Find the biggest interval that has all its members in list in O(n) [duplicate]

Answer 1

That would be linear considering dictionaries built with average O(1) hash tables.

L = [1,3,5,7,4,6,10]

a_to_b = {}
b_to_a = {}

for i in L:
    if i+1 in a_to_b and i-1 in b_to_a:
        new_a = b_to_a[i-1]
        new_b = a_to_b[i+1]
        a_to_b[new_a] = new_b
        b_to_a[new_b] = new_a
        continue
    if i+1 in a_to_b:
        a_to_b[i] = a_to_b[i+1]
        b_to_a[a_to_b[i]] = i
    if i-1 in b_to_a:
        b_to_a[i] = b_to_a[i-1]
        a_to_b[b_to_a[i]] = i
    if not (i+1 in a_to_b or i-1 in b_to_a):
        a_to_b[i] = i
        b_to_a[i] = i

max_a_b = max_a = max_b = 0
for a,b in a_to_b.iteritems():
    if b-a > max_a_b:
        max_a = a
        max_b = b
        max_a_b = b-a

print max_a, max_b