Solution to problem 10 in Python

src/P10.py

@@ -0,0 +1,136 @@
+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()