I knew the general approach: label each island with a unique id using union-find or BFS, store each island's size, then for every 0-cell check its distinct neighboring island ids and sum their sizes plus one.
First, identify all existing islands and label each cell with a unique island ID, storing the size of each island. Then, for each water cell, compute the potential island size by summing the sizes of distinct neighboring islands plus one, and track the maximum. This achieves O(n^2) time by processing each cell a constant number of times.
Pro tip: Explicitly state the time and space complexity and discuss trade-offs, such as using BFS vs. DFS for labeling and handling edge cases like all water or all land. This shows you consider efficiency and robustness, which is crucial for data science roles.
Traverse the grid and use BFS/DFS to assign a unique ID to each island, recording the size of each island in a hash map or array.
For each water cell (0), examine its four neighbors, collect the distinct island IDs, and sum their sizes plus one to get the potential island size if flipped.
Keep a running maximum of the potential island sizes and also consider the case where no flip is made (i.e., the largest existing island).
If the grid has no water, return the size of the largest island; if all water, return 1 (since flipping one cell creates an island of size 1).
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This tripped me up more than it should have.
Clarify that flipping a 1 to 0 can only decrease the number of islands by at most 1, and the minimum is achieved by flipping a 1 that is a cut vertex (articulation point) in the graph of land cells. Then describe an algorithm to find such a cell efficiently, such as using Tarjan's bridge-finding algorithm on the grid graph, and analyze its time and space complexity.
Pro tip: Mention that if no such cut vertex exists, the number of islands remains the same, and highlight that this problem is equivalent to finding articulation points in a grid graph, which can be solved in linear time.
Explain that flipping a 1 to 0 can either keep the number of islands the same or decrease it by 1, but never increase it. The goal is to find a 1 whose removal decreases the island count.
Represent the grid as a graph where each land cell is a vertex and edges connect adjacent land cells. The number of islands is the number of connected components.
A land cell whose removal decreases the number of connected components is a cut vertex (articulation point). If such a cell exists, flipping it reduces the island count by 1; otherwise, the count remains unchanged.
Use Tarjan's algorithm (DFS with discovery and low-link values) to find articulation points in O(V+E) time, where V is the number of land cells and E is the number of edges (at most 4V).
If any articulation point exists, the minimum number of islands is original_count - 1; otherwise, it is original_count. Return this value.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
First, clarify that the flips are on distinct cells and can be done in any order, meaning you can flip a 0 to 1 and a 1 to 0, potentially removing a cell from one island and adding it to another. Then, model the problem as maximizing the sum of sizes of two islands connected by a flipped 0, minus the size of the island that loses a cell if the 1-to-0 flip disconnects it. Prove that the optimal strategy is to choose the two largest islands that can be connected by a single 0, and ensure the 1-to-0 flip does not reduce the total, or if it does, choose a 1 that is not part of those islands.
Pro tip: Emphasize that the order of flips matters: flipping 1-to-0 first might disconnect an island, but flipping 0-to-1 first could merge islands; however, since the flips are on distinct cells, the net effect is equivalent to removing one 1 and adding one 0, so you can analyze the final configuration directly.
Confirm that the grid is binary, flips are on distinct cells, and order does not affect the final set of flipped cells. State that the goal is to maximize the largest island after both flips.
Observe that the two flips together change one 0 to 1 and one 1 to 0, so the total number of 1s remains the same. The largest island can increase by merging two islands via the 0-to-1 flip, but may decrease if the 1-to-0 flip removes a cell from an island.
Consider all pairs of islands that can be connected by flipping a single 0 to 1. For each pair, compute the merged size. Then, consider the effect of the 1-to-0 flip: if the flipped 1 is not in the merged island, the merged size is achieved; if it is, the merged island loses one cell.
Show that the maximum possible largest island is either the sum of the two largest islands that can be connected by a 0, or that sum minus one if the only available 1-to-0 flip is within those islands. Argue that any other choice yields a smaller or equal result.
Discuss cases where no 0 can connect two islands, or where flipping a 1 disconnects an island into smaller pieces. Provide a counterexample if the optimal strategy fails under certain conditions.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Start by clarifying the problem context and constraints, then systematically address each edge case: all-zeros/all-ones grids, recursion depth, and large-scale adaptation. For each, explain the algorithmic fix (e.g., early exit, iterative DFS, chunking/union-find) and tie it back to Dropbox's data scale and reliability needs.
Pro tip: Mention that you'd validate edge cases with unit tests and use iterative or union-find approaches to avoid recursion limits, showing production-ready thinking. Also, discuss trade-offs between chunking and union-find in terms of memory and complexity, demonstrating you can adapt to ambiguous requirements.
Ask about grid size, memory limits, and whether the grid is static or streaming. Confirm the definition of 'connected' and the expected output (e.g., number of islands).
For all-zeros, return 0 immediately; for all-ones, return 1 (or the appropriate count). Explain that these can be detected with a quick scan or by checking the first and last elements.
Replace recursive DFS with an iterative stack-based approach or use union-find to avoid stack overflow for large n. Mention that Python's default recursion limit (~1000) is easily exceeded.
Propose chunking (process subgrids and merge boundaries), streaming (process row by row with union-find), or union-find labeling (disjoint set with path compression) to handle memory and scale.
Compare approaches: chunking adds complexity at boundaries, streaming requires careful state management, union-find is efficient but needs memory for parent array. Emphasize testing with edge cases and performance benchmarks.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.