Start by defining a Quadtree node class with fields for value, isLeaf, and children. Then implement a recursive build function that checks if all pixels in the current region are identical; if so, create a leaf node, otherwise split into four quadrants and recurse. Analyze time and space complexity, noting O(N^2) time and O(N^2) space in the worst case.
Pro tip: Mention that the quadtree is useful for image compression and that the recursion depth is O(log N), which is efficient for large N. Also, discuss how to handle non-power-of-2 sizes by padding, showing awareness of practical constraints.
Create a class with fields: val (the pixel value if leaf), isLeaf (boolean), and children (array of four Quadtree nodes). Optionally include topLeftRow, topLeftCol, and size for clarity.
Write a function build(grid, row, col, size) that checks if all pixels in the region are the same. If yes, return a leaf node with that value; otherwise, split into four quadrants of size/2 and recursively build each child.
Iterate through the region to check if all values are identical. Optimize by early termination when a mismatch is found.
If size == 1, always return a leaf node. Ensure recursion terminates and correctly assigns children.
Discuss time complexity: O(N^2) in worst case (all pixels different) and O(1) if all same. Space complexity: O(N^2) worst case. Mention edge cases like N=1 or non-power-of-2 (if allowed).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Model the problem as finding nodes that are not part of any cycle and cannot reach a cycle. Use a reverse graph and topological sort (Kahn's algorithm) to iteratively remove nodes with out-degree zero in the original graph, or use DFS with cycle detection to mark unsafe nodes.
Pro tip: Clarify edge cases upfront: empty graph, self-loops, and multiple components. Mention that the reverse graph approach avoids recursion depth issues and is O(V+E).
Restate that a safe node is one from which every path eventually reaches a terminal node (out-degree 0). Nodes that can reach a cycle are unsafe.
Decide between DFS with cycle detection (three-color marking) or reverse graph + topological sort (Kahn's algorithm). Both are O(V+E).
For DFS: mark nodes in current recursion stack as unsafe; propagate unsafe status to predecessors. For Kahn's: build reverse graph, compute out-degrees, enqueue nodes with out-degree 0, and process.
After processing, gather all safe nodes and sort them in ascending order as required.
State time and space complexity O(V+E). Discuss handling of empty graph, self-loops, and disconnected components.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.