Q: Given a circular linked list. Find the longest sequence of numbers, where a sequence starts with any number, for example 3, and ends when you see that number again, another 3,
Imagine the circular linked list
3 8 9 7 2 1 3 4 6 [3] same as the first element i.e 3
The longest sequence would be 3 8 9 7 2 1, the other candidate being 3, 4, 6
Finding for instance, starting at 8 and getting to the same 8 wouldn't count as a valid sequence.
Courtesy : CareerCup
A:
We build a hashtable of the last encountered index of the elements in the list. If the element is seen again, we calculate the difference in the new and the previous index and as long as it is less than the total count of elements in the list and more than the previously encountered difference, we update the difference and note down the start. After all the updates are made, we have a start index and the highest difference between two occurrences of same element.

void PrintLongestSequence(List<int> numbers)
{
int start = 0;
int length = 0;
var h = new Hashtable<int,int>();
for (int i = 0; i < numbers.Count; i++)
{
    if (h.Contains(numbers[i]))
    {
         var old = h[numbers[i]];
         var new = i;
         var difference = new  - old;
         if (difference > length)
         {
               start = old;
               length = difference;
         }
     }
     h[numbers[i]] = i;
}
Console.WriteLine();
for (int i = start; i < start+length; i++)
    Console.Write("{0} ", numbers[start]);
Console.WriteLine();
}