Clarify that 'level by level' means a breadth-first traversal, then implement it using a queue. For each level, process all nodes currently in the queue, collect their values, and enqueue their children. After coding, state that the time complexity is O(n) and space complexity is O(w) where w is the maximum width of the tree.
Pro tip: At Meta, interviewers value clean, bug-free code and clear communication. Before coding, briefly discuss edge cases (empty tree, skewed tree) and mention that the queue-based BFS naturally handles level separation by tracking the queue size at each iteration.
Confirm that 'level by level' means printing nodes in breadth-first order, with each level on a new line or as separate lists. Ask if the output should be a list of lists or just printed values.
Select a queue (e.g., collections.deque in Python) to perform BFS. Explain that a queue ensures nodes are processed in the order they are discovered, which naturally groups nodes by level.
Describe the steps: initialize an empty queue and enqueue the root. While the queue is not empty, record the current queue size (number of nodes at this level), then iterate that many times: dequeue a node, add its value to the current level list, and enqueue its left and right children if they exist. After the inner loop, append the level list to the result.
Write clean code with meaningful variable names. Handle the edge case of an empty tree by returning an empty list. Use a list to store the result if needed.
State that time complexity is O(n) because each node is visited once. Space complexity is O(w) where w is the maximum width of the tree, due to the queue holding at most one level's worth of nodes. In the worst case (perfect binary tree), w = n/2, so O(n).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
The base class was fine, I got through it pretty quickly with a hashmap.
Start by clarifying requirements and designing a simple hash map-based solution with O(1) add, get, and remove operations. Then address follow-ups by introducing a running total for efficient total count, generalizing with generics, and discussing trade-offs of different data structures and concurrency approaches.
Pro tip: Proactively discuss the trade-offs between simplicity and scalability, and mention how you would test the class for correctness and performance under concurrent access.
Ask questions to confirm expected operations, data types, concurrency needs, and performance goals. This ensures you design the right solution.
Propose a hash map (dictionary) to store strings and their counts, with methods for add, get, and remove. Analyze time and space complexity.
For total count, suggest maintaining a running total variable. For arbitrary types, use generics. For swappable data structures, discuss alternatives like trees or tries and their trade-offs.
Explain how to make the class thread-safe using locks, concurrent data structures, or atomic operations, and the performance implications.
Conclude by comparing approaches, highlighting when to choose each based on requirements like read/write ratio, memory, and concurrency.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
This one tripped me up at first because my instinct was a naive recursive approach that recomputes sums repeatedly.
Use a post-order traversal to compute subtree sums and node counts bottom-up. At each node, check if its value equals the average of its subtree (sum / count) and count matches. This yields O(n) time and O(h) space.
Pro tip: Mention that you avoid floating-point division by comparing value * count == sum to prevent precision issues. Also, clarify that the average is computed as an integer division? Actually, the problem likely expects exact equality, so use multiplication.
Confirm that the average is the sum of all node values in the subtree divided by the number of nodes, and that we need exact equality (no rounding). Ask if the tree can be empty or have negative values.
Select a post-order traversal (left, right, root) because we need subtree information before processing the current node.
Write a function that returns (sum, count) for the subtree rooted at a node. For a null node, return (0,0). For a leaf, sum=node.val, count=1.
After computing left and right subtree sums and counts, compute total sum and count. If node.val * total_count == total_sum, increment a global counter.
Explain that each node is visited once, so time is O(n). Space is O(h) for recursion stack, where h is tree height.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Recognize that the number of pieces obtained from a given cut length L is monotonic: as L increases, the total pieces decrease. Therefore, binary search on L between 1 and the maximum wood length, checking feasibility by summing floor(length / L) for all pieces. Return the largest feasible L, or 0 if even L=1 yields fewer than k pieces.
Pro tip: Clarify edge cases upfront: if k is 0, return the maximum length (or handle as specified); if the sum of all lengths is less than k, return 0 immediately. Also, use integer division and avoid floating-point to prevent precision issues.
Restate the problem: given an array of wood lengths and integer k, find the largest integer L such that sum(floor(length_i / L)) >= k. If no such L exists, return 0. Confirm edge cases: k=0, empty array, k larger than total possible pieces.
Observe that as L increases, the total number of pieces decreases. So binary search on L from 1 to max(wood lengths). The answer is the largest L that satisfies the condition.
For a given L, compute total pieces by summing length // L for each wood piece. If total >= k, L is feasible; otherwise, not. Use integer division to avoid floating-point errors.
Initialize low=1, high=max_length, ans=0. While low <= high: mid = (low+high)//2. If feasible(mid), update ans=mid and low=mid+1; else high=mid-1. Return ans.
Time complexity: O(n log(max_length)), space O(1). Test with cases like k=0, k=1, k greater than total pieces, and all wood lengths equal.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.