Explain that you would perform a DFS traversal, but leverage the BST property to prune branches that cannot contain values in the range. Specifically, if the current node's value is less than the low bound, skip the left subtree; if greater than the high bound, skip the right subtree. Otherwise, include the node's value and recurse on both children.
Pro tip: Emphasize that pruning reduces the time complexity to O(n) in the worst case but often much less, and that the space complexity is O(h) for the recursion stack, where h is the tree height. Mention that this is optimal because you must visit each node in the range at least once.
Restate the problem: given a BST and a range [low, high], return the sum of all node values within the range. Ask about edge cases: empty tree, low > high, or range not overlapping with any node.
Describe a recursive function that takes a node and returns the sum. At each node, compare its value with low and high to decide whether to explore left, right, both, or neither, using the BST property to prune.
If node.val < low, then all values in the left subtree are also < low, so skip left and recurse only on right. If node.val > high, skip right and recurse only on left. Otherwise, include node.val and recurse on both children.
Time: O(n) worst-case (e.g., all nodes in range), but pruning can reduce visits. Space: O(h) for recursion stack, where h is tree height; O(n) worst-case for skewed tree, O(log n) for balanced tree.
Mention that an iterative stack-based DFS can avoid recursion depth issues, but recursion is simpler. Also note that if the tree is balanced, the average time is O(log n + k) where k is number of nodes in range.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Start by clarifying the problem constraints and edge cases, then explain the recursive post-order traversal that returns the node if it matches p or q, otherwise recurses on left and right subtrees. For the iterative approach, describe using a parent pointer map and a set to track ancestors, then compare time and space complexities of both methods.
Pro tip: Emphasize that the recursive solution is elegant but may cause stack overflow for skewed trees, while the iterative solution avoids recursion but uses extra space for the parent map; mention that in an interview, you should discuss trade-offs and possibly implement the one that best fits the constraints.
Confirm that the tree is not a BST, nodes p and q are guaranteed to exist, and we need the lowest common ancestor (LCA). Discuss edge cases like one node being the ancestor of the other.
Explain the post-order traversal: if current node is null or equals p or q, return current node. Recurse left and right; if both return non-null, current node is LCA; otherwise return the non-null child.
Describe using a stack for DFS to build a parent pointer map (child -> parent). Then, traverse from p to root using a set to store ancestors, and traverse from q upwards until finding a node in the set.
Compare: Recursive: O(n) time, O(h) space (recursion stack). Iterative: O(n) time, O(n) space (parent map and set). Mention that h can be n in worst case.
Highlight that recursive is simpler and uses less space for balanced trees, but iterative avoids stack overflow for deep trees. Choose based on constraints and mention potential optimizations.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
This one tripped me up more than I expected for what sounds like a two-pointer problem.
Clarify the problem constraints and edge cases, then present the O(n) solution for pre-sorted inputs by scanning and maintaining the minimum departure price seen so far. For unsorted inputs, explain that sorting both lists by date (O(n log n)) enables the same linear scan, and discuss trade-offs like space complexity and stability.
Pro tip: Mention that if the lists are already sorted, you can avoid sorting and achieve O(n) time; otherwise, sorting is necessary. Also, note that you can optimize space by not storing all valid pairs, just the best one.
Ask about input sizes, whether dates are unique, if prices can be negative, and if multiple flights share the same date. Confirm that return date must be strictly after departure date.
Use two pointers or a single pass: iterate through return flights in date order, and for each, consider the minimum departure price among all departures with date < return date. Maintain this minimum as you go.
Sort both departure and return lists by date (O(n log n)), then apply the same linear scan. Alternatively, sort one list and use binary search on the other, but that would be O(n log n) as well.
State time and space complexity for both scenarios. Discuss whether sorting in-place is allowed, and if additional space for sorting is acceptable.
Walk through a simple example, including cases where no valid pair exists, or where the optimal pair uses the earliest departure and latest return.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
You need a post-order traversal that returns (sum, count) pairs up the tree, compute the floor average at each node, compare to node value, and accumulate a counter.
