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