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

Posted by Hooked on Stack Overflow See other posts from Stack Overflow or by Hooked
Published on 2010-05-04T21:41:32Z Indexed on 2010/05/04 21:48 UTC
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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?

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