2021-12-01

Day 01: Sonar Sweep

Description:
--- Day 1: Sonar Sweep ---

You're minding your own business on a ship at sea when the overboard alarm goes off! You rush to see if you can help. Apparently, one of the Elves tripped and accidentally sent the sleigh keys flying into the ocean!

Before you know it, you're inside a submarine the Elves keep ready for situations like this. It's covered in Christmas lights (because of course it is), and it even has an experimental antenna that should be able to track the keys if you can boost its signal strength high enough; there's a little meter that indicates the antenna's signal strength by displaying 0-50 stars.

Your instincts tell you that in order to save Christmas, you'll need to get all fifty stars by December 25th.

Collect stars by solving puzzles. Two puzzles will be made available on each day in the Advent calendar; the second puzzle is unlocked when you complete the first. Each puzzle grants one star. Good luck!

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 Two ---

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?

Input:
100
125
124
127
141
145
160
161
178
185
193
196
195
200
212
216
222
244
274
276
279
307
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325
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336
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772
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812
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1033
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1094
1102
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9274
9262
9261
9270
9284
9285
9286
9289
9290
9291
9293
9282
9285
9286
9289
9287
9289
9291
9289
9290
9299
9302
9316
9317
9342
9337
9339
9340
9341
9342
9371
9373
9374
9373
9374
9376
9384
9390
9423
9424
9427
9429
9430
9431
9432
9434
9465
9473
9474
9478
9482
9483
9485
9469
9471
9441
9457
9467
9462
9474
9475
9481
9475
9476
9479
9482
9484
9493
9500
9506
9509
9510
9511
9524
9529
9558
9571
9576
9579
9581
9582
9587
9594
9618
9638
9646
9649
9650
9649
9655
9646
9648
9652
9665
9659
9697
9703
9733
9750
9751
9753
9754
9755
9761
9762
9766
9784
9789
9766
9767
9768
9771
9785
9789
9792
9807
9821
9823
9830
9844
9847
9857
9858
9860
9873
9870
9873
9887
9880
9878
9891
9893
9894
9896
9902
9911
9920
9931
9963
9972
9974
9973
9974
9976
9990
9989
9996
9995
10024
10025
10026
10028
10040
10041
10040
10041
10044

``````package advent

import (
"math"
"strconv"
)

func Day1Part1(ctx *util.AOCContext) (string, error) {

incCount := 0

val := int64(math.MaxInt64)
for _, v := range ctx.InputLines() {
num, err := strconv.ParseInt(v, 10, 64)
if err != nil {
return "", err
}

if val < num {
incCount++
}
val = num
}

return strconv.Itoa(incCount), nil
}

func Day1Part2(ctx *util.AOCContext) (string, error) {

incCount := 0

input := ctx.InputLines()

prev := int64(math.MaxInt64)

sw0 := int64(0)
sw1 := int64(0)
sw2 := int64(0)

for i, v := range input {
num, err := strconv.ParseInt(v, 10, 64)
if err != nil {
return "", err
}

sw0 = sw1
sw1 = sw2
sw2 = num

val := sw0 + sw1 + sw2

if i > 2 && prev < val {
incCount++
//fmt.Printf("%d < %d + %d + %d\n", prev, sw0, sw1, sw2)
}
prev = val
}

return strconv.Itoa(incCount), nil
}
``````
Result Part 1: 1715
Result Part 2: 1739

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