mybest speedy solution for Xs and Os Referee by nennogabriel - python coding challenges

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mybest solution in Speedy category for Xs and Os Referee by nennogabriel

def checkio(rows):
    #based on Veky clear solution and gyhun_dash speed.
    #the clear from veky is faster then gyhun,  
    #butcombine bouth reduce to half of time
    inline = ''.join(rows)
    diags = [inline[0:9:4], inline[2:7:2]]
    cols  = list(map(''.join, zip(*rows)))
    lines = rows + list(cols) + list(diags)
    return 'X' if ('XXX' in lines) else 'O' if ('OOO' in lines) else 'D'

if __name__ == '__main__':
    #These "asserts" using only for self-checking and not necessary for auto-testing
    assert checkio([
        "X.O",
        "XX.",
        "XOO"]) == "X", "Xs wins"
    assert checkio([
        "OO.",
        "XOX",
        "XOX"]) == "O", "Os wins"
    assert checkio([
        "OOX",
        "XXO",
        "OXX"]) == "D", "Draw"
    assert checkio([
        "O.X",
        "XX.",
        "XOO"]) == "X", "Xs wins again"

July 23, 2014

Comments:

veky on Aug. 4, 2014, 11:43 p.m.
Too many "list"s (in cols). You don't need () around '___' in lines. You don't need 0 and 9 (len) in slices: inline[::4]. If you're really obsessed with speed, try inlining diags and cols. Also try extending rows directly instead of copying.

nennogabriel on Aug. 5, 2014, 3:22 a.m.
on my tests this was a little bit faster than your clear code. but.. ok.. I will try harder well you use : r.extend("".join(r)[next(p):next(p):next(p)] for _ in range(5)) but for a shorter code you can write: r.extend("".join(r)[n(p):n(p):n(p)] for n in [next]*5) that is my golf shot for you.

veky on Aug. 6, 2014, 8:43 a.m.
Of course. That wasn't really golfing, I just tried to unify the cases and to show all of these are just various slices. DOX and XOX are for golf.

michal.lipinski on Dec. 27, 2015, 8:29 p.m.
rly nice

mmartsinenko on Dec. 9, 2024, 11:30 a.m.
This clear and concise solution generates rows, columns, and diagonals in a Pythonic way using zip and slicing. It matches my solution in efficiency and is quite readable. +1

List, list and list again

0 %

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