(* Copyright (C) 2008 Mauricio Fernandez <mfp@acm.org> http//eigenclass.org *)
(* Solution to the coding challenge at
 * http://beust.com/weblog/archives/000491.html *)
open Printf

let count = ref 0 and prev = ref 0 and maxj = ref 0 and base = ref 0

let handle num =
  incr count;
  let jump = num - !prev in
    if jump > !maxj then (maxj := jump; base := !prev);
    prev := num

let permutations len =
  let rec loop n len l1 = function
      d::(e::tl as tl1) ->
        let l = d :: l1 in
        let ll = e :: l in
        aux (n * 10 + d) (len - 1) l1 tl1;
        aux (n * 10 + e) (len - 1) l tl;
        loop n len ll tl
    | d::tl -> aux (n * 10 + d) (len - 1) l1 tl; loop n len (d::l1) tl
    | [] -> ()
  and aux n len l1 l2 = match len with
      0 -> handle n
    | len -> loop n len [] (List.rev_append l1 l2)
  in for i = 1 to 9 do
    let l = List.filter ((<>) i) (Array.to_list (Array.init 10 (fun i -> i))) in
      aux i (len - 1) [] l
  done

let () =
  let report () =
    printf "Found %d numbers.\n" !count;
    printf "Max jump: %d (%d -- %d).\n" !maxj !base (!base + !maxj)

  in try
    for i = 1 to 10 do
      permutations i
    done;
    report ()
  with Exit -> report ()