giles
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Two-pass partition DP for max profit with at most 2 transactions:
left[i] = max single-trans profit in prices[0..i]
(forward scan tracking running min)
right[i] = max single-trans profit in prices[i..n-1]
(backward scan tracking running max)
answer = max over i of (left[i] + right[i])
For [3; 3; 5; 0; 0; 3; 1; 4]:
optimal partition i = 2:
left[2] = sell@5 after buy@3 = 2
right[2] = sell@4 after buy@0 in [2..7] = 4
total = 6
Tests parallel forward + backward passes on parallel DP arrays,
mixed ref + array state, for downto + for ascending scans on
the same data.
190 baseline programs total.
30 lines
829 B
OCaml
30 lines
829 B
OCaml
let max_profit_two prices =
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let n = Array.length prices in
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if n < 2 then 0
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else begin
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let left = Array.make n 0 in
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let min_p = ref prices.(0) in
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for i = 1 to n - 1 do
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if prices.(i) < !min_p then min_p := prices.(i);
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let p = prices.(i) - !min_p in
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left.(i) <- if p > left.(i - 1) then p else left.(i - 1)
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done;
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let right = Array.make n 0 in
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let max_p = ref prices.(n - 1) in
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for i = n - 2 downto 0 do
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if prices.(i) > !max_p then max_p := prices.(i);
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let p = !max_p - prices.(i) in
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right.(i) <- if p > right.(i + 1) then p else right.(i + 1)
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done;
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let best = ref 0 in
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for i = 0 to n - 1 do
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let total = left.(i) + right.(i) in
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if total > !best then best := total
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done;
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!best
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end
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;;
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max_profit_two [| 3; 3; 5; 0; 0; 3; 1; 4 |]
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