I started by thinking about it left-to-right and almost got the dependency direction backwards.
Model each block as a node in a dependency graph where an edge points from a block to the block immediately to its right in the same row, since a block depends on that right neighbor being removed first. To find a removable block, identify a node with no outgoing edges (i.e., the rightmost present block in any row). For the full removal sequence, repeatedly remove such nodes and update dependencies, which is essentially a topological sort on the reversed graph.
Pro tip: Clarify that the dependency graph is a forest of disjoint paths (one per row), so the problem reduces to maintaining the rightmost present block per row; this simplifies the solution and avoids overcomplicating with general graph algorithms.
Create a graph where each block is a node, and add a directed edge from block (r, c) to block (r, c+1) if both exist, representing that (r, c) depends on (r, c+1) being removed first.
A block is removable if it has no outgoing edges (i.e., no block to its right in the same row). Initially, these are the rightmost blocks in each row.
To simulate the full sequence, repeatedly pick a removable block, remove it, and update the graph: the block immediately to its left (if any) may become removable. Use a queue or stack to manage candidates.
After removing a block, check its left neighbor; if it exists and has no right neighbor, add it to the set of removable blocks. Continue until no blocks remain.
Each block is removed once, and each removal involves O(1) updates, so the total time is O(N) where N is the number of blocks, assuming we can access neighbors efficiently.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.