The BFS part was fine, I've done island counting before.
Use BFS to traverse each island, but since the grid cannot be modified, maintain a separate visited matrix (or set) to track explored cells. Iterate through each cell; when an unvisited land cell is found, increment the island count and BFS to mark all connected land cells as visited.
Pro tip: Explicitly discuss the space-time tradeoff: using a visited matrix costs O(m*n) extra space but preserves the input, which is often required in production systems. Mention that if memory is a concern, you could use a hash set of encoded coordinates, but a boolean matrix is more efficient.
Confirm grid dimensions, connectivity (4-directional), and that the grid must remain unmodified. Discuss edge cases like empty grid, all water, or all land.
Use a 2D boolean array (or set) for visited tracking and a queue for BFS. Explain why a separate visited structure is necessary.
Loop through each cell. If it's land and unvisited, increment island count and start BFS from that cell.
For each dequeued cell, check its 4 neighbors. If a neighbor is land and unvisited, mark it visited and enqueue it.
State time complexity O(m*n) and space complexity O(m*n) due to visited matrix and queue. Mention that DFS is an alternative but BFS avoids recursion depth issues.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Use DFS/BFS to identify each island and record its cells. For each island, normalize its shape by translating it so that its top-left-most cell (or minimum row and column) is at (0,0), then store the normalized shape in a hash set. The count of distinct shapes is the size of the set.
Pro tip: Mention that you can encode the normalized shape as a string or tuple of coordinates to use as a hash key, and emphasize that translation normalization is key to ignoring absolute positions.
Traverse the grid using DFS or BFS to find all connected components of land cells (1s). For each island, collect the coordinates of its cells.
For each island, find the minimum row and column among its cells. Subtract these from all cell coordinates to translate the island so its top-left-most cell is at (0,0).
Convert the normalized list of coordinates into a hashable form, such as a sorted tuple of (row, col) pairs or a string encoding.
Insert each canonical representation into a hash set. The number of distinct shapes is the size of the set.
State that time complexity is O(R*C) for traversal and normalization, and space complexity is O(R*C) for storing visited cells and shapes.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.