The idea here is to pass throught the array v and checking for how long we can iterate starting from an index i while there are only 2 numbers in the on going iteration.
While we iterate starting from index i, we also keep tracking how many repeated numbers there are behind index i. So, consider a table t where:
t[i] = 1; // if i = 0 || v[i] != v[i-1]
t[i] = t[i - 1] + 1; // if i > 0 && v[i] == v[i-1]In this way, we can start from i and iterate until we are finding only 2 repeated numbers in the sequence. If the sequence ends on j (such that the number in position j was the number that broke the sequence), then the sequence length of 2 repeated numbers is:
j - i + t[i] - 1;Where (j - i) is the length of the sequence, plus the amount of repeated numbers before i that are the same as v[i], -1 because we are counting v[i] twice.
The solution is the greatest sequence length we found doing this calculations.
This code passes once on each number in the given array of size n, plus another array of size n for the counting. The code runs in O(n) and requires O(n) space.