Redundant Connection
Recognize the pattern
Brute force idea
If you approach Redundant Connection in the most literal way possible, you get this: For each edge, remove it and check if the remaining graph is still a tree. Rebuilds the graph for every edge. It is a fair place to begin because it matches the surface of the question, yet it does not capture the deeper structure that makes the problem simpler.
Better approach
The real unlock in Redundant Connection comes when you notice this: Union-Find: process edges one by one. If union(u,v) finds they're already connected, this edge creates a cycle — it's the redundant edge. Instead of recomputing the world every time, you preserve just enough context to let the next decision become obvious.
Key invariant
The compass for Redundant Connection is this: In a tree with n nodes, there are exactly n-1 edges. The graph has n edges, so exactly one is redundant. That edge connects two nodes already in the same component — detected by Union-Find. As long as that statement keeps holding, you can trust the steps built on top of it.
Watch out for
A common way to get lost in Redundant Connection is this: Returning the first cycle edge found — the problem asks for the LAST such edge in the input order. Process all edges; the last one that causes a union to fail is the answer. Most mistakes here are not about syntax; they come from losing track of what your state, pointer, or structure is supposed to mean.