First clear solution for Loading Cargo by brownie57 - python coding challenges

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First solution in Clear category for Loading Cargo by brownie57

def checkio(data):
    from itertools import combinations as C
    ans = sum(data)
    h = ans/2
    for i in range(len(data)):
        c = list(C(data, i))
        for j in c:
            if abs(h-sum(j)) < ans:
                ans = abs(h-sum(j))
    return ans*2


#These "asserts" using only for self-checking and not necessary for auto-testing
if __name__ == '__main__':
    assert checkio([10, 10]) == 0, "1st example"
    assert checkio([10]) == 10, "2nd example"
    assert checkio([5, 8, 13, 27, 14]) == 3, "3rd example"
    assert checkio([5, 5, 6, 5]) == 1, "4th example"
    assert checkio([12, 30, 30, 32, 42, 49]) == 9, "5th example"
    assert checkio([1, 1, 1, 3]) == 0, "6th example"

Aug. 10, 2018

Comments:

ibonyun on June 10, 2021, 10:59 p.m.
Your lines 6 and 7: <pre class='brush: python'> c = list(C(data, i)) for j in c: ... </pre> This generates all possible combinations up front, stores them all in a list, and then iterates over the entire list testing each combination. Instead, you should be doing this: <pre class='brush: python'> for j in C(data, i): ... </pre> This generates one combination at a time, testing each combination before generating the next. Because there's no need to store them, it's more memory efficient. And because we're testing as we go, it allows us to exit early if we think we've found an ideal combination which will make it twice as fast on average. That's not a huge difference, but it's such an easy optimization to make.

List, list and list again

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