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    <p>This problem is the topic of column 7 in Jon Bentleys celebrated "Programming Pearls". The solution is an algorithm with a run time of O(n), where n is the length of the vector x.</p> <p>An R implementation can solve a vector of millions of elements in less than 3 seconds:</p> <pre><code>x &lt;- rnorm(1e6) system.time(m &lt;- maxsub(x)) </code></pre> <p>If you want the negative sums, too, call maxsub(-x). It's easy to change the code such that the function will return the indices. The two ranges cannot overlap, but one can be a part of the other.</p> <pre><code>maxsub &lt;- function(x) { if (!is.numeric(x)) stop("Argument 'x' must be a numeric vector.") m1 &lt;- m2 &lt;- 0.0 for (i in 1:length(x)) { m2 &lt;- max(m2 + x[i], 0.0) m1 &lt;- max(m1, m2) } return(m1) } </code></pre> <p>Though the code looks so simple, it took quite a long time until someone came up with this solution, as Bentley reports. A corresponding algorithm for the two-(or higher)dimensional case has not yet been found.</p> <p>Update: Here is a version that also returns the indices. It looks complicated, but is not. It simply keeps track of the indices in each of the two steps above.</p> <pre><code>maxsub &lt;- function(x, inds = FALSE) { if (!is.numeric(x)) stop("Argument 'x' must be a numeric vector.") n &lt;- length(x) if (!inds) { m1 &lt;- m2 &lt;- 0.0 for (i in 1:n) { m2 &lt;- max(m2 + x[i], 0.0) m1 &lt;- max(m1, m2) } return(m1) } else { m1 &lt;- m2 &lt;- 0 p1 &lt;- p2 &lt;- 0 q1 &lt;- q2 &lt;- 1 for (i in 1:n) { if (m2 &gt; -x[i]) { m2 &lt;- m2 + x[i] q2 &lt;- i if (m2 &gt; m1) { m1 &lt;- m2 p1 &lt;- q1; p2 &lt;- q2 } } else { m2 &lt;- 0 q1 &lt;- q2 &lt;- i+1 } } return(list(sum = m1, inds = c(p1, p2))) } } </code></pre> <p>A Fortran version bounded to an R package does solve a million elements in 0.015 secs.</p>
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    1. USHans W.
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    1. COThanks for the reference. Any ideas about what to do about the potential inclusion of one range in the other?
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    2. CO@Hans: I have to say I don't yet fully understand the underlying logic - could you perhaps give the necessary modifications for returning the indices, that will make things clearer, I hope. Thank you for sharing this obviously highly efficient and elegant algorithm!
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    3. CO...BTW: I think it is column 8, not 7.
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