C++ std::map or std::set - efficiently insert duplicates

Question

I have a bunch of data full of duplicates and I want to eliminate the duplicates. You know, e.g. [1, 1, 3, 5, 5, 5, 7] becomes [1, 3, 5, 7].

It looks like I can use either std::map or std::set to handle this. However I'm not sure whether it's faster to (a) simply insert all the values into the container, or (b) check whether they already exist in the container and only insert if they don't - are inserts very efficient? Even if there's a better way... can you suggest a fast way to do this?

Another question - if the data I'm storing in them isn't as trivial as integers, and instead is a custom class, how does the std::map manage to properly store (hash?) the data for fast access via operator[]?

Answer 1

std::map doesn't use hashing. std::unordered_map does, but that's C++11. std::map and std::set both use a comparator that you provide. The class templates have defaults for this comparator, which boils down to an operator< comparison, but you can provide your own.

If you don't need both a key and a value to be stored (looks like you don't) you should just use a std::set, as that's more appropriate.

The Standard doesn't say what data structures maps and sets use under the hood, only that certian actions have certain time complexities. In reality, most implementations I'm aware of use a tree.

It makes no difference time-complexity-wise if you use operator[] or insert, but I would use insert or operator[] before I did a search followed by an insert if the item isn't found. The later would imply two seperate searches to insert an item in to the set.

Answer 2

An insert() on any of the associated containers does a find() to see if the object exists and then inserts the object. Simply inserting the elements into an std::set<T> should get rid of the duplicates reasonably efficiently.

Depending on the size of your set and the ratio of duplicates to unique values, it may be faster to put the objects into std::vector<T>, std::sort() then, and then use std::unique() together with std::vector<T>::erase() to get rid of the duplicates.

Answer 3

How many times should you do it?

If insert is usual:

//*/
std::set<int> store;
/*/
// for hash:
std::unordered_set<int> store;
//*/
int number;

if ( store.insert(number).second )
{
  // was not in store
}

If you fill once:

std::vector<int> store;
int number;

store.push_back(number);
std::sort(store.begin(),store.end());
store.erase(std::unique(store.begin(),store.end()),store.end() );

// elements are unique

Answer 4

Assuming the common implementation strategy for std::map and std::set, i.e. balanced binary search trees, both insertion and lookup have to do a tree traversal to find the spot where the key should be. So failed lookup followed by insertion would be roughly twice as slow as just inserting.

how does the std::map manage to properly store (hash?) the data for fast access via operator[]?

By means of a comparison function that you specify (or std::less, which works if you overload operator< on your custom type). In any case, std::map and std::set are not hash tables.

Answer 5

std::set and std::map are both implemented as red black tree as far as I know. And probably using only insertion would be faster (then both because you would be doubling the lookup time).

Also map and set use operator <. As long as your class has defined operator < It would be able to use them as keys.