Longest distinct consecutive list in Python

Question

I have a list:

a = [2, 3, 5, 6, 6, 7, 10, 11, 13, 14, 15, 16, 16, 17, 18, 20, 21]

Is it possible to make a function that shows the longest list of distinct, consecutive elements?

Please, show how to do it

In this case the answer should be:

13, 14, 15, 16, 17, 18

Answer 1

There is probably a more pythonic way, but I can't think of it right now, so here a very basic solution:

def longestDistinctConsecutiveList ( lst ):
    lst = list( set( lst ) ) # get rid of duplicated elements
    lst.sort() # sort

    s, l = 0, 0
    for i in range( len( lst ) ):
        for j in range( i, len( lst ) ):
            if lst[j] - lst[i] == len( lst[i:j] ) > l:
                l = 1 + a[j] - a[i]
                s = i
    return lst[s:s+l]

Answer 2

Assuming your list is sorted:

>>> from itertools import groupby
>>> z = zip(a, a[1:])
>>> tmp = [list(j) for i, j in groupby(z, key=lambda x: (x[1] - x[0]) <= 1)]
>>> max(tmp, key=len)
[(13, 14), (14, 15), (15, 16), (16, 16), (16, 17), (17, 18)]
>>> list(range(_[0][0], _[-1][-1]+1))
[13, 14, 15, 16, 17, 18]

ETA: fixed last step;

Answer 3

The simplest thing to do seems to be to loop through the list once, building any sequences you can find and then print the longest one.

a = [2, 3, 5, 6, 6, 7, 10, 11, 13, 14, 15, 16, 16, 17, 18, 20, 21]

seqlist = [] # List of Sequences
seq = []     # Current Sequence
last = -1

for item in a:
   # Start a new sequence if the gap from the last item is too big
   if item - last > 1:
       seqlist.append(seq)
       seq = []

   # only add item to the sequence if it's not the same as the last
   if item != last:
        seq.append(item)

   last = item

# Print longest sequence found
print max(seqlist)