feat(post): clarify constant time upper bound

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Barrett Ruth 2024-11-09 16:06:16 -05:00
parent 1d9979fd00
commit 4f64a105c8

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@ -176,7 +176,7 @@
2^k-1=x\))). \(x\) and \(n\) have \(lg(x)\) and \(lg(n)\) bits
respectively, so the solution is \(O(lg(x) + lg(n))\in O(log(xn))\).
\(1\leq x,n\leq 1e8\), so this runtime is bounded by
\(O(log(1e8^2))=O(log(1e16))=O(16)\).
\(O(log(1e8^2))=O(log(1e16))=O(16)=O(1)\).
</p>
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