giles
c8206e718a
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Sort, then compare two candidates: p1 = product of three largest values p2 = product of two smallest (potentially negative) values and the largest For [-10;-10;1;3;2]: sorted = [-10;-10;1;2;3] p1 = 3 * 2 * 1 = 6 p2 = (-10) * (-10) * 3 = 300 max = 300 Tests List.sort + Array.of_list + arr.(n-i) end-walk + candidate-pick via if-then-else. 104 baseline programs total.
12 lines
288 B
OCaml
12 lines
288 B
OCaml
let max_prod3 xs =
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let sorted = List.sort compare xs in
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let arr = Array.of_list sorted in
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let n = Array.length arr in
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let p1 = arr.(n - 1) * arr.(n - 2) * arr.(n - 3) in
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let p2 = arr.(0) * arr.(1) * arr.(n - 1) in
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if p1 > p2 then p1 else p2
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;;
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max_prod3 [-10; -10; 1; 3; 2]
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