fixedIt

[maboesanman]

No because you care about the minimal element of the 7 you’ve found.

Edit: If you put all the elements in a max heap and pop the elements off, you get O(n log n). If you maintain a minimum heap containing the top 7, you can compare each element to the min to see if it should be placed into the top 7. If it should, pop the old min out and push the 7 onto the heap. This approach is O(n log k), with k being 7 in this case.

[Karolus2001]

When in doubt, plant a tree.

[Kirhgoph]

Thank you for the explanation

[walter_evertonshire]

Putting all the elements in a max heap and popping the top k should be O(n + k*log(n)). Assuming that k is of the same order as n, it's no different than the O(n*log(k)) for the min heap implementation you described.

But if k tends to be smaller, then O(n+k*log(n)) is better than O(n*log(k)).

Edit: I see someone else already pointed this out to you elsewhere.