Interview Question for Senior Software Developer at Microsoft:

set seen;

for( element in array):
  if( seen.contains( element ) ):
    array.erase( element )

Runtime:
* For a tree based set, O( n * log n )
* For a hash table based set, O( n ) assuming no collisions and constant hashing

Improvements: If array.erase isn't efficient (it's an array, it won't be) then keep a collection of pointers or indices into the array to erase in a more efficient manner than 1 at a time.