Question

For instance, if I have a list

[1,4,2,3,5,4,5,6,7,8,1,3,4,5,9,10,11]

This algorithm should return [1,2,3,4,5,6,7,8,9,10,11].

To clarify, the longest list should run forwards. I was wondering what is an algorithmically efficient way to do this (preferably not O(n^2))?

Also, I’m open to a solution not in python since the algorithm is what matters.

Thank you.

Answer 1

Here is a simple one-pass O(n) solution:

s = [1,4,2,3,5,4,5,6,7,8,1,3,4,5,9,10,11,42]
maxrun = -1
rl = {}
for x in s:
    run = rl[x] = rl.get(x-1, 0) + 1
    print x-run+1, 'to', x
    if run > maxrun:
        maxend, maxrun = x, run
print range(maxend-maxrun+1, maxend+1)

The logic may be a little more self-evident if you think in terms of ranges instead of individual variables for the endpoint and run length:

rl = {}
best_range = xrange(0)
for x in s:
    run = rl[x] = rl.get(x-1, 0) + 1
    r = xrange(x-run+1, x+1)
    if len(r) > len(best_range):
        best_range = r
print list(best_range)