2021 Day 01

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Akumatic
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# 2021 Day 01: Sonar Sweep
Copyright (c) Eric Wastl
#### [Direct Link](https://adventofcode.com/2021/day/01)
## Part 1
As the submarine drops below the surface of the ocean, it automatically performs a sonar sweep of the nearby sea floor. On a small screen, the sonar sweep report (your puzzle input) appears: each line is a measurement of the sea floor depth as the sweep looks further and further away from the submarine.
For example, suppose you had the following report:
```
199
200
208
210
200
207
240
269
260
263
```
This report indicates that, scanning outward from the submarine, the sonar sweep found depths of `199`, `200`, `208`, `210`, and so on.
The first order of business is to figure out how quickly the depth increases, just so you know what you're dealing with - you never know if the keys will get carried into deeper water by an ocean current or a fish or something.
To do this, count **the number of times a depth measurement increases** from the previous measurement. (There is no measurement before the first measurement.) In the example above, the changes are as follows:
```
199 (N/A - no previous measurement)
200 (increased)
208 (increased)
210 (increased)
200 (decreased)
207 (increased)
240 (increased)
269 (increased)
260 (decreased)
263 (increased)
```
In this example, there are **`7`** measurements that are larger than the previous measurement.
**How many measurements are larger than the previous measurement?**
## Part 2
Considering every single measurement isn't as useful as you expected: there's just too much noise in the data.
Instead, consider sums of a **three-measurement sliding window**. Again considering the above example:
```
199 A
200 A B
208 A B C
210 B C D
200 E C D
207 E F D
240 E F G
269 F G H
260 G H
263 H
```
Start by comparing the first and second three-measurement windows. The measurements in the first window are marked `A` (`199`, `200`, `208`); their sum is `199 + 200 + 208 = 607`. The second window is marked `B` (`200`, `208`, `210`); its sum is `618`. The sum of measurements in the second window is larger than the sum of the first, so this first comparison **increased**.
Your goal now is to count **the number of times the sum of measurements in this sliding window increases** from the previous sum. So, compare `A` with `B`, then compare `B` with `C`, then `C` with `D`, and so on. Stop when there aren't enough measurements left to create a new three-measurement sum.
In the above example, the sum of each three-measurement window is as follows:
```
A: 607 (N/A - no previous sum)
B: 618 (increased)
C: 618 (no change)
D: 617 (decreased)
E: 647 (increased)
F: 716 (increased)
G: 769 (increased)
H: 792 (increased)
```
In this example, there are **`5`** sums that are larger than the previous sum.
Consider sums of a three-measurement sliding window. **How many sums are larger than the previous sum?**

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# SPDX-License-Identifier: MIT
# Copyright (c) 2021 Akumatic
#
# https://adventofcode.com/2021/day/01
def read_file() -> list:
with open(f"{__file__.rstrip('code.py')}input.txt", "r") as f:
return [int(line) for line in f.read().strip().split("\n")]
def part1(vals: list) -> int:
return sum(vals[i+1] > vals[i] for i in range(len(vals) - 1))
def part2(vals: list) -> int:
return sum(vals[i+3] + vals[i+2] + vals[i+1] > vals[i+2] + vals[i+1] + vals[i]
for i in range(len(vals) - 3))
if __name__ == "__main__":
vals = read_file()
print(f"Part 1: {part1(vals)}")
print(f"Part 2: {part2(vals)}")

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Part 1: 1139
Part 2: 1103

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# SPDX-License-Identifier: MIT
# Copyright (c) 2021 Akumatic
from code import part1, part2
def test():
vals = [199,200,208,210,200,207,240,269,260,263]
assert part1(vals) == 7
print("Passed Part 1")
assert part2(vals) == 5
print("Passed Part 2")
if __name__ == "__main__":
test()