# --- Day 7: The Treachery of Whales ---

# A giant whale has decided your submarine is its next meal, and it's much
# faster than you are. There's nowhere to run!

# Suddenly, a swarm of crabs (each in its own tiny submarine - it's too deep
# for them otherwise) zooms in to rescue you! They seem to be preparing to
# blast a hole in the ocean floor; sensors indicate a massive underground cave
# system just beyond where they're aiming!

# The crab submarines all need to be aligned before they'll have enough power
# to blast a large enough hole for your submarine to get through. However, it
# doesn't look like they'll be aligned before the whale catches you! Maybe you
# can help?

# There's one major catch - crab submarines can only move horizontally.

# You quickly make a list of the horizontal position of each crab (your puzzle
# input). Crab submarines have limited fuel, so you need to find a way to make
# all of their horizontal positions match while requiring them to spend as
# little fuel as possible.

# For example, consider the following horizontal positions:

# 16,1,2,0,4,2,7,1,2,14

# This means there's a crab with horizontal position 16, a crab with horizontal
# position 1, and so on.

# Each change of 1 step in horizontal position of a single crab costs 1 fuel.
# You could choose any horizontal position to align them all on, but the one
# that costs the least fuel is horizontal position 2:

#     Move from 16 to 2: 14 fuel
#     Move from 1 to 2: 1 fuel
#     Move from 2 to 2: 0 fuel
#     Move from 0 to 2: 2 fuel
#     Move from 4 to 2: 2 fuel
#     Move from 2 to 2: 0 fuel
#     Move from 7 to 2: 5 fuel
#     Move from 1 to 2: 1 fuel
#     Move from 2 to 2: 0 fuel
#     Move from 14 to 2: 12 fuel

# This costs a total of 37 fuel. This is the cheapest possible outcome; more
# expensive outcomes include aligning at position 1 (41 fuel), position 3 (39
# fuel), or position 10 (71 fuel).

# Determine the horizontal position that the crabs can align to using the least
# fuel possible. How much fuel must they spend to align to that position?

with open("files/P7.txt") as f:
    positions = [int(num) for num in f.read().strip().split(",")]


def part_1() -> None:
    fuel = []
    for pos in positions:
        sum = 0
        for elem in positions:
            sum += abs(elem - pos)
        fuel.append(sum)

    print(f"It is needed {min(fuel)} units of fuel")


# --- Part Two ---

# The crabs don't seem interested in your proposed solution. Perhaps you
# misunderstand crab engineering?

# As it turns out, crab submarine engines don't burn fuel at a constant rate.
# Instead, each change of 1 step in horizontal position costs 1 more unit of
# fuel than the last: the first step costs 1, the second step costs 2, the
# third step costs 3, and so on.

# As each crab moves, moving further becomes more expensive. This changes the
# best horizontal position to align them all on; in the example above, this
# becomes 5:

#     Move from 16 to 5: 66 fuel
#     Move from 1 to 5: 10 fuel
#     Move from 2 to 5: 6 fuel
#     Move from 0 to 5: 15 fuel
#     Move from 4 to 5: 1 fuel
#     Move from 2 to 5: 6 fuel
#     Move from 7 to 5: 3 fuel
#     Move from 1 to 5: 10 fuel
#     Move from 2 to 5: 6 fuel
#     Move from 14 to 5: 45 fuel

# This costs a total of 168 fuel. This is the new cheapest possible outcome;
# the old alignment position (2) now costs 206 fuel instead.

# Determine the horizontal position that the crabs can align to using the least
# fuel possible so they can make you an escape route! How much fuel must they
# spend to align to that position?

# TIL :D
def triangular_cost(steps: int) -> int:
    return steps * (steps + 1) // 2


def part_2() -> None:
    fuel = []
    for pos in positions:
        sum = 0
        for elem in positions:
            sum += triangular_cost(abs(elem - pos))
        fuel.append(sum)

    print(f"It is needed {min(fuel)} units of fuel")


if __name__ == "__main__":
    part_1()
    part_2()