Look at this solution:


Is the lower bound $m\log n$ because we are only looking at the lower bound for union by rank only? If we make $n$ MAKE-SET operations, then there would be $\log n$ UNION operations, and then $m - 2n + 1$ FIND-SET operations. The lower bound seems larger to me but what am I missing?

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    $\begingroup$ Welcome to the site! So that your question is accessible to search engines and the visually impaired, it would be very helpful if you could transcribe the image to text. Also, you need to give credit to the source of that answer. $\endgroup$ Commented Apr 14, 2017 at 20:13

1 Answer 1


You are asking two questions.

Is the lower bound only for this specific implementation?

Yes. If you also use path compression, the running time will be $o(m\log n)$.

The lower bound seems to large to me.

Your mistake is that you assume that there are only $\log n$ UNION operations, whereas there are $2^{\lfloor \log_2 n \rfloor}-1 = \Theta(n)$, as the answer clearly .


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