Stretch: mutable arrays and two-pass loops

For score_guess, the repeated-letter fix.

Array.make and mutation

Array.make n v creates a new array of length n with every element set to v. Unlike a list, an array's elements can be overwritten in place with arr.(i) <- new_value.

let counters = Array.make 5 0
let () = counters.(2) <- 9
(* counters = [| 0; 0; 9; 0; 0 |] *)

This is the tool for "start with a blank slate of a known size, then fill it in as you go" -- exactly what a per-letter remaining-claims budget needs: 26 counters, one per letter of the alphabet, all starting from count_letter target letter and counting down as matches claim them.

for loops

A for loop runs its body once for each integer in a range, purely for side effects (mutating an array, printing, and so on) -- there is no accumulator to thread through by hand, unlike List.fold_left.

let () =
  for i = 0 to 4 do
    Printf.printf "%d " i
  done
(* prints: 0 1 2 3 4 *)

for loops pair naturally with arrays: for i = 0 to Array.length a - 1 do ... done visits every valid index in order, letting the body read and write a.(i) directly.

Why two separate passes?

Sometimes a computation genuinely can't be done correctly in one pass over the data, because a later element's correct treatment depends on a decision made about an earlier element that hasn't happened yet if everything runs in one interleaved walk.

Scoring a guess's repeated letters is exactly this: every Correct match must be decided and have its claim on the target's letters spent before any Present match is handed out, or a letter that's Correct in one position could get double-counted as Present for a duplicate elsewhere. Splitting into two for loops over the same range -- one pass that only looks for Correct, a second pass that only looks for Present -- makes that ordering explicit instead of trying to interleave both decisions in a single walk.

let scores_two_pass data =
  let n = Array.length data in
  let result = Array.make n 0 in
  (* pass 1: something that must happen everywhere first *)
  for i = 0 to n - 1 do
    if data.(i) > 100 then result.(i) <- 1
  done;
  (* pass 2: everything else, now that pass 1 has fully finished *)
  for i = 0 to n - 1 do
    if result.(i) = 0 && data.(i) > 0 then result.(i) <- 2
  done;
  result