38 lines
851 B
Python
38 lines
851 B
Python
|
from collections import defaultdict
|
||
|
|
|
||
|
|
with open("input.txt") as f:
|
||
|
|
all_crabs = [int(i) for i in f.read().strip().split(",")]
|
||
|
|
|
||
|
|
|
||
|
|
# group all crabs by starting position
|
||
|
|
crabs = defaultdict(int)
|
||
|
|
for crab in all_crabs:
|
||
|
|
crabs[crab] += 1
|
||
|
|
|
||
|
|
|
||
|
|
crabs = dict(crabs)
|
||
|
|
tries = []
|
||
|
|
|
||
|
|
# pre-calcuate crab movement costs
|
||
|
|
crab_cost = []
|
||
|
|
cost = 0
|
||
|
|
for i in range(0, max(all_crabs) + 1):
|
||
|
|
cost = cost + i
|
||
|
|
crab_cost.append(cost)
|
||
|
|
|
||
|
|
# apparently the equation `v = k * (k+1) / 2` is equivalent to the list
|
||
|
|
# generated above, however, this approach using a pre-computed list is 10-15%
|
||
|
|
# faster in my testing.
|
||
|
|
|
||
|
|
for i in range(min(all_crabs), max(all_crabs) + 1):
|
||
|
|
moves = 0
|
||
|
|
for crab, count in crabs.items():
|
||
|
|
more = crab_cost[abs(i - crab)] * count
|
||
|
|
moves += more
|
||
|
|
|
||
|
|
tries.append((i, moves, ))
|
||
|
|
|
||
|
|
|
||
|
|
tries.sort(key=lambda x: x[1])
|
||
|
|
print(tries[0][1])
|