/* 

  Find the nth permutation (Code golf) in Picat.

  https://codegolf.stackexchange.com/questions/114883/i-give-you-nth-permutation-you-give-me-n
  """
  I give you Nth permutation, you give me N

  Input: a sequence of uppercase letters (ASCII [65;90]) which is the Nth* lexicographical 
  permutation of the multiset of its characters

  *permutations are numbered from 0 or 1 upwards

  Output: base-10 integer N

  Rulez
  There might be duplicates (that's how this challenge differs from this one)
  The characters are ordered by their ASCII value
  In case of an input of length less than or equal to 1, the input is the first permutation 
  and the result is 0 or 1 respectively

  First permutation is that in which the leftmost character has the lowest value, the rightmost 
  character has the highest value, and the sequence of characters between the first and the 
  last character is the first permutation of the multiset of its characters (recursive definition!)

  Shortest entry wins
  Example
  Input AAB produces output 0
  Input ABA produces output 1
  Input BAA produces output 2
  Input ZZZ produces output 0
  Input DCBA produces output 23

  EDIT
  Extra kudos to the one who can come up with a solution which doesn't produce all permutations 
  and then search for the input. That's some challenge.
  """

  This program was created by Hakan Kjellerstrand, hakank@gmail.com
  See also my Picat page: http://www.hakank.org/picat/

*/

import util. % Don't forget to count this! 12 chars

main => go.

go ?=>

  Tests = ["AAB",
           "ABA",
           "BAA",
           "ZZZ",
           "DCBA"],
  foreach(T in Tests)
    % 74 chars + 12 (for "import util.") = total 86 chars
    % P=permutations(T).sort_remove_dups,[I-1:I in 1..P.len,P[I]==T].min.println,

    % Ungolfed 
    % P=permutations(T).sort.remove_dups,
    % [I-1:I in 1..P.len,P[I]==T].min.println

    % Or as 1-based: total 84 chars
    % P=permutations(T).sort_remove_dups,[I:I in 1..P.len,P[I]==T].min.println    

    % Shorter using find_first_of/2: 58 chars + 12 chars =  70 chars
    % Or without println/1: 49 + 12 = 61 chars
    permutations(T).sort_remove_dups.find_first_of(T).println,
    
  end,
  
  nl.
go => true.