module Angabe1 where

{- 1. Vervollstaendigen Sie gemaess Angabentext!
   2. Vervollständigen Sie auch die vorgegebenen Kommentaranfänge!
   3. Loeschen Sie keine Deklarationen aus diesem Rahmenprogramm, auch nicht die Modulanweisug!
   4. Achten Sie darauf, dass `Gruppe' Leserechte fuer Ihre Abgabedatei hat!
-}


type Nat0                = Int
type Zeichenreihe        = String
type Teilzeichenreihe    = String
type IstTeilzeichenreihe = Bool
type Zerlegungszeuge     = (Zeichenreihe,Zeichenreihe,Zeichenreihe)
type Zerlegungszeugen    = [Zerlegungszeuge]


-- Aufgabe A.1
{-
    Solution:
    - if both strings are the same, True
    - if the second string is empty, True
    - if the second string is longer than the first, False
    - otherwise, compare the first characters of both strings and call the function recursively again.
      If the second string is contained in the first string, it will get equal to [] eventually and the function will return True.
      If it is not, the first string will get smaller than the second eventually and the function will return False
-}

ist_tzr :: Zeichenreihe -> Teilzeichenreihe -> IstTeilzeichenreihe

ist_tzr z t
    | t == z = True
    | t == [] = True
    | length t > length z = False
    | otherwise =
        if head z == head t then
            ist_tzr (tail z) (tail t)
        else
            ist_tzr (tail z) t

-- Aufgabe A.2
{-
    Solution:
    - the base cases are as follows:
        - both strings are empty, return a triple with empty strings
        - the second string is empty, return a triple with empty strings as first two elements and the whole first string as a third.
          The problem description states that the split doesn't matter, so we take this because it is easy and cheap to do :)
        - the second string is not a subset of the first string, return the "error triple"
    - for handling the other cases:
        - we find the position of the last subset string in the first string
        - we take all of the characters until this position - this is the first element of the triple
        - we then take all of the characters after this position summed with the length of the second string - this is the third element of the triple
-}

tzr_zeuge :: Zeichenreihe -> Teilzeichenreihe -> Zerlegungszeuge
tzr_zeuge [] [] =  ("", "", "")
tzr_zeuge z [] = ("", "", z)
tzr_zeuge z t
    | not (ist_tzr z t) = ("", t ++ t, "")
    | otherwise =
        let pos = (find_tzr_position z t 0) - 1
            pre = take pos z
            suf = drop (pos + (length t)) z
        in (pre, t, suf)

-- find the index of the last occurrence of t in z
find_tzr_position :: Zeichenreihe -> Teilzeichenreihe -> Nat0 -> Nat0
find_tzr_position (z:zs) t n
    | not (ist_tzr (z:zs) t) = n
    | otherwise = find_tzr_position zs t (n + 1)

-- Aufgabe A.3
{- Knapp, aber gut nachvollziehbar geht tzr_zeugen folgendermassen vor:
   ...
-}

tzr_zeugen :: Zeichenreihe -> Teilzeichenreihe -> Zerlegungszeugen
tzr_zeugen [] [] = []
tzr_zeugen z t
    | not (ist_tzr z t) = []
    | wieOft z t == 1 = [tzr_zeuge z t]
    | otherwise = get_all_substring_variants z t []

get_all_substring_variants :: Zeichenreihe -> Teilzeichenreihe -> Zerlegungszeugen -> Zerlegungszeugen
get_all_substring_variants z t r
    | not (ist_tzr z t) = r
    | otherwise =
        let
            pos = (find_tzr_position z t 0) - 1
            new_res = r ++ [(tzr_zeuge z t)]
        in
            get_all_substring_variants (take pos z) t new_res


-- Aufgabe A.4
{-
    Solution:
    - we handle the case when the second string is empty - the result is always the length of the first string + 1
    - if the second string is not a substring of the first one, return 0 as a result
    - otherwise, count how many times the second string occurs in the first string by counting how many times the first string can be "cut" on the position of the second string
-}

wieOft :: Zeichenreihe -> Teilzeichenreihe -> Nat0
wieOft z [] = (length z) + 1
wieOft z t
    | not (ist_tzr z t) = 0
    | otherwise = count_sub_string z t 0

--- find how many times a substring occurs in a string
count_sub_string :: Zeichenreihe -> Teilzeichenreihe -> Nat0 -> Nat0
count_sub_string z t n
    | not (ist_tzr z t) = n
    | otherwise =
        let
            pos = (find_tzr_position z t 0) - 1
        in count_sub_string (take pos z) t (n + 1)