giles
0231bb46a6
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Classic trapped-rain-water two-pass DP:
left_max[i] = max(heights[0..i]) (forward sweep)
right_max[i] = max(heights[i..n-1]) (downto sweep)
water = sum over i of (min(left_max[i], right_max[i])
- heights[i])
For [0; 1; 0; 2; 1; 0; 1; 3; 2; 1; 2; 1]: water = 6.
Tests dual sweep (forward + downto), array of running maxes,
inline-if rhs of <- for running-max update (uses iter-236 fix
for <- accepting if/match RHS).
203 baseline programs total.
26 lines
768 B
OCaml
26 lines
768 B
OCaml
let trap heights =
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let n = Array.length heights in
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if n < 3 then 0
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else begin
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let left_max = Array.make n 0 in
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let right_max = Array.make n 0 in
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left_max.(0) <- heights.(0);
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for i = 1 to n - 1 do
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left_max.(i) <- if heights.(i) > left_max.(i - 1) then heights.(i) else left_max.(i - 1)
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done;
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right_max.(n - 1) <- heights.(n - 1);
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for i = n - 2 downto 0 do
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right_max.(i) <- if heights.(i) > right_max.(i + 1) then heights.(i) else right_max.(i + 1)
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done;
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let water = ref 0 in
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for i = 0 to n - 1 do
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let min_lr = if left_max.(i) < right_max.(i) then left_max.(i) else right_max.(i) in
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water := !water + min_lr - heights.(i)
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done;
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!water
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end
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;;
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trap [| 0; 1; 0; 2; 1; 0; 1; 3; 2; 1; 2; 1 |]
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