Find missing numbers in a contiguous sequence and express them as a string of ranges.
For example,
124569 should output 3,7-8
String GetMissingNumbers(List<int> numbers)
{
string ret = string.empty;
numbers.sort();
if ( numbers.Count() <= 1) return ret;
for (int i = 1; i < numbers.Count(); i++)
{
if (numbers[i] - numbers[i-1] == 2)
   ret += (numbers[i-1] + 1).toString() + ",";
if (numbers[i] - numbers[i-1] > 2)
  ret += (numbers[i-1] + 1).toString() + "-" + (numbers[i]-1).toString() + ",";
}
return ret;
}
The above is O(n)
We can make it O(log n) with below:

void GetMissing(List<int> numbers, ref List<string> result, int start, int end)
{
if (start < end)
{
mid = (start + end)/2;
GetMissing(numbers, ref result. start, mid);
GetMissing(numbers, ref result, mid+1, end);
Merge(numbers, result, start, mid, end);
}
}
void Merge(List<int> numbers, ref List<string> result, int start, int mid, int end)
{
// constant time operation
if (numbers[mid] - numbers[start] == mid - start &&
     numbers[end] - numbers[mid+1] == end-(mid+1) &&
     numbers[mid+1] - numbers[mid] <= 1) return; // assume sequence is already sorted
if (numbers[mid+1] - numbers[mid] == 2)
   result.Add (numbers[mid] + 1).toString();
if (numbers[mid+1] - numbers[mid] > 2)
  result.Add(numbers[mid] + 1).toString() + "-" + (numbers[mid+1]-1).toString() ;
}

result.Join(",")