How to see if the list contains consecutive numbers

Question

I want to test if a list contains consecutive integers and no repetition of numbers. For example, if I have

l = [1, 3, 5, 2, 4, 6]

It should re

Answer 1

With sorting

In Python 3, I use this simple solution:

def check(lst):
    lst = sorted(lst)
    if lst:
        return lst == list(range(lst[0], lst[-1] + 1))
    else:
        return True

Note that, after sorting the list, its minimum and maximum come for free as the first (lst[0]) and the last (lst[-1]) elements. I'm returning True in case the argument is empty, but this decision is arbitrary. Choose whatever fits best your use case.

In this solution, we first sort the argument and then compare it with another list that we know that is consecutive and has no repetitions.

Without sorting

In one of the answers, the OP commented asking if it would be possible to do the same without sorting the list. This is interesting, and this is my solution:

def check(lst):
    if lst:
        r = range(min(lst), max(lst) + 1) # *r* is our reference
        return (
            len(lst) == len(r)
            and all(map(lst.__contains__, r))
            # alternative: all(x in lst for x in r)
            # test if every element of the reference *r* is in *lst*
        )
    else:
        return True

In this solution, we build a reference range r that is a consecutive (and thus non-repeating) sequence of ints. With this, our test is simple: first we check that lst has the correct number of elements (not more, which would indicate repetitions, nor less, which indicates gaps) by comparing it with the reference. Then we check that every element in our reference is also in lst (this is what all(map(lst.__contains__, r)) is doing: it iterates over r and tests if all of its elements are in lts).