> I would also like to know what the first student asked: Hey, looking back on lecture material, for kruskals algorithm the complexity is O((n+m)log(n)), where n = vertices and m = edges. > > It states that all the MST union operations result in complexity of nlog(n), but I can't find where this is explained, as appearance of log(n) always suggests some halving shenanigans that to my knowledge aren't expanded upon in the lectures.//// > this is what I found in the lecture but couldn't find anything else on getting closer to nlogn: > > Partition-Based > Implementation >  Partition-based version of > Kruskal’s Algorithm >  Cluster merges as unions >  Cluster locations as finds >  Running time O((n + m) log n) >  Priority Queue operations: O(m log n) >  Union-Find operations: O(n log n) > > Thank you