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in O(n) determine the balanced position with zero trials! solution in Clear category for Can Balance by piter239
from typing import Iterable
from itertools import count
from operator import mul
def can_balance(weights: Iterable) -> int:
torque = sum(map(mul, weights, count())) # compute initial torque at pos 0
s = sum(weights) # each step to the right subtracts exacly the sum
# if torque divisible by sum - balance can be found!
return torque // s if torque % s == 0 else -1
if __name__ == '__main__':
print("Example:")
print(can_balance([6, 1, 10]))
print(can_balance([6, 1, 10, 5, 4]))
# These "asserts" are used for self-checking and not for an auto-testing
assert can_balance([6, 1, 10, 5, 4]) == 2
assert can_balance([10, 3, 3, 2, 1]) == 1
assert can_balance([7, 3, 4, 2, 9, 7, 4]) == -1
assert can_balance([42]) == 0
print("Coding complete? Click 'Check' to earn cool rewards!")
April 22, 2020
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