import numpy as np

# --- Day 10: Adapter Array ---

# Patched into the aircraft's data port, you discover weather forecasts of a
# massive tropical storm. Before you can figure out whether it will impact your
# vacation plans, however, your device suddenly turns off!

# Its battery is dead.

# You'll need to plug it in. There's only one problem: the charging outlet near
# your seat produces the wrong number of jolts. Always prepared, you make a
# list of all of the joltage adapters in your bag.

# Each of your joltage adapters is rated for a specific output joltage (your
# puzzle input). Any given adapter can take an input 1, 2, or 3 jolts lower
# than its rating and still produce its rated output joltage.

# In addition, your device has a built-in joltage adapter rated for 3 jolts
# higher than the highest-rated adapter in your bag. (If your adapter list were
# 3, 9, and 6, your device's built-in adapter would be rated for 12 jolts.)

# Treat the charging outlet near your seat as having an effective joltage
# rating of 0.

# Since you have some time to kill, you might as well test all of your
# adapters. Wouldn't want to get to your resort and realize you can't even
# charge your device!

# If you use every adapter in your bag at once, what is the distribution of
# joltage differences between the charging outlet, the adapters, and your
# device?

# For example, suppose that in your bag, you have adapters with the following
# joltage ratings:

# 16
# 10
# 15
# 5
# 1
# 11
# 7
# 19
# 6
# 12
# 4

# With these adapters, your device's built-in joltage adapter would be rated
# for 19 + 3 = 22 jolts, 3 higher than the highest-rated adapter.

# Because adapters can only connect to a source 1-3 jolts lower than its
# rating, in order to use every adapter, you'd need to choose them like this:

#     The charging outlet has an effective rating of 0 jolts, so the only
# adapters that could connect to it directly would need to have a joltage
# rating of 1, 2, or 3 jolts. Of these, only one you have is an adapter rated 1
# jolt (difference of 1).
#     From your 1-jolt rated adapter, the only choice is your 4-jolt rated
# adapter (difference of 3).
#     From the 4-jolt rated adapter, the adapters rated 5, 6, or 7 are valid
# choices. However, in order to not skip any adapters, you have to pick the
# adapter rated 5 jolts (difference of 1).
#     Similarly, the next choices would need to be the adapter rated 6 and then
# the adapter rated 7 (with difference of 1 and 1).
#     The only adapter that works with the 7-jolt rated adapter is the one
# rated 10 jolts (difference of 3).
#     From 10, the choices are 11 or 12; choose 11 (difference of 1) and then
# 12 (difference of 1).
#     After 12, only valid adapter has a rating of 15 (difference of 3), then
# 16 (difference of 1), then 19 (difference of 3).
#     Finally, your device's built-in adapter is always 3 higher than the
# highest adapter, so its rating is 22 jolts (always a difference of 3).

# In this example, when using every adapter, there are 7 differences of 1 jolt
# and 5 differences of 3 jolts.

# Here is a larger example:

# 28
# 33
# 18
# 42
# 31
# 14
# 46
# 20
# 48
# 47
# 24
# 23
# 49
# 45
# 19
# 38
# 39
# 11
# 1
# 32
# 25
# 35
# 8
# 17
# 7
# 9
# 4
# 2
# 34
# 10
# 3

# In this larger example, in a chain that uses all of the adapters, there are
# 22 differences of 1 jolt and 10 differences of 3 jolts.

# Find a chain that uses all of your adapters to connect the charging outlet to
# your device's built-in adapter and count the joltage differences between the
# charging outlet, the adapters, and your device. What is the number of 1-jolt
# differences multiplied by the number of 3-jolt differences?

with open("files/P10.txt", "r") as f:
    inputs = [int(num) for num in f.read().strip().split("\n")]


def part_1() -> None:
    # Add zero (charging outlet) and max+3 (device own built-in adapter)
    joltages = sorted(inputs + [0, max(inputs) + 3])

    joltages_diffs = np.diff(joltages)

    print(
        f"The difference is {np.sum(joltages_diffs == 1) * np.sum(joltages_diffs == 3)}"
    )


if __name__ == "__main__":
    part_1()