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Sweep-line algorithm via separately-sorted starts / ends arrays:
while i < n do
if starts[i] < ends[j] then begin busy++; rooms = max; i++ end
else begin busy--; j++ end
done
intervals: (0,30) (5,10) (15,20) (10,25) (5,12)
(20,35) (0,5) (8,18)
At time 8, meetings (0,30), (5,10), (5,12), (8,18) are all active
simultaneously -> answer = 4.
Tests local helper bound via let (`let bubble a = ...`) for
in-place sort, dual-pointer sweep on parallel ordered event streams.
193 baseline programs total.
41 lines
931 B
OCaml
41 lines
931 B
OCaml
let min_rooms intervals =
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let n = List.length intervals in
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let arr = Array.of_list intervals in
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let starts = Array.make n 0 in
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let ends = Array.make n 0 in
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for i = 0 to n - 1 do
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let (s, e) = arr.(i) in
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starts.(i) <- s;
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ends.(i) <- e
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done;
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let bubble a =
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for i = 0 to n - 1 do
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for j = 0 to n - 2 - i do
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if a.(j) > a.(j + 1) then begin
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let t = a.(j) in
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a.(j) <- a.(j + 1);
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a.(j + 1) <- t
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end
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done
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done
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in
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bubble starts;
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bubble ends;
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let rooms = ref 0 in
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let busy = ref 0 in
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let i = ref 0 and j = ref 0 in
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while !i < n do
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if starts.(!i) < ends.(!j) then begin
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busy := !busy + 1;
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if !busy > !rooms then rooms := !busy;
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i := !i + 1
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end else begin
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busy := !busy - 1;
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j := !j + 1
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end
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done;
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!rooms
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;;
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min_rooms [(0, 30); (5, 10); (15, 20); (10, 25); (5, 12); (20, 35); (0, 5); (8, 18)]
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