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by
entrope
1 days ago
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comments
by
moi2388
1 days ago
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[-]
My algorithm is O(n+n+..n) which is O(n), since there we also ignore addition fortunately :D
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by
AlotOfReading
1 days ago
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parent
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It would only be O(n) if the number of additions was constant. Here it varies with the size of the multiplier, giving us O(n*m).
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