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Two-phase palindrome-partition DP for the minimum-cuts variant:
Phase 1: is_pal[i][j] palindrome table via length-major fill
(single chars, then pairs, then expand inward).
Phase 2: cuts[i] = 0 if s[0..i] is itself a palindrome,
= min over j of (cuts[j-1] + 1)
where s[j..i] is a palindrome.
min_cut "aabba" = 1 ("a" | "abba")
Tests two sequential 2D DPs sharing the same is_pal matrix,
inline begin/end branches inside the length-major fill, mixed
bool and int 2D arrays.
181 baseline programs total.
36 lines
864 B
OCaml
36 lines
864 B
OCaml
let min_cut s =
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let n = String.length s in
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if n <= 1 then 0
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else begin
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let is_pal = Array.init n (fun _ -> Array.make n false) in
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for i = 0 to n - 1 do is_pal.(i).(i) <- true done;
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for len = 2 to n do
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for i = 0 to n - len do
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let j = i + len - 1 in
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if s.[i] = s.[j] then begin
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if len = 2 then is_pal.(i).(j) <- true
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else is_pal.(i).(j) <- is_pal.(i + 1).(j - 1)
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end
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done
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done;
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let cuts = Array.make n 0 in
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for i = 0 to n - 1 do
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if is_pal.(0).(i) then cuts.(i) <- 0
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else begin
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let best = ref i in
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for j = 1 to i do
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if is_pal.(j).(i) then begin
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let c = cuts.(j - 1) + 1 in
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if c < !best then best := c
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end
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done;
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cuts.(i) <- !best
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end
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done;
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cuts.(n - 1)
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end
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;;
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min_cut "aabba"
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