I was studying Union Find, and according to Wikipedia, there are 2 types of union: union by rank and Union by size. My question is, what is the runtime difference between the two (if any)?

Intuitively, it feels like Union by size would always be better, since each time we merge, we are increasing the rank of every node in one tree by 1, and to minimize overall runtime, we want to increase that one tree to be the one with smaller number of nodes, even though the final tree height might be greater.

But does this make a big difference in runtime?


1 Answer 1


If you combine union by rank or union by size with e.g. path compression the amortized complexity is the same [$O(m\alpha(m,n))$]. But notice that Wikipedia uses union by rank in order to prove the upper bound $O(m\log^*(n))$ because for proof purposes the union by rank algorithm is simpler to handle. On the other hand if you are implementing such a data structure you usually want to allow the user to access the sizes of the sets in the structure and thus use union by size method in order to keep track of the sizes and don't waste space for an extra array.

Its worth mentioning if we don't combine them with e.g. path compression both work in amortized $O(log(n))$.

This is a good lecture on this topic. It contains references to all the papers proving bounds for the problem. It also contains explanation of what $O(m\log^*(n))$ and $O(m\alpha(m,n))$ mean.

  • 1
    $\begingroup$ Thank you for the explanation. Just out of curiosity, is there such thing as union by rank and size? Each Node could be attributed a value based on its distance from the tree, and the tree keeps track of all the total Node values. $\endgroup$
    – timg
    Commented Jul 11, 2020 at 17:07
  • $\begingroup$ I haven't heard of it but you can of course use union by rank and still store the sizes of the sets. Generally the performance stays the same as long as you add the set with a lower depth to the one with higher depth and do path compression; so you can get creative. $\endgroup$
    – plshelp
    Commented Jul 11, 2020 at 19:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.