Took me a minute to even parse what 'closed' meant here.
Use a graph traversal algorithm (DFS or BFS) to identify all land cells connected to the grid's edges and mark them as non-closed. Then, iterate through the grid to count the remaining unvisited land components, which are the closed islands.
Pro tip: Clarify edge cases upfront, such as empty grid or all water, and discuss time/space complexity. Mention that you can optimize by modifying the grid in-place to avoid extra space.
Confirm that closed islands are land regions not touching any edge. Discuss edge cases: empty grid, all water, all land, single row/column.
Decide between DFS (recursive or iterative) and BFS. Consider recursion depth for large grids; iterative DFS or BFS may be safer.
Traverse all land cells on the four edges and perform DFS/BFS to mark all connected land as visited (e.g., change 1 to 0 or use a visited set).
Iterate through the grid; for each unvisited land cell, increment count and traverse to mark the entire island as visited.
State time complexity O(m*n) and space complexity O(m*n) for visited set or O(1) if modifying grid in-place. Discuss trade-offs.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.