Disclaimer: Since the solution is based on hashtables, the running times are expected, not worst case.
This O(n) solution depends on the integers being unique. If they are not unique, make a hashset with O(1) insertion and membership lookup, and just skip the numbers already encountered, as you go through the list.
Make a O(1) lookup/insert hashmap where the values are the beginnings of ranges, and the keys are the numbers that fit at the end of those ranges. For a value v and a key k, this means that the range starting from v and ending with k-1 inclusive is located at key k.
Go through the list of numbers. For each number n check if the map has a value v at key n. This corresponds to there being a range starting from v that would allow n at the end. If there is, move v to key n+1 and delete the entry at key n. If there isn't any range, insert n at key n+1.
Since the numbers are unique, none of the ranges overlap in the end, but there may be some contiguous ones. Run through the key/value pairs of the map. For each key k and value v, if the map has a value v1 at key k1 = v, then it means that there is a range from v1 to k-1. Insert v1 at k, and delete the entry k1/v1.
Go through the k/v entries of the map to find the largest range [v,k-1] of size k-v, using a running maximum.
For your example:
setup:
l = [1,3,5,7,4,6,10]
m = {}
iteration:
process 1 : m = {2->1}
process 3 : m = {2->1, 4->3}
process 5 : m = {2->1, 4->3, 6->5}
process 7 : m = {2->1, 4->3, 6->5, 8->7}
process 4 : m = {2->1, 5->3, 6->5, 8->7}
process 6 : m = {2->1, 5->3, 7->5, 8->7}
process 10 : m = {2->1, 5->3, 7->5, 8->7, 11->10}
concatenation of contiguous ranges:
initial: m = {2->1, 5->3, 7->5, 8->7, 11->10}
first concatenation: m = {2->1, 7->3, 8->7, 11->10}, k=7, v=5, k1=5, v1=3
second concatenation: m = {2->1, 8->3, 11->10}, k=8, v=7, k1=7, v1=3
result:
largest range : [3,7] of size 5
