= transfer
# We are opposite of a house, it will change from B to F
if transfer < 0:
self.front_consumption -= self.L/2 * transfer
self.front_consumption += self.L/2 * transfer
last_position = position
yield (position, self.back_consumption + self.front_consumption)
def update_solution(self, position, weight):
(best_position, best_weight) = self.solution
if weight > best_weight:
self.solution = (position, weight)
def _initialize_front(self, front_houses):
weight = 0
consumption = 0
for position, population in front_houses:
distance = self.distance(self.starting_position, position)
consumption += distance * population
weight += population
return (weight, consumption)
def _initialize_back(self, back_houses):
weight = back_houses[0][1]
consumption = 0
for position, population in back_houses[1:]:
distance = self.distance(self.starting_position, position)
consumption += distance * population
weight += population
return (weight, consumption)
O(L)
def hippo_island(houses):
return PlantBuilder(houses).solution
class PlantBuilder:
def __init__(self, houses):
self.houses = houses
self.front_count = len(houses) // 2
self.back_count = len(houses) - self.front_count - 1
(self.back_weight, self.back_consumption) = self.initialize_back()
(self.front_weight, self.front_consumption) = self.initialize_front()
self.solution = (0, self.back_weight + self.front_weight)
self.run()
def run(self):
for (position, weight) in self.consumptions():
self.update_solution(position, weight)
def consumptions(self):
for position in range(1, len(self.houses)):
self.remove_current_position_from_front(position)
self.add_house_furthest_from_back_to_front(position)
self.remove_furthest_house_from_back(position)
self.add_house_at_last_position_to_back(position)
yield (position, self.back_consumption + self.front_consumption)
def add_house_at_last_position_to_back(self, position):
self.back_weight += self.houses[position - 1]
self.back_consumption += self.back_weight
def remove_furthest_house_from_back(self, position):
house_position = position - self.back_count - 1
distance = self.back_count
self.back_weight -= self.houses[house_position]
self.back_consumption -= distance * self.houses[house_position]
def add_house_furthest_from_back_to_front(self, position):
house_position = position - self.back_count - 1
distance = self.front_count
self.front_weight += self.houses[house_position]
self.front_consumption += distance * self.houses[house_position]
def remove_current_position_from_front(self, position):
self.front_consumption -= self.front_weight
self.front_weight -= self.houses[position]
def update_solution(self, position, weight):
(best_position, best_weight) = self.solution
if weight > best_weight:
self.solution = (position, weight)
def initialize_front(self):
weight = 0
consumption = 0
for distance in range(1, self.front_count + 1):
consumption += distance * self.houses[distance]
weight += self.houses[distance]
return (weight, consumption)
def initialize_back(self):
weight = 0
consumption = 0
for distance in range(1, self.back_count + 1):
consumption += distance * self.houses[-distance]
weight += self.houses[-distance]
return (weight, consumption)
Result:
>>> hippo_island([0, 3, 0, 1, 0, 0, 0, 0, 0, 0, 1, 2])
(7, 33)
