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Classic word-break DP — for each position i, check whether any
dictionary word ends at i with a prior reachable position:
dp[i] = exists w in dict with wl <= i and
dp[i - wl] && s.sub (i - wl) wl = w
Dictionary: apple, pen, pine, pineapple, cats, cat, and, sand, dog
Inputs:
applepenapple yes (apple pen apple)
pineapplepenapple yes (pineapple pen apple)
catsanddog yes (cats and dog)
catsandog no (no segmentation reaches the end)
applesand yes (apple sand)
Tests bool-typed Array, String.sub primitive, nested List.iter
over the dict inside for-loop over end positions, closure capture
of the outer dp.
179 baseline programs total.
28 lines
727 B
OCaml
28 lines
727 B
OCaml
let word_break s words =
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let n = String.length s in
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let dp = Array.make (n + 1) false in
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dp.(0) <- true;
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for i = 1 to n do
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List.iter (fun w ->
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let wl = String.length w in
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if i >= wl && dp.(i - wl) then begin
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let prefix = String.sub s (i - wl) wl in
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if prefix = w then dp.(i) <- true
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end
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) words
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done;
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if dp.(n) then 1 else 0
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let count_ok strings words =
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let count = ref 0 in
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List.iter (fun s ->
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count := !count + word_break s words
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) strings;
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!count
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;;
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let dict = ["apple"; "pen"; "pine"; "pineapple"; "cats"; "cat"; "and"; "sand"; "dog"] in
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let inputs = ["applepenapple"; "pineapplepenapple"; "catsanddog"; "catsandog"; "applesand"] in
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count_ok inputs dict
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