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+# --- Day 7: The Treachery of Whales ---
+
+# A giant whale has decided your submarine is its next meal, and it's much
+# faster than you are. There's nowhere to run!
+
+# Suddenly, a swarm of crabs (each in its own tiny submarine - it's too deep
+# for them otherwise) zooms in to rescue you! They seem to be preparing to
+# blast a hole in the ocean floor; sensors indicate a massive underground cave
+# system just beyond where they're aiming!
+
+# The crab submarines all need to be aligned before they'll have enough power
+# to blast a large enough hole for your submarine to get through. However, it
+# doesn't look like they'll be aligned before the whale catches you! Maybe you
+# can help?
+
+# There's one major catch - crab submarines can only move horizontally.
+
+# You quickly make a list of the horizontal position of each crab (your puzzle
+# input). Crab submarines have limited fuel, so you need to find a way to make
+# all of their horizontal positions match while requiring them to spend as
+# little fuel as possible.
+
+# For example, consider the following horizontal positions:
+
+# 16,1,2,0,4,2,7,1,2,14
+
+# This means there's a crab with horizontal position 16, a crab with horizontal
+# position 1, and so on.
+
+# Each change of 1 step in horizontal position of a single crab costs 1 fuel.
+# You could choose any horizontal position to align them all on, but the one
+# that costs the least fuel is horizontal position 2:
+
+#     Move from 16 to 2: 14 fuel
+#     Move from 1 to 2: 1 fuel
+#     Move from 2 to 2: 0 fuel
+#     Move from 0 to 2: 2 fuel
+#     Move from 4 to 2: 2 fuel
+#     Move from 2 to 2: 0 fuel
+#     Move from 7 to 2: 5 fuel
+#     Move from 1 to 2: 1 fuel
+#     Move from 2 to 2: 0 fuel
+#     Move from 14 to 2: 12 fuel
+
+# This costs a total of 37 fuel. This is the cheapest possible outcome; more
+# expensive outcomes include aligning at position 1 (41 fuel), position 3 (39
+# fuel), or position 10 (71 fuel).
+
+# Determine the horizontal position that the crabs can align to using the least
+# fuel possible. How much fuel must they spend to align to that position?
+
+with open("files/P7.txt") as f:
+    positions = [int(num) for num in f.read().strip().split(",")]
+
+
+def part_1() -> None:
+    fuel = []
+    for pos in positions:
+        sum = 0
+        for elem in positions:
+            sum += abs(elem - pos)
+        fuel.append(sum)
+
+    print(f"It is needed {min(fuel)} units of fuel")
+
+
+# --- Part Two ---
+
+# The crabs don't seem interested in your proposed solution. Perhaps you
+# misunderstand crab engineering?
+
+# As it turns out, crab submarine engines don't burn fuel at a constant rate.
+# Instead, each change of 1 step in horizontal position costs 1 more unit of
+# fuel than the last: the first step costs 1, the second step costs 2, the
+# third step costs 3, and so on.
+
+# As each crab moves, moving further becomes more expensive. This changes the
+# best horizontal position to align them all on; in the example above, this
+# becomes 5:
+
+#     Move from 16 to 5: 66 fuel
+#     Move from 1 to 5: 10 fuel
+#     Move from 2 to 5: 6 fuel
+#     Move from 0 to 5: 15 fuel
+#     Move from 4 to 5: 1 fuel
+#     Move from 2 to 5: 6 fuel
+#     Move from 7 to 5: 3 fuel
+#     Move from 1 to 5: 10 fuel
+#     Move from 2 to 5: 6 fuel
+#     Move from 14 to 5: 45 fuel
+
+# This costs a total of 168 fuel. This is the new cheapest possible outcome;
+# the old alignment position (2) now costs 206 fuel instead.
+
+# Determine the horizontal position that the crabs can align to using the least
+# fuel possible so they can make you an escape route! How much fuel must they
+# spend to align to that position?
+
+# TIL :D
+def triangular_cost(steps: int) -> int:
+    return steps * (steps + 1) // 2
+
+
+def part_2() -> None:
+    fuel = []
+    for pos in positions:
+        sum = 0
+        for elem in positions:
+            sum += triangular_cost(abs(elem - pos))
+        fuel.append(sum)
+
+    print(f"It is needed {min(fuel)} units of fuel")
+
+
+if __name__ == "__main__":
+    part_1()
+    part_2()