Use Union-Find (Disjoint Set Union) to dynamically maintain connected components as land is added. For each new land position, increment the island count, then union with any adjacent existing land, decrementing the count for each successful union. This yields near O(k α(n)) time for k additions.
Pro tip: Mention that Union-Find with union by rank and path compression is optimal here, and discuss how you'd handle edge cases like duplicate positions or out-of-bounds coordinates. Also, note that a naive BFS/DFS per addition would be O(k * m*n) and is inefficient.
Confirm that the grid starts empty and positions are added one by one, and that you need to return the number of islands after each addition. Ask about grid size limits and whether duplicate positions can occur.
Select Union-Find (Disjoint Set Union) with path compression and union by rank/size to efficiently track connected components. Explain why it's better than BFS/DFS for incremental updates.
Initialize a parent array for all cells (or a hash map for sparse grids) and a count of islands. For each added land, mark it as land, increment island count, then check its four neighbors: if a neighbor is land and not already connected, union them and decrement island count.
If a position is already land, skip processing and return the current count. Ensure boundary checks for neighbors. Consider using a 2D-to-1D index mapping for efficiency.
State time complexity: O(k α(m*n)) where k is number of additions, and space O(m*n). Walk through a small example to verify correctness, and discuss potential optimizations like early termination if no neighbors.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.