Find first sublist that has all numbers from 1 to

Question

I need to find the first sublist having all numbers from 1 to K and return it and its length.

Sorry in advance for the bad editing

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So I check if 1 to K is in sublist if not then I delete the element (Hd) from

NumList ,append it to our result(SubList) and recursively call the function

with the tail as our new List to check.

findFirstSubListHavingAllColors( NumList, [Hd|Tl] ,SubList, X):-
   (   oneToKinSub(SubList,NumList)
   ->  length(SubList,X)
   ;   delete(NumList,Hd,NumList1),
       append(SubList,[Hd],SubList1),
       findFirstSubListHavingAllColors(NumList1,Tl,SubList1,_)
   ).

oneToKinSub(_,[]).
oneToKinSub(SubString,[Hd|Tl]) :-
   member(Hd,SubString),
   oneToKinSub(SubString,Tl).

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For instance if

NumList =[1,2,3]

[Hd|Tl] =[1,3,1,3,1,3,3,2,2,1]

the expected result should be SubList=[1,3,1,3,1,3,3,2] and X= 8

Answer 1

You may use append/3 and subtract/3 to obtain the first sublist that contains all items:

findFirstSubListHavingAllColors( NumList, List ,SubList, Len):-
  once((
   append(SubList, _, List),        % get SubList
   subtract(NumList, SubList, []),  % test whether NumList is contained in SubList
   length(SubList, Len)
  )).

once/1 here is to avoid getting other (wrong) solutions on backtracking.

Answer 2

Here is another solution using aggegate_all/3

firstSubList(NumList,  In, Out) :-
    aggregate_all(max(E),(member(X, NumList), once(nth1(E, In, X))), Len),
    length(Out, Len),
    append(Out, _, In).
Once/1 is used because numbers may appears many times in the list !