This was the warmup and I treated it like one, which was fine.
Treat the grid as a graph where each '1' is a node connected to its 4-directional neighbors. Use DFS or BFS to explore each connected component, keeping track of the maximum size found. Alternatively, use Union-Find to merge adjacent '1's and track component sizes.
Pro tip: Mention that you can modify the grid in-place to mark visited cells (e.g., set to '0') to avoid extra space, but clarify that this mutates the input. Also, discuss trade-offs between DFS (recursion depth risk) and BFS (queue memory) for large grids.
Confirm grid dimensions, connectivity definition (4-directional), and whether the grid can be modified. Ask about edge cases like empty grid or no 1s.
Select DFS, BFS, or Union-Find based on constraints. For large grids, BFS avoids recursion limits; Union-Find is efficient for dynamic connectivity but may be overkill.
Iterate through each cell; when a '1' is found, explore its component using the chosen method, counting the size. Mark visited cells to avoid revisiting.
After each component traversal, update the global maximum size. Return the maximum after processing all cells.
State time complexity O(m*n) and space complexity O(m*n) for visited tracking (or O(1) if in-place). Discuss handling of edge cases like all 0s or all 1s.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Explain that you can use a binary search on the smaller array to partition both arrays such that the left half contains exactly half of the total elements, and the max of the left half is ≤ the min of the right half. Then compute the median from the boundary elements. This yields O(log(min(m,n))) time and O(1) space.
Pro tip: Mention edge cases like empty arrays, one array much smaller than the other, and even/odd total lengths, and clarify that you're binary searching on the smaller array to minimize complexity.
Confirm that the arrays are sorted, can be of different sizes, and may be empty. State that the goal is to find the median without merging, ideally in logarithmic time.
Explain that you need to partition both arrays into left and right halves such that all elements in the left half are ≤ all elements in the right half, and the left half has exactly (m+n+1)/2 elements.
Perform binary search on the smaller array to find the correct partition index. For each mid, compute the corresponding partition in the other array and check if the max of the left half ≤ min of the right half.
Once the correct partition is found, if the total number of elements is odd, the median is the maximum of the left half. If even, it's the average of the max of the left half and the min of the right half.
State that time complexity is O(log(min(m,n))) and space is O(1). Discuss handling empty arrays and ensuring indices are within bounds.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Felt like a relief after the median question.
Start by clarifying the graph representation (adjacency list) and the need to handle disconnected components by iterating over all vertices. Then present BFS and DFS (both iterative and recursive) with a shared visited set, and analyze time and space complexity.
Pro tip: Emphasize that for large-scale ML graphs (e.g., social networks at Meta), iterative BFS is preferred to avoid recursion depth limits and to enable distributed processing, while recursive DFS is simpler but risks stack overflow.
Confirm the graph is undirected, possibly disconnected, and represented as an adjacency list. Discuss input size and whether recursion depth is a concern.
Explain that you will maintain a global visited set and iterate over all vertices to ensure disconnected components are covered. For each unvisited vertex, launch a traversal.
Describe the iterative BFS using a queue: enqueue the start vertex, mark visited, then process neighbors level by level. Mention that BFS naturally handles disconnected graphs via the outer loop.
Show both recursive DFS (using call stack) and iterative DFS (using an explicit stack). Highlight that iterative DFS may need to push neighbors in reverse order to mimic recursion order, and that both handle disconnected graphs with the outer loop.
State that time complexity is O(V + E) for both BFS and DFS. Space complexity is O(V) for visited set and queue/stack, plus recursion depth for recursive DFS. Discuss when to choose BFS vs DFS based on memory and graph structure.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.