ocaml: phase 5.1 lcs.ml baseline (LCS of "ABCBDAB" and "BDCAB" = 4)

Classic 2D DP for longest common subsequence, optimized to use
two rolling 1D arrays (prev / curr) for O(min(m,n)) space:

  for i = 1 to m do
    for j = 1 to n do
      if s1.[i-1] = s2.[j-1] then curr.(j) <- prev.(j-1) + 1
      else if prev.(j) >= curr.(j-1) then curr.(j) <- prev.(j)
      else curr.(j) <- curr.(j-1)
    done;
    for j = 0 to n do prev.(j) <- curr.(j) done
  done;
  prev.(n)

  lcs "ABCBDAB" "BDCAB" = 4

Two valid LCS witnesses: BCAB and BDAB.

Avoids Array.make_matrix (not in our runtime) by manual rolling.

142 baseline programs total.

lib/ocaml/baseline/expected.json

@@ -82,6 +82,7 @@
   "kadane.ml": 6,
   "kmp.ml": 5,
   "lambda_calc.ml": 7,
+  "lcs.ml": 4,
   "majority_vote.ml": 4,
   "levenshtein.ml": 11,
   "memo_fib.ml": 75025,

lib/ocaml/baseline/lcs.ml

@@ -0,0 +1,23 @@
+let lcs s1 s2 =
+  let m = String.length s1 in
+  let n = String.length s2 in
+  let prev = Array.make (n + 1) 0 in
+  let curr = Array.make (n + 1) 0 in
+  for i = 1 to m do
+    for j = 1 to n do
+      if s1.[i - 1] = s2.[j - 1] then
+        curr.(j) <- prev.(j - 1) + 1
+      else if prev.(j) >= curr.(j - 1) then
+        curr.(j) <- prev.(j)
+      else
+        curr.(j) <- curr.(j - 1)
+    done;
+    for j = 0 to n do
+      prev.(j) <- curr.(j)
+    done
+  done;
+  prev.(n)
+
+;;
+
+lcs "ABCBDAB" "BDCAB"