The boundary condition tripped me up more than I expected.
Precompute the distance to the nearest obstacle in each of the four directions for every cell using four directional passes (left, right, up, down). Then iterate through all cells and check if the tuple of distances matches the target signature, ensuring no direction has an infinite distance (i.e., no obstacle before the boundary).
Pro tip: Clarify upfront that the robot's position must have obstacles in all four directions; if any direction lacks an obstacle, the cell is automatically invalid. This shows attention to edge cases and avoids wasted computation.
Confirm the grid dimensions, obstacle representation, and that the target signature is a tuple of four distances (left, right, up, down). Clarify that a valid position requires an obstacle in every direction before the grid boundary.
Perform four separate passes over the grid: left-to-right for left distances, right-to-left for right distances, top-to-bottom for up distances, and bottom-to-top for down distances. Use a large sentinel value (e.g., infinity) when no obstacle is encountered before the boundary.
Iterate through each cell, skip if any of the four distances is the sentinel (meaning no obstacle in that direction). Otherwise, compare the tuple of distances to the target signature and record the cell if they match.
The approach runs in O(R*C) time and O(R*C) space, which is optimal for this problem. Discuss potential optimizations like early termination or using a single pass with state tracking if memory is a concern.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
This one I actually liked, which is a weird thing to say about a problem that humbled me a bit.
Recognize that a string can be rearranged into a palindrome if at most one character has an odd frequency. For each query node, traverse from that node up to the root, maintaining a frequency parity bitmask (e.g., 26-bit for lowercase letters). Count how many prefixes (from node to ancestor) have a bitmask with at most one set bit. Optimize by precomputing bitmasks from root to each node and using a hash map to count valid ancestor bitmasks for each query.
Pro tip: Mention that if queries are numerous, you can preprocess the tree to answer each query in O(depth) or even O(1) with offline processing, but always clarify the constraints first to choose the right trade-off.
Confirm the definition of a path segment (node to ancestor inclusive), the character set (e.g., lowercase English), and the number of queries and tree size to determine the required efficiency.
Explain that a multiset of characters can form a palindrome iff at most one character has an odd count. Represent the parity of character counts as a bitmask.
Perform a DFS from the root, computing for each node the XOR bitmask of characters along the path from the root to that node. This allows O(1) computation of the bitmask for any ancestor-descendant path.
For a query node, the path to an ancestor has bitmask = mask[node] XOR mask[parent(ancestor)]. Count ancestors where this bitmask has at most one set bit. Use a hash map during DFS to maintain counts of bitmasks from root to current node, enabling O(1) per ancestor check.
Discuss time complexity: O(N + Q * depth) naive, or O(N + Q) with offline processing. Handle edge cases: single node, all same characters, empty tree.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.