c++ Algorithm to Compare various length vectors and isolate "unique", sort of

Loop through the vectors and for each vector, map the count of unique values occurring in it. unordered_map<int, int> would suffice for this, let's call it M.

Also maintain a set<unordered_map<int, int>>, say S, ordered by the size of unordered_map<int, int> in decreasing order.

Now we will have to compare contents of M with the contents of unordered_maps in S. Let's call M', the current unordered_map in S being compared with M. M will be a subset of M' only when the count of all the elements in M is less than or equal to the count of their respective elements in M'. If that's the case then it's a duplicate and we'll not insert. For any other case, we'll insert. Also notice that if the size of M is greater than the size of M', M can't be a subset of M'. That means we can insert M in S. This can be used as a pre-condition to speed things up. Maintain the indices of vectors which weren't inserted in S, these are the duplicates and have to be deleted from vector_list in the end.

Time Complexity: O(N*M) + O(N^2*D) + O(N*log(N)) = O(N^2*D) where N is the number of vectors in vector_list, M is the average size of the vectors in vector_list and D is the average size of unordered_map's in S. This is for the worst case when there aren't any duplicates. For average case, when there are duplicates, the second complexity will come down.

Edit: The above procedure will create a problem. To fix that, we'll need to make unordered_maps of all vectors, store them in a vector V, and sort that vector in decreasing order of the size of unordered_map. Then, we'll start from the biggest in this vector and apply the above procedure on it. This is necessary because, a subset, say M1 of a set M2, can be inserted into S before M2 if the respective vector of M1 comes before the respective vector of M2 in vector_list. So now we don't really need S, we can compare them within V itself. Complexity won't change.

Edit 2: The same problem will occur again if sizes of two unordered_maps are the same in V when sorting V. To fix that, we'll need to keep the contents of unordered_maps in some order too. So just replace unordered_map with map and in the comparator function, if the size of two maps is the same, compare element by element and whenever the keys are not the same for the very first time or are same but the M[key] is not the same, put the bigger element before the other in V.

Edit 3: New Time Complexity: O(N*M*log(D)) + O(N*D*log(N)) + O(N^2*D*log(D)) = O(N^2*D*log(D)). Also you might want to pair the maps with the index of the respective vectors in vector_list so as to know which vector you must delete from vector_list when you find a duplicate in V.

IMPORTANT: In sorted V, we must start checking from the end just to be safe (in case we choose to delete a duplicate from vector_list as well as V whenever we encounter it). So for the last map in V compare it with the rest of the maps before it to check if it is a duplicate.

Example:

vector_list = {
  {1, 2, 3},
  {2, 3, 1},
  {3, 2, 2},
  {4, 2, 3, 2, 5},
  {1, 2, 3, 4, 6, 2},
  {2, 3, 4, 5, 6},
  {1, 5}
}

Creating maps of respective vectors:

V = {
  {1->1, 2->1, 3->1},
  {1->1, 2->1, 3->1},
  {2->2, 3->1},
  {2->2, 3->1, 4->1, 5->1},
  {1->1, 2->2, 3->1, 4->1, 6->1},
  {2->1, 3->1, 4->1, 5->1, 6->1},
  {1->1, 5->1}
}

After sorting:

V = {
  {1->1, 2->2, 3->1, 4->1, 6->1},
  {2->1, 3->1, 4->1, 5->1, 6->1},
  {2->2, 3->1, 4->1, 5->1},
  {1->1, 2->1, 3->1},
  {1->1, 2->1, 3->1},
  {1->1, 5->1},
  {2->2, 3->1}
}

After deleting duplicates:

V = {
  {1->1, 2->2, 3->1, 4->1, 6->1},
  {2->1, 3->1, 4->1, 5->1, 6->1},
  {2->2, 3->1, 4->1, 5->1},
  {1->1, 5->1}
}

Edit 4: I tried coding it up. Running it a 1000 times on a list of 100 vectors, the size of each vector being in range [1-250], the range of the elements of vector being [0-50] and assuming the input is available for all the 1000 times, it takes around 2 minutes on my machine. It goes without saying that there is room for improvement in my code (and my machine).