Prolog: Is there a way to find the longest consecutive number sequence in a list and get its length?

Question

Given a list of 5 random natural numbers (the number's range is from 1 to 13), i.e. [1,6,11,12,13], how can I find out the longest consecutive number sequence (i.e. [11,12,13]) that its length must be larger or equal to 3, and also get its length?

Answer 1

You can traverse the list keeping track of the current sequence and its length along with the maximum sequence and length found so far.

After processing all the list you can check whether the maximum length is >= 3 and select the proper output:

longest_sequence([First|L], Len1, Seq) :-
    longest_sequence(L, 1, [First], 0, [], Len, RSeq),
    (   Len>=3
    ->  reverse(RSeq, Seq),
        Len1=Len
    ;   Seq=[],
        Len1=0
    ).
  
longest_sequence([], CurLen, CurSeq, MaxLen, MaxSeq, Len, Seq):-
  (CurLen > MaxLen -> Len-Seq=CurLen-CurSeq ; Len-Seq=MaxLen-MaxSeq).
longest_sequence([CurItem|L], CurLen, [LastItem|CurSeq], MaxLen, MaxSeq, Len, Seq):-
  (
   succ(LastItem, CurItem) ->
    (
      succ(CurLen, CurLen1),
      CurSeq1=[CurItem,LastItem|CurSeq],
      MaxLen1-MaxSeq1=MaxLen-MaxSeq
    ) ;
    (
      (CurLen > MaxLen -> MaxLen1-MaxSeq1=CurLen-[LastItem|CurSeq] ; MaxLen1-MaxSeq1=MaxLen-MaxSeq),
      CurLen1=1,
      CurSeq1=[CurItem]
    )
   ),
   longest_sequence(L, CurLen1, CurSeq1, MaxLen1, MaxSeq1, Len, Seq).

Sample runs:

?- longest_sequence([1,6,11,12,13], Len, Seq).
Len = 3,
Seq = [11, 12, 13].

?- longest_sequence([1,6,11,12,13,2,3,4,5], Len, Seq).
Len = 4,
Seq = [2, 3, 4, 5].

?- longest_sequence([1,2], Len, Seq).
Len = 0,
Seq = [].

Answer 2

def longestSeq(n):
    seqMax = 0
    counter = 0
    i = 0
    while i+1 < len(n):
        curr = n[i+1]
        last = n[i]

        
        sub = curr - last

        if sub == 1:
            if counter == 0:
                counter = 2 
            else:
                counter += 1 
        else:
            if counter > 0:
               if seqMax < counter:
                   seqMax = counter
                   counter = 0 
        i += 1 

    if seqMax < counter:
        seqMax = counter

    if seqMax >= 3:
        return seqMax
    else:
        return 0