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Diferente intre continut:
return s
==
Each query takes on average $k/2 + n/k + k/2 = k + n/k$ time. This is minimized for $k = sqrt(n)$. So we get a $O(sqrt(n))$ time complexity query.
Each query takes less than $k + n/k + k = 2k + n/k$ time. For $k = sqrt(n)$ we get a $O(sqrt(n))$ time complexity query.
This trick also works for other associative operations, like: min, gcd, product etc.
Diferente intre securitate:
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