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First solution in Clear category for Most Efficient Cutting by derrickding95
def most_efficient_cutting(bom):
bom = sorted(bom, reverse=True)
queue = [[6000-bom.pop(0), bom, 0]]
result = []
while queue:
[current, rest, wasted] = queue.pop(0)
if not rest:
result.append(wasted+current)
continue
if all(current < i for i in rest):
queue.append([6000-rest.pop(0), rest, wasted+current])
else:
for (i, next_) in enumerate(rest):
if next_ <= current:
queue.append([current-next_, rest[:i]+rest[i+1:], wasted])
return min(result)
#These "asserts" using only for self-checking and not necessary for auto-testing
if __name__ == '__main__':
assert(most_efficient_cutting([3000, 2200, 2000, 1800, 1600, 1300]) == 100)
assert(most_efficient_cutting([4000, 4000, 4000]) == 6000), "wasted: 3x2000"
assert(most_efficient_cutting([1]) == 5999), "5999"
assert(most_efficient_cutting([3001, 3001]) == 5998), "2x2999"
assert(most_efficient_cutting([3000, 2200, 1900, 1800, 1600, 1300]) == 200), "2x5900"
assert(most_efficient_cutting([3000]) == 3000)
assert(most_efficient_cutting([3000, 2200, 2000, 1800, 1600, 1400]) == 0)
print('"Run" is good. How is "Check"?')
Jan. 1, 2020
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