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--- 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 [1m[37mthree-measurement sliding window[0m. 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 [1m[37mincreased[0m.
Your goal now is to count [1m[37mthe number of times the sum of measurements in this sliding
window increases[0m 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 ([1m[37mincreased[0m)
C: 618 (no change)
D: 617 (decreased)
E: 647 ([1m[37mincreased[0m)
F: 716 ([1m[37mincreased[0m)
G: 769 ([1m[37mincreased[0m)
H: 792 ([1m[37mincreased[0m)
In this example, there are [1m[37m5[0m sums that are larger than the previous sum.
Consider sums of a three-measurement sliding window. [1m[37mHow many sums are larger than the
previous sum?[0m
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