The base problem I handled fine, BFS with a column offset tracked per node, dump into a sorted map.
Start by clarifying the problem and edge cases, then explain a BFS/DFS solution using a column index, ensuring correct ordering by row and left-to-right. Analyze time and space complexity, then discuss the follow-up by proposing a balanced BST or segment tree with column indices for dynamic insertions and queries.
Pro tip: Mention that using a hash map with min/max column tracking avoids sorting, and for the follow-up, highlight the trade-off between update and query times, suggesting a balanced BST keyed by column for O(log n) operations.
Confirm the problem details: vertical order traversal, ordering within same column and row, and output format. Ask about tree size, balance, and whether node values are unique.
Propose a BFS or DFS traversal that assigns a column index to each node (root at 0, left child -1, right child +1). Use a map from column to list of (row, value) and sort by row, then value for left-to-right order.
State time complexity: O(n log n) due to sorting within columns, or O(n) if using BFS with level-order and tracking min/max columns. Space complexity: O(n) for storing nodes and the map.
For streaming insertions and real-time queries, suggest a balanced BST (e.g., AVL or Red-Black) keyed by column, where each node stores a list of values sorted by row. Insertions are O(log n) and queries O(log n + k) for k results.
Compare the static solution (simple, O(n log n)) with the dynamic solution (more complex, but supports updates). Mention alternative data structures like segment trees or skip lists, and their pros/cons.
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Asked for both BFS and DFS which I liked, felt like they actually wanted to see if you understood the tradeoffs rather than just memorizing one approach.
Clarify the problem definition first: the right-side view consists of the rightmost node at each depth, with missing children simply skipped. Then present both BFS (level-order traversal, taking the last node per level) and DFS (pre-order traversal prioritizing right subtree, recording the first node seen at each depth) solutions, and compare their time/space complexities.
Pro tip: Mention that BFS naturally handles tie-breaking by processing levels left-to-right and taking the last node, while DFS requires a depth check to ensure only the first node encountered at each depth is recorded. Also note that both approaches are O(n) time, but BFS may use O(w) space (w = max width) while DFS uses O(h) space (h = height), which matters for skewed trees.
Define what 'right side view' means: the set of nodes visible when the tree is viewed from the right, which are the rightmost nodes at each depth. Confirm that missing children are simply absent and do not affect visibility.
Use a queue to perform level-order traversal. For each level, process all nodes and record the value of the last node (the rightmost one). Enqueue left and right children as usual, skipping nulls.
Use pre-order traversal, visiting the right child before the left. Keep track of the current depth and a result list; if the depth equals the result size, append the node's value (first visit at this depth).
For BFS, tie-breaking is inherent: the last node in each level is the rightmost. For DFS, tie-breaking is handled by visiting right first, so the first node seen at a depth is the rightmost. Missing children are simply not enqueued or visited.
Both solutions run in O(n) time. BFS uses O(w) space where w is the maximum width of the tree, while DFS uses O(h) space where h is the height. Choose based on tree shape: BFS may be memory-heavy for wide trees, DFS for deep trees.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Use a two-pointer technique starting from both ends of the sorted array. Move pointers inward based on the sum compared to the target, and when a match is found, count all duplicates of each value to account for multiple pairs efficiently. This achieves O(n) time and O(1) space.
Pro tip: Clarify that the array is sorted and that indices are unique even if values are duplicated. Emphasize that counting duplicates correctly is key to handling edge cases without extra space.
Set left pointer to 0, right pointer to n-1, and a counter for the number of pairs.
Compute sum = nums[left] + nums[right]. If sum < T, increment left; if sum > T, decrement right.
When sum == T, count how many times nums[left] and nums[right] appear. If the values are different, add count_left * count_right to the result and move both pointers past their duplicates. If the values are the same, add count_left * (count_left - 1) / 2 and break.
After the loop, return the total count of unique index pairs.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Start by clarifying the problem and edge cases, then present a BFS solution using a deque to alternate direction per level. Follow with a DFS solution that tracks level and inserts nodes at the correct position based on level parity, and compare their tradeoffs in terms of time, space, and code complexity.
Pro tip: Mention that BFS is more intuitive for level-order traversal and avoids recursion depth issues, but DFS can be more memory-efficient for skewed trees; showing awareness of these practical considerations demonstrates maturity.
Restate the problem to ensure understanding: level-order traversal with alternating directions per level. Discuss edge cases like empty tree, single node, and skewed trees.
Explain the BFS approach: use a queue to process nodes level by level, and a deque to collect nodes in the current level. Alternate between appending left-to-right and right-to-left based on level parity.
Explain the DFS approach: recursively traverse the tree, passing the level number. For each node, insert its value into the result list at the appropriate level, either at the end or beginning based on level parity.
Discuss time and space complexity: both are O(n) time, but BFS uses O(w) space where w is max width, while DFS uses O(h) space for recursion stack. Also compare code readability and potential stack overflow in DFS for deep trees.
Summarize which approach is preferable in different scenarios, e.g., BFS for balanced trees or when level order is natural, DFS for memory-constrained environments with deep trees.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.