"""
--- 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?


--- Part Two ---
To completely determine whether you have enough adapters, you'll need to figure out how many
different ways they can be arranged.
Every arrangement needs to connect the charging outlet to your device.
The previous rules about when adapters can successfully connect still apply.

The first example above (the one that starts with 16, 10, 15) supports the following arrangements:

```
(0), 1, 4, 5, 6, 7, 10, 11, 12, 15, 16, 19, (22)
(0), 1, 4, 5, 6, 7, 10, 12, 15, 16, 19, (22)
(0), 1, 4, 5, 7, 10, 11, 12, 15, 16, 19, (22)
(0), 1, 4, 5, 7, 10, 12, 15, 16, 19, (22)
(0), 1, 4, 6, 7, 10, 11, 12, 15, 16, 19, (22)
(0), 1, 4, 6, 7, 10, 12, 15, 16, 19, (22)
(0), 1, 4, 7, 10, 11, 12, 15, 16, 19, (22)
(0), 1, 4, 7, 10, 12, 15, 16, 19, (22)
```

(The charging outlet and your device's built-in adapter are shown in parentheses.)
Given the adapters from the first example, the total number of arrangements that connect the
charging outlet to your device is 8.

The second example above (the one that starts with 28, 33, 18) has many arrangements. Here are a few:

```
(0), 1, 2, 3, 4, 7, 8, 9, 10, 11, 14, 17, 18, 19, 20, 23, 24, 25, 28, 31,
32, 33, 34, 35, 38, 39, 42, 45, 46, 47, 48, 49, (52)

(0), 1, 2, 3, 4, 7, 8, 9, 10, 11, 14, 17, 18, 19, 20, 23, 24, 25, 28, 31,
32, 33, 34, 35, 38, 39, 42, 45, 46, 47, 49, (52)

(0), 1, 2, 3, 4, 7, 8, 9, 10, 11, 14, 17, 18, 19, 20, 23, 24, 25, 28, 31,
32, 33, 34, 35, 38, 39, 42, 45, 46, 48, 49, (52)

(0), 1, 2, 3, 4, 7, 8, 9, 10, 11, 14, 17, 18, 19, 20, 23, 24, 25, 28, 31,
32, 33, 34, 35, 38, 39, 42, 45, 46, 49, (52)

(0), 1, 2, 3, 4, 7, 8, 9, 10, 11, 14, 17, 18, 19, 20, 23, 24, 25, 28, 31,
32, 33, 34, 35, 38, 39, 42, 45, 47, 48, 49, (52)

(0), 3, 4, 7, 10, 11, 14, 17, 20, 23, 25, 28, 31, 34, 35, 38, 39, 42, 45,
46, 48, 49, (52)

(0), 3, 4, 7, 10, 11, 14, 17, 20, 23, 25, 28, 31, 34, 35, 38, 39, 42, 45,
46, 49, (52)

(0), 3, 4, 7, 10, 11, 14, 17, 20, 23, 25, 28, 31, 34, 35, 38, 39, 42, 45,
47, 48, 49, (52)

(0), 3, 4, 7, 10, 11, 14, 17, 20, 23, 25, 28, 31, 34, 35, 38, 39, 42, 45,
47, 49, (52)

(0), 3, 4, 7, 10, 11, 14, 17, 20, 23, 25, 28, 31, 34, 35, 38, 39, 42, 45,
48, 49, (52)
```

In total, this set of adapters can connect the charging outlet to your device in 19208 distinct arrangements.

You glance back down at your bag and try to remember why you brought so many adapters; there must
be more than a trillion valid ways to arrange them!
Surely, there must be an efficient way to count the arrangements.

What is the total number of distinct ways you can arrange the adapters to connect the charging outlet to
your device?
"""
from collections import Counter
from typing import List

adapters = []


def clean_input(input_: str) -> List[int]:
    return sorted([int(v) for v in input_.splitlines()])


def cache_counts(f):
    counts = Counter()

    def inner(rating, adapter_idx):
        if not counts[rating]:
            counts[rating] = f(rating, adapter_idx)

        return counts[rating]
    return inner


@cache_counts
def calc_distinct_adapters(rating, adapter_idx: int = 0):
    count = 0
    for ind, adapter in enumerate(adapters[adapter_idx:]):
        if adapter-rating > 3:
            break

        count += calc_distinct_adapters(rating=adapter, adapter_idx=adapter_idx+ind+1)
    else:
        return 1
    return count


def main():
    global adapters
    adapters = clean_input("""30
73
84
136
132
117
65
161
49
68
139
46
21
127
109
153
163
160
18
22
131
146
62
113
172
150
171
98
93
130
170
59
1
110
2
55
37
44
148
102
40
28
35
43
56
169
33
5
141
83
15
105
142
36
116
11
45
82
10
17
159
140
12
108
29
72
121
52
91
166
88
97
118
99
124
149
16
9
143
104
57
79
123
58
96
24
162
23
92
69
147
156
25
133
34
8
85
76
103
122""")
    adapters.append(adapters[-1]+3)

    return calc_distinct_adapters(rating=0, adapter_idx=0)


if __name__ == '__main__':
    import timeit
    print(timeit.timeit("main()", setup="from __main__ import main")/1_000_000)
