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Laaksonen Antti H S 2017-07-08 19:48:37 +03:00
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@ -457,8 +457,8 @@ After this change, the time complexity is $O(n^2)$.
Surprisingly, it is possible to solve the problem Surprisingly, it is possible to solve the problem
in $O(n)$ time\footnote{In \cite{ben86}, this linear-time algorithm in $O(n)$ time\footnote{In \cite{ben86}, this linear-time algorithm
is attributed to J. B. Kadene, and the algorithm is sometimes is attributed to J. B. Kadane, and the algorithm is sometimes
called \index{Kadene's algorithm} \key{Kadene's algorithm}.}, which means called \index{Kadane's algorithm} \key{Kadane's algorithm}.}, which means
that just one loop is enough. that just one loop is enough.
The idea is to calculate, for each array position, The idea is to calculate, for each array position,
the maximum sum of a subarray that ends at that position. the maximum sum of a subarray that ends at that position.

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@ -28,6 +28,6 @@ You can always send feedback on the book to
\texttt{ahslaaks@cs.helsinki.fi}. \texttt{ahslaaks@cs.helsinki.fi}.
\begin{flushright} \begin{flushright}
Helsinki, June 2017 \\ Helsinki, July 2017 \\
Antti Laaksonen Antti Laaksonen
\end{flushright} \end{flushright}