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Classic distinct-subsequences 2D DP: dp[i][j] = dp[i-1][j] + (s[i-1] = t[j-1] ? dp[i-1][j-1] : 0) dp[i][0] = 1 (empty t is a subseq of any prefix of s) count_subseq "rabbbit" "rabbit" = 3 The three witnesses correspond to which 'b' in "rabbbit" is dropped (positions 2, 3, or 4 zero-indexed of the run of bs). Complements subseq_check.ml (just tests presence); this one counts distinct embeddings. Tests 2D DP with Array.init n (fun _ -> Array.make m 0), base row initialization, mixed string + array indexing. 175 baseline programs total.
21 lines
445 B
OCaml
21 lines
445 B
OCaml
let count_subseq s t =
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let m = String.length s in
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let n = String.length t in
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let dp = Array.init (m + 1) (fun _ -> Array.make (n + 1) 0) in
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for i = 0 to m do
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dp.(i).(0) <- 1
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done;
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for i = 1 to m do
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for j = 1 to n do
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if s.[i - 1] = t.[j - 1] then
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dp.(i).(j) <- dp.(i - 1).(j) + dp.(i - 1).(j - 1)
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else
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dp.(i).(j) <- dp.(i - 1).(j)
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done
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done;
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dp.(m).(n)
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;;
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count_subseq "rabbbit" "rabbit"
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