Kadane's algorithm in O(n):
let max_subarray xs =
let max_so_far = ref min_int in
let cur = ref 0 in
List.iter (fun x ->
cur := max x (!cur + x);
max_so_far := max !max_so_far !cur
) xs;
!max_so_far
For [-2;1;-3;4;-1;2;1;-5;4] the optimal subarray is [4;-1;2;1] = 6.
Exercises min_int (iter 94), max as global, ref / ! / :=, and
List.iter with two side-effecting steps in one closure body.
46 baseline programs total.
giles
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