I went straight to DFS because it's the most natural fit and I can code it fast.
Use a graph traversal algorithm (DFS or BFS) to explore each unvisited land cell, marking all connected land cells as visited to count one island. Iterate through the grid, and for each unvisited '1', increment the island count and perform a traversal to mark its entire connected component.
Pro tip: Mention that you can optimize space by modifying the grid in-place (e.g., changing '1' to '0') to mark visited cells, avoiding a separate visited matrix. Also, discuss handling edge cases like empty grid or all water.
Confirm the definition of an island: a group of connected '1's using 4-directional adjacency. Ask about grid size limits and whether modifying the input is allowed.
Decide between DFS (recursive or iterative) and BFS. Consider recursion depth for large grids; iterative DFS or BFS may be safer.
Loop through each cell. When encountering an unvisited '1', increment the island count and launch a traversal to mark all connected land cells as visited.
During traversal, mark cells as visited by changing '1' to '0' or using a separate visited set. Ensure you don't revisit cells.
State time complexity O(m*n) since each cell is visited once, and space complexity O(m*n) in worst case for recursion stack or queue.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
This caught me a little flat-footed because I'd already mentally closed out the problem.
Start by acknowledging that the core traversal remains the same, but instead of incrementing a counter, you collect the size of each island. Then sort the collected sizes in descending order and return the sorted list. Emphasize that the time complexity is still O(R*C) for traversal plus O(K log K) for sorting, where K is the number of islands.
Pro tip: Mention that you can avoid sorting by using a max-heap or counting sort if island sizes are bounded, but for general cases, sorting is fine. Also, clarify that you should modify the existing function signature to return a list of integers rather than an integer.
Confirm that the input is a 2D grid of 0s and 1s, and that islands are connected 4-directionally. Ask if the output should be a list of integers sorted descending, and if there are any constraints on grid size or number of islands.
Decide between DFS, BFS, or Union-Find. For simplicity and clarity, DFS or BFS is usually preferred. Explain that you will traverse each cell, and when you find an unvisited '1', you explore the entire island to compute its size.
During traversal, maintain a count of cells in the current island. Instead of incrementing a global counter, append the size to a list after the island is fully explored. Ensure you mark visited cells to avoid double-counting.
After processing all cells, sort the list of island sizes in descending order. You can use built-in sorting (e.g., Arrays.sort with reverse order or Collections.sort).
Return the sorted list. State that time complexity is O(R*C + K log K) where K is number of islands, and space complexity is O(R*C) for visited set or recursion stack.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Start by clarifying the problem and its constraints, then explain each approach (DFS, BFS, Union-Find) in terms of how it works and its complexity. Compare them based on the problem's characteristics, such as graph size, density, and whether dynamic updates are needed, and conclude with a recommendation for when to use each.
Pro tip: Emphasize that the choice of algorithm depends on the specific problem constraints and requirements, and mention that Union-Find is particularly useful for dynamic connectivity problems. Also, note that BFS is optimal for unweighted shortest paths, while DFS is simpler for traversal and backtracking.
Ask questions to understand the problem: Is it a graph traversal, connectivity, or shortest path problem? What are the input size, graph density, and whether the graph is static or dynamic?
Describe DFS: it explores as far as possible along each branch before backtracking. Mention its time complexity O(V+E) and space complexity O(V) due to recursion stack or explicit stack.
Describe BFS: it explores level by level using a queue. Mention its time complexity O(V+E) and space complexity O(V) for the queue, and that it finds shortest paths in unweighted graphs.
Describe Union-Find (Disjoint Set Union): it maintains disjoint sets and supports union and find operations. Mention its near-constant time complexity with path compression and union by rank, and space complexity O(V).
Compare the approaches: DFS/BFS are for traversal and pathfinding, Union-Find for connectivity and dynamic updates. Discuss trade-offs: BFS for shortest path, DFS for memory efficiency in deep graphs, Union-Find for incremental connectivity.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.