Determining if an unordered vector<T> has all unique elements

Question

Profiling my cpu-bound code has suggested I that spend a long time checking to see if a container contains completely unique elements. Assuming that I have some large container of unsorted elements (with < and = defined), I have two ideas on how this might be done:

The first using a set:

template <class T>
bool is_unique(vector<T> X) {
  set<T> Y(X.begin(), X.end());
  return X.size() == Y.size();
}

The second looping over the elements:

template <class T>
bool is_unique2(vector<T> X) {
  typename vector<T>::iterator i,j;
  for(i=X.begin();i!=X.end();++i) {
    for(j=i+1;j!=X.end();++j) {
      if(*i == *j) return 0;
    }
  }
  return 1;
}

I've tested them the best I can, and from what I can gather from reading the documentation about STL, the answer is (as usual), it depends. I think that in the first case, if all the elements are unique it is very quick, but if there is a large degeneracy the operation seems to take O(N^2) time. For the nested iterator approach the opposite seems to be true, it is lighting fast if X[0]==X[1] but takes (understandably) O(N^2) time if all the elements are unique.

Is there a better way to do this, perhaps a STL algorithm built for this very purpose? If not, are there any suggestions eek out a bit more efficiency?

Answer 1

The standard library has std::unique, but that would require you to make a copy of the entire container (note that in both of your examples you make a copy of the entire vector as well, since you unnecessarily pass the vector by value).

template <typename T>
bool is_unique(std::vector<T> vec)
{
    std::sort(vec.begin(), vec.end());
    return std::unique(vec.begin(), vec.end()) == vec.end();
}

Whether this would be faster than using a std::set would, as you know, depend :-).

Answer 2

Is it infeasible to just use a container that provides this "guarantee" from the get-go? Would it be useful to flag a duplicate at the time of insertion rather than at some point in the future? When I've wanted to do something like this, that's the direction I've gone; just using the set as the "primary" container, and maybe building a parallel vector if I needed to maintain the original order, but of course that makes some assumptions about memory and CPU availability...

Answer 3

For one thing you could combine the advantages of both: stop building the set, if you have already discovered a duplicate:

template <class T>
bool is_unique(const std::vector<T>& vec)
{
    std::set<T> test;
    for (typename std::vector<T>::const_iterator it = vec.begin(); it != vec.end(); ++it) {
        if (!test.insert(*it).second) {
            return false;
        }
    }
    return true;
}

BTW, Potatoswatter makes a good point that in the generic case you might want to avoid copying T, in which case you might use a std::set<const T*, dereference_less> instead.

---

You could of course potentially do much better if it wasn't generic. E.g if you had a vector of integers of known range, you could just mark in an array (or even bitset) if an element exists.

Answer 4

You can use std::unique, but it requires the range to be sorted first:

template <class T>
bool is_unique(vector<T> X) {
  std::sort(X.begin(), X.end());
  return std::unique(X.begin(), X.end()) == X.end();
}

std::unique modifies the sequence and returns an iterator to the end of the unique set, so if that's still the end of the vector then it must be unique.

This runs in nlog(n); the same as your set example. I don't think you can theoretically guarantee to do it faster, although using a C++0x std::unordered_set instead of std::set would do it in expected linear time - but that requires that your elements be hashable as well as having operator == defined, which might not be so easy.

Also, if you're not modifying the vector in your examples, you'd improve performance by passing it by const reference, so you don't make an unnecessary copy of it.

Answer 5

Your first example should be O(N log N) as set takes log N time for each insertion. I don't think a faster O is possible.

The second example is obviously O(N^2). The coefficient and memory usage are low, so it might be faster (or even the fastest) in some cases.

It depends what T is, but for generic performance, I'd recommend sorting a vector of pointers to the objects.

template< class T >
bool dereference_less( T const *l, T const *r )
 { return *l < *r; } 

template <class T>
bool is_unique(vector<T> const &x) {
    vector< T const * > vp;
    vp.reserve( x.size() );
    for ( size_t i = 0; i < x.size(); ++ i ) vp.push_back( &x[i] );
    sort( vp.begin(), vp.end(), ptr_fun( &dereference_less<T> ) ); // O(N log N)
    return adjacent_find( vp.begin(), vp.end(),
           not2( ptr_fun( &dereference_less<T> ) ) ) // "opposite functor"
        == vp.end(); // if no adjacent pair (vp_n,vp_n+1) has *vp_n < *vp_n+1
}

or in STL style,

template <class I>
bool is_unique(I first, I last) {
    typedef typename iterator_traits<I>::value_type T;
    …

---

And if you can reorder the original vector, of course,

template <class T>
bool is_unique(vector<T> &x) {
    sort( x.begin(), x.end() ); // O(N log N)
    return adjacent_find( x.begin(), x.end() ) == x.end();
}

Answer 6

Well, your first one should only take N log(N), so it's clearly the better worse case scenario for this application.

