The brute force instinct here is a trap you have to talk yourself out of.
Model the problem as a graph where each stone is an edge connecting its row and column nodes. The maximum number of stones that can be removed equals the total number of stones minus the number of connected components in this graph, because each component must leave at least one stone. Use union-find or DFS to count components efficiently.
Pro tip: Clarify that the answer is n minus the number of connected components, and mention that this is a classic problem (e.g., LeetCode 947). Showing awareness of the graph transformation and its proof demonstrates strong algorithmic maturity.
Restate the problem: stones can be removed if they share a row or column with another remaining stone. Ask about input size and whether coordinates are bounded.
Treat each row and column as nodes, and each stone as an edge connecting its row and column. This transforms the removal condition into connectivity of edges.
In each connected component, you can remove all but one stone. Thus, maximum removable stones = total stones - number of connected components.
Use union-find (disjoint set union) to efficiently count connected components, or DFS/BFS on the graph. Union-find is often simpler and faster for this problem.
Write code to build the graph, union row and column nodes for each stone, then count unique components. Test with edge cases like no stones, all stones isolated, or all connected.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.