I got the basic idea pretty quick: root the tree at node 0, count reversals needed, then reroot by adjusting the count as you move the root along each edge.
First, compute the answer for an arbitrary root using a DFS/BFS that counts edges that need reversal. Then, use rerooting DP: when moving the root from parent to child, the answer changes by +1 if the edge is directed parent→child (since it becomes reversed) and -1 if directed child→parent. Propagate these deltas to get all answers in O(n).
Pro tip: Clarify that the tree is undirected in structure but edges have directions; the goal is to orient all edges toward the root. Mention that the rerooting technique generalizes to many tree problems and is a common Uber interview pattern.
Treat the given directed edges as an undirected tree with a direction attribute. For a fixed root, the cost is the number of edges whose direction points away from the root (i.e., need reversal to point toward the root).
Pick an arbitrary root (e.g., node 0) and run a DFS/BFS to count how many edges are directed away from it. This gives the answer for that root in O(n).
When moving the root from u to its neighbor v, the edge (u,v) flips its contribution: if it was directed u→v, it now needs reversal (+1); if v→u, it no longer needs reversal (-1). All other edges' contributions remain unchanged.
Perform a second DFS from the initial root, updating the cost using the transition rule. Store the cost for each node as you visit it.
After the second DFS, you have the minimum reversals for every node as root. Return them in order.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.