An anadrome is a string that forms another word when read in reverse. For example, “top” and “pot” are anadromes of each other. This algorithm takes a set of strings and returns a set of all the pairs of anadromes in the input set.


(** Anadromes Problem *)

module StrSet = Set.Make(String)

let read_line_seq ch =
  let rec repeat () =
    match input_line ch with
    | s -> Seq.Cons (s, repeat)
    | exception End_of_file -> Nil
  in repeat

let string_rev s =
  let last = pred (String.length s) in
  String.init (succ last) (fun i -> s.[last - i])

let get_anadromes set =
  let aux s =
    let r = string_rev s in
    if s < r && StrSet.mem r set
    then Some (s, r)
    else None
  Seq.filter_map aux (StrSet.to_seq set)

let () = read_line_seq stdin |> Seq.filter (fun s -> String.length s > 6)
  |> String.lowercase_ascii |> StrSet.of_seq |> get_anadromes
  |> Seq.iter (fun (s, r) -> Printf.printf "%9s | %s
" s r)

The algorithm reads lines from standard input until end-of-file is reached. It filters out lines that are too short to have anadromes (less than 7 characters), transforms all the letters to lowercase using String.lowercase_ascii, and stores the resulting strings in a StrSet. Then it calls get_anadromes on the set, which returns a sequence of pairs of anadromes. Finally, it prints each pair as a line with its two elements separated by a vertical bar.

Step-by-step Explanation

  1. Define a module StrSet that provides a set implementation for strings.
  2. Define a function read_line_seq that returns a sequence of lines read from a channel until end-of-file is reached. Use Seq.Cons to build the sequence recursively, and a match expression to handle end-of-file as an exception.
  3. Define a function string_rev that takes a string s and returns its reverse. Use String.length to find the index of the last character, String.init to initialize a new string with the same length as s, and a lambda function to fill each position with the corresponding character from s.
  4. Define a function get_anadromes that takes a set set of strings and returns a sequence of all the pairs of anadromes in set. Use Seq.filter_map to apply the auxiliary function aux to each element of set and keep only the ones that return Some (i.e., anadromes).
  5. Define an auxiliary function aux that takes a string s and checks whether its reverse is in set. If it is, return Some (s, r) where r is the reverse of s. If it isn’t, return None.
  6. In the main program, read lines from standard input using read_line_seq, filter out the ones that are too short using Seq.filter, transform them to lowercase using and String.lowercase_ascii, build a StrSet with StrSet.of_seq, call get_anadromes on it, and iterate over the resulting sequence with Seq.iter, printing each pair as a line with Printf.printf.

Complexity Analysis

Let n be the number of strings in the input set, and m be the maximum length of a string.

The time complexity of String.length is O(1), the time complexity of String.init is O(m), and the time complexity of the lambda function is O(1), so the time complexity of string_rev is O(m).

The time complexity of StrSet.mem is logarithmic in the size of the set, so the time complexity of aux is O(m log n).

The time complexity of Seq.filter_map is proportional to the size of the sequence, which is at most n, and the time complexity of get_anadromes is proportional to the number of anadrome pairs in the set, which is at most n/2, so the time complexity of the main algorithm is O(nm log n).

The space complexity of the algorithm is dominated by the StrSet data structure, which stores up to n strings of up to length m, so the space complexity is O(nm). The other data structures used (sequences, pairs, variables) have constant size.