BFS with a hashmap from column index to list of values.
Use a BFS traversal while tracking each node's x-coordinate, storing nodes in a map from x to list of values. Since BFS processes nodes level by level, nodes within each column are naturally ordered top-to-bottom, and left-to-right order is preserved by enqueueing left child before right child. Finally, sort the x-coordinates and output the lists.
Pro tip: Clarify that BFS ensures the required ordering without extra sorting within columns, and mention that if the tree is very deep, an iterative BFS avoids recursion stack overflow.
Restate the problem: group nodes by vertical column (x-coordinate), order top-to-bottom and left-to-right. Consider edge cases: empty tree, single node, skewed tree.
Use BFS (level-order) with a queue storing (node, x). Use a hash map (e.g., defaultdict(list)) to map x to list of node values. Track min and max x to avoid sorting later.
While queue not empty, pop node, append its value to map[x]. Enqueue left child with x-1 and right child with x+1. Update min_x and max_x.
Iterate from min_x to max_x, appending map[x] to result list. Return result.
Time: O(n) for traversal and O(n) for output, where n is number of nodes. Space: O(n) for queue and map.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Explain that you can avoid storing the entire traversal by using a BFS-like level-order traversal with a queue, but instead of collecting all nodes, you process and emit each column as soon as it's complete. Alternatively, use a DFS with a hash map of column indices to lists, but that still stores all nodes; so the streaming approach requires a different strategy like iterative deepening or a queue-based level-order traversal that tracks column boundaries. Emphasize the trade-off between memory and time, and propose a solution that uses O(width) memory by processing nodes column by column using a queue of (node, column) pairs and emitting when the column index changes.
Pro tip: Mention that in a real interview, you should clarify whether the tree is static or dynamic, and whether you can assume the tree fits in memory; then discuss how to handle infinite streams or very large trees by using a bounded queue and external sorting. Also, note that the column-by-column output can be achieved by a modified BFS that processes nodes in order of their column index, but that requires a priority queue if columns are not contiguous.
Ask whether the tree is static, whether memory is strictly limited, and whether the output must be in exact column order. Confirm that 'streaming' means emitting columns one at a time without storing all nodes.
Explain that BFS stores O(width) nodes, which is already better than O(n), but to emit column by column you need to know when a column ends. DFS with a map stores O(n) in the worst case, so it's not streaming.
Use a queue of (node, column) pairs, process level by level, and maintain a min-heap or sorted structure of columns to emit in order. Alternatively, use a two-pass approach: first find min and max column indices, then for each column, do a BFS that only emits nodes in that column, but that's O(n * width) time.
Compare time vs. memory: the two-pass approach uses O(width) memory but O(n * width) time; the heap-based approach uses O(width) memory and O(n log width) time. Mention that if columns are contiguous, a simple queue with a column offset can work.
Recommend the heap-based BFS: enqueue root with column 0, use a min-heap keyed by column to process nodes in column order, and emit a column when the next node has a different column. This uses O(width) memory and O(n log width) time.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Easy change, just sort within each (row, col) bucket by value.
First, clarify the problem context: this is likely a vertical order traversal where nodes in the same row and column are currently ordered by traversal order. To sort by node value instead, modify the comparator used when grouping nodes by column and row. Then, explain how to implement this change in code, ensuring that the sorting is stable and efficient.
Pro tip: Mention that changing the tie-breaking rule may affect the overall time complexity, so it's important to consider the trade-offs and whether the problem constraints allow for it. Also, note that this change might require adjusting the data structures used to collect nodes.
Restate the problem: nodes are grouped by column and row, and within the same row and column, they are currently sorted by traversal order. Confirm that the goal is to sort by node value instead.
Locate the part of the algorithm where nodes with the same row and column are ordered. This is typically when adding nodes to a list for a specific (row, column) pair.
Change the sorting key from traversal order to node value. If using a list, sort it by node value after collecting all nodes for that (row, column). If using a priority queue, adjust the comparator.
Discuss how this change affects time and space complexity. For example, sorting each group adds O(k log k) per group, where k is the number of nodes in that group.
Mention testing with cases where multiple nodes share the same row and column, and ensure the output is sorted by value as expected.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Talked through memory usage and worst-case column spread (a completely skewed tree gives you n distinct columns).
Acknowledge that scaling to 100,000 nodes shifts the focus from correctness to resource constraints like recursion depth, memory, and time complexity. Discuss how your approach handles these by converting recursion to iteration, using memory-efficient data structures, and analyzing asymptotic behavior. Conclude by identifying which parts of the algorithm remain efficient and which require adaptation.
Pro tip: Mention that at 100,000 nodes, recursion depth can cause stack overflow in languages like Python or Java, so an iterative solution with an explicit stack is often necessary. Also, highlight that constant factors and memory locality can matter as much as Big-O at this scale.
