First speedy solution for Xs and Os Referee by MichalBiderman127330 - python coding challenges

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First solution in Speedy category for Xs and Os Referee by MichalBiderman127330

def checkio(G):
    for i in [0,1,2]:
        if G[i][2] == G[i][1] == G[i][0] != '.':
            return G[i][0]
        if G[2][i] == G[1][i] == G[0][i] != '.':
            return G[1][i]
    if G[1][1] != '.':
        if G[0][0] == G[1][1] == G[2][2] or G[0][2] == G[1][1] == G[2][0]:
            return G[1][1]
    return "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"

Oct. 29, 2015

Comments:

mmartsinenko on Dec. 9, 2024, 11:36 a.m.
<p>This solution uses a nested loop to check rows and columns and explicitly checks diagonals afterward. While readable, it performs redundant checks on individual cells, leading to slightly reduced performance compared to my solution. The solution is straightforward but not as compact or efficient.</p>

List, list and list again

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