However, you should be able to get a better best case if you check as you add things to the set:

template <class T>
bool is_unique3(vector<T> X) {
  set<T> Y;
  typename vector<T>::const_iterator i;
  for(i=X.begin(); i!=X.end(); ++i) {
    if (Y.find(*i) != Y.end()) {
      return false;
    }
    Y.insert(*i);
  }
  return true;
}

This should have O(1) best case, O(N log(N)) worst case, and average case depends on the distribution of the inputs.

Answer 7

You must sort the vector if you want to quickly determine if it has only unique elements. Otherwise the best you can do is O(n^2) runtime or O(n log n) runtime with O(n) space. I think it's best to write a function that assumes the input is sorted.

template<class Fwd>
bool is_unique(In first, In last)
{
    return asjacent_find(first, last) == last;
}

then have the client sort the vector, or a make a sorted copy of the vector. This will open a door for dynamic programming. That is, if the client sorted the vector in the past then they have the option to keep and refer to that sorted vector so they can repeat this operation for O(n) runtime.

Answer 8

If the type T You store in Your vector is large and copying it is costly, consider creating a vector of pointers or iterators to Your vector elements. Sort it based on the element pointed to and then check for uniqueness.

You can also use the std::set for that. The template looks like this

template <class Key,class Traits=less<Key>,class Allocator=allocator<Key> > class set

I think You can provide appropriate Traits parameter and insert raw pointers for speed or implement a simple wrapper class for pointers with < operator.

Don't use the constructor for inserting into the set. Use insert method. The method (one of overloads) has a signature

pair <iterator, bool> insert(const value_type& _Val);

By checking the result (second member) You can often detect the duplicate much quicker, than if You inserted all elements.

Answer 9

Using the current C++ standard containers, you have a good solution in your first example. But if you can use a hash container, you might be able to do better, as the hash set will be n*O(1) instead of n*O(log n) for a standard set. Of course everything will depend on the size of n and your particular library implementation.

Answer 10

In the (very) special case of sorting discrete values with a known, not too big, maximum value N.
You should be able to start a bucket sort and simply check that the number of values in each bucket is below 2.

bool is_unique(const vector<int>& X, int N)
{
  vector<int> buckets(N,0);
  typename vector<int>::const_iterator i;
  for(i = X.begin(); i != X.end(); ++i)
    if(++buckets[*i] > 1)
      return false;
  return true;
}

The complexity of this would be O(n).

Answer 11

If I may add my own 2 cents.

First of all, as @Potatoswatter remarked, unless your elements are cheap to copy (built-in/small PODs) you'll want to use pointers to the original elements rather than copying them.

Second, there are 2 strategies available.

1. Simply ensure there is no duplicate inserted in the first place. This means, of course, controlling the insertion, which is generally achieved by creating a dedicated class (with the vector as attribute).
2. Whenever the property is needed, check for duplicates

I must admit I would lean toward the first. Encapsulation, clear separation of responsibilities and all that.

Anyway, there are a number of ways depending on the requirements. The first question is:

• do we have to let the elements in the vector in a particular order or can we "mess" with them ?

If we can mess with them, I would suggest keeping the vector sorted: Loki::AssocVector should get you started. If not, then we need to keep an index on the structure to ensure this property... wait a minute: Boost.MultiIndex to the rescue ?

Thirdly: as you remarked yourself a simple linear search doubled yield a O(N²) complexity in average which is no good.

If < is already defined, then sorting is obvious, with its O(N log N) complexity. It might also be worth it to make T Hashable, because a std::tr1::hash_set could yield a better time (I know, you need a RandomAccessIterator, but if T is Hashable then it's easy to have T* Hashable to ;) )

But in the end the real issue here is that our advises are necessary generic because we lack data.

• What is T, do you intend the algorithm to be generic ?
• What is the number of elements ? 10, 100, 10.000, 1.000.000 ? Because asymptotic complexity is kind of moot when dealing with a few hundreds....
• And of course: can you ensure unicity at insertion time ? Can you modify the vector itself ?