Ask about the tree's properties (balanced vs. skewed), available memory, and language/runtime. This shows you consider practical factors before diving into solutions.
State the Big-O for your algorithm (e.g., O(n) time, O(h) space) and explain how it behaves for n=100,000. Discuss worst-case scenarios like a skewed tree causing O(n) recursion depth.
Point out specific issues: stack overflow from recursion, memory overhead from storing nodes, and performance hits from pointer chasing or cache misses.
Describe modifications: iterative traversal with explicit stack, Morris traversal for O(1) space, or using arrays for complete trees. Explain trade-offs like increased code complexity.
Estimate actual memory usage (e.g., 100,000 nodes * ~40 bytes = 4MB) and time (e.g., 100,000 operations is trivial for modern CPUs). Mention testing with large inputs.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Classic stack-based approach: track indices of unmatched open parens, and do a single pass marking unmatched closes as you go.
Use a two-pass stack-based approach: first pass removes unmatched closing parentheses, second pass removes unmatched opening parentheses. Alternatively, use a counter-based method to mark invalid parentheses in O(n) time and O(n) space. Explain the algorithm step-by-step, prove its correctness by showing it removes exactly the minimum number, and analyze time and space complexity.
Pro tip: Mention that the problem can be solved with a single pass using a stack to track indices of unmatched parentheses, then mark them for removal. This demonstrates deeper understanding and efficiency, and you can discuss trade-offs between stack and counter approaches.
Confirm that the string contains only lowercase letters and parentheses, and that we need to remove the minimum number to make it balanced. Ask if multiple valid answers are acceptable (yes, any).
Describe a two-pass approach: first, scan left to right, using a stack to track indices of '(' and removing unmatched ')'. Second, scan right to left, removing unmatched '(' (or use a counter). Alternatively, use a single pass with a stack to mark invalid parentheses.
Choose a sample string like 'a)b(c)d' and demonstrate how the algorithm processes it, showing which parentheses are removed and the resulting balanced string.
Explain that every unmatched closing parenthesis must be removed, and every unmatched opening parenthesis must be removed. The algorithm removes exactly these, so it's minimal. Argue that no balanced string can be obtained by removing fewer parentheses.
State that time complexity is O(n) because each character is processed at most twice, and space complexity is O(n) for the stack or boolean array. Discuss edge cases: empty string, no parentheses, all parentheses, nested parentheses.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Yes, two counters: one for open parens seen but not yet matched, one for closes that had nothing to match.
Clarify the exact problem variant (e.g., remove minimum invalid parentheses to make string valid, or just check balance) and then outline a single-pass algorithm using a stack or counter. Emphasize that you can achieve O(n) time and O(1) space for the check, or O(n) space for removal, and discuss trade-offs.
Pro tip: Mention that you can solve the check in one pass with two counters (open and close) without a stack, but for removal you need to track indices; this shows you understand the space-time trade-off and can adapt to constraints.
Ask whether the goal is to check if parentheses are balanced, or to remove minimum invalid parentheses to make the string valid. Also confirm the character set (only parentheses or other chars).
For checking balance, use a counter: increment on '(', decrement on ')', and if counter goes negative, it's invalid. For removal, use a stack to track indices of unmatched parentheses.
State that the check is O(n) time and O(1) space, while removal is O(n) time and O(n) space due to the stack. Discuss if O(1) space removal is possible (e.g., two-pass with counters).
Mention empty string, all open or all close, and strings with other characters. Ensure the algorithm correctly handles these.
Compare stack vs. counter approaches, and mention that for removal, a two-pass approach can achieve O(1) space if we only need to remove minimum parentheses (but not if we need to return the string).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
I said yes in theory but fumbled the actual mechanics a bit.
First, clarify that in-place edits mean modifying the input array directly, so we can use two pointers to overwrite characters without extra space. Then, explain the algorithm: one pointer reads, one writes, and we handle edge cases like consecutive duplicates or removal. Finally, discuss trade-offs: O(1) space but O(n) time, and note that in-place may not be suitable if the input must be preserved.
Pro tip: Mention that in-place edits can be risky if the input is immutable or shared, so always confirm with the interviewer whether mutating the input is acceptable. Also, highlight that O(1) space often comes at the cost of increased code complexity and potential bugs, so balance clarity with optimization.
Confirm that in-place edits are allowed and that the input is mutable (e.g., a character array). Ask if the final length needs to be returned or if the string can be truncated.
Use a read pointer to scan the original string and a write pointer to overwrite characters. This allows O(1) extra space by reusing the input array.
Define the condition for writing (e.g., skip duplicates, remove specific characters). Ensure the write pointer does not overtake the read pointer and handle empty or single-character inputs.
State that time complexity remains O(n) as each character is processed once, and space complexity is O(1) since only pointers are used.
Mention that in-place modification destroys the original input, which may be undesirable. Compare with using extra space for clarity or if the input is immutable.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.