Use a post-order DFS that returns the sum and count of nodes in each subtree, allowing you to compute the average and floor in O(1) per node. At each node, after processing children, compute the floor of the average and compare it to the node's value, incrementing a global counter if they match. This yields an O(n) time and O(h) space solution.
Pro tip: Emphasize that the floor operation must be handled carefully for negative numbers (e.g., using integer division or math.floor) and clarify that the average is computed as a floating-point number before flooring. Also, mention that you can avoid floating-point precision issues by comparing sum >= value * count and sum < (value+1) * count for non-negative values, but for general values, use floor division.
Confirm that the tree can have negative values, that the average is computed as a real number, and that floor means the greatest integer less than or equal to the average. Ask about input size to justify O(n).
Define a function that returns a pair (sum, count) for the subtree rooted at a given node. Use post-order traversal to compute these values from children.
At each node, compute average = sum / count, then floor_avg = floor(average). If node.val == floor_avg, increment a global counter.
Discuss how to handle negative numbers correctly (e.g., using integer division or math.floor) and avoid floating-point precision issues by using integer arithmetic when possible.
State that the algorithm visits each node once, so time is O(n) and space is O(h) for recursion stack. Walk through a small example to verify correctness.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
The base problem is a flood fill and I coded BFS without thinking twice.
Start by clarifying the problem: the grid is binary, and we need all cells 4-directionally connected to the start with the same value. Then, present a BFS or DFS solution to find the component, and for the follow-up, collect border neighbors by checking adjacent cells outside the component. Finally, discuss trade-offs between BFS and DFS and analyze time and space complexity.
Pro tip: Mention that BFS is generally safer for large grids to avoid recursion depth limits, but DFS with an explicit stack can be equally memory-efficient. Also, note that the border neighbors can be collected during the same traversal by checking neighbors of visited cells.
Confirm the definition of 'connected' (4-directional) and 'border neighbors' (cells adjacent to the component but not part of it). Ask about grid size, mutability, and whether the start cell is guaranteed to be within bounds.
Decide between BFS (queue) and DFS (stack or recursion). Explain that both work, but BFS is iterative and avoids recursion limits, while DFS may be simpler to code.
Traverse from the start cell, visiting only cells with the same value, and mark them as visited (e.g., by changing value or using a visited set). Collect all component cells.
For each cell in the component, check its 4 neighbors. If a neighbor is within bounds, not in the component, and not already added, add it to the border set.
State that time complexity is O(N) where N is the number of cells in the grid (or O(R*C)), and space complexity is O(N) for the queue/stack and visited set. Compare BFS vs DFS: BFS uses queue, DFS uses stack/recursion; both have same complexity, but BFS may use more memory for wide components, DFS may risk stack overflow.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
The heap-order check is easy to do with any traversal.
Start by clarifying the two properties: completeness (all levels filled except possibly the last, which is left-packed) and heap-order (parent >= children). Then propose a single BFS traversal that simultaneously checks for gaps (a node after a null child) and verifies parent-child ordering, achieving O(n) time and O(n) space. Finally, explain why DFS cannot detect completeness violations because it lacks level-order context.
Pro tip: Mention that you can check completeness without extra space by counting nodes and verifying each node's index is less than n, but BFS is more intuitive and still O(n). Also, note that if the tree is not complete, the heap-order check alone is insufficient, so both must be checked together.
Confirm that a valid max-heap requires both completeness and heap-order. Discuss edge cases: empty tree, single node, and trees with duplicate values (heap-order allows equality).
Use a queue to traverse level by level. Track whether a null child has been seen; if a non-null node appears after a null, the tree is not complete. Also, for each node, ensure its value is >= its children's values.
Write the BFS code, ensuring each node is enqueued once. Explain that time complexity is O(n) and space complexity is O(n) in the worst case (queue size proportional to number of nodes at a level).
Illustrate with an example: a tree where a node is missing a left child but has a right child. DFS might visit the right child and not notice the gap, whereas BFS processes level by level and detects the missing left child immediately.
Conclude that BFS is the natural fit for level-order checks. Mention alternative approaches like index-based recursion (O(n) time, O(h) space) but note BFS is simpler for combined checks.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.