BFS with a flag to flip direction each level.
Use BFS with a queue to process the tree level by level, tracking the current level number to determine direction. For each level, collect node values in a list, then reverse the list if the level is odd (or even, depending on 0-indexing) before adding to the result.
Pro tip: Mention that you can avoid reversing the list by using a deque and appending to the front or back based on the direction, which is more efficient. Also, clarify the indexing convention (0-indexed or 1-indexed) to avoid off-by-one errors.
Confirm the definition of zigzag traversal: first level left-to-right, second right-to-left, third left-to-right, etc. Ask about edge cases like empty tree or single node.
Explain that level-order traversal naturally fits BFS. Use a queue to process nodes level by level, and a variable to track the current level index.
For each level, determine the direction based on level parity. Collect node values in a list, then reverse if needed, or use a deque to insert at front/back efficiently.
While processing nodes, enqueue their left and right children for the next level. Ensure the queue size is captured before processing to separate levels.
State time complexity O(n) and space O(n) for the queue and result. Discuss edge cases: empty tree, skewed tree, and large tree.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Model the courses and prerequisites as a directed graph, then detect cycles using either DFS with recursion stack or Kahn's topological sort algorithm. Clearly explain the graph representation, the cycle detection method, and analyze time and space complexity.
Pro tip: Mention that this is a classic topological sort problem and that Kahn's algorithm is often preferred in interviews because it's iterative and avoids recursion depth issues. Also, discuss how to handle edge cases like duplicate prerequisites or disconnected components.
Confirm that prerequisites form a directed graph where an edge from course A to course B means A must be taken before B. Ask about input format (e.g., number of courses, list of pairs) and constraints.
Decide between DFS with recursion stack or Kahn's algorithm (BFS-based topological sort). Briefly explain the chosen approach and why it's suitable.
Walk through the implementation: for DFS, track visited and recursion stack; for Kahn's, compute in-degrees and process nodes with zero in-degree. Handle disconnected components.
State time complexity O(V+E) and space complexity O(V+E). Discuss edge cases: empty input, self-loops, duplicate edges, and large graphs.
Validate the solution with a simple acyclic case (e.g., 2 courses, 1 prerequisite) and a cyclic case (e.g., 2 courses with mutual prerequisites).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Sliding window with a hashmap tracking character counts.
Use a sliding window with two pointers to maintain a window that contains at most K distinct characters. Expand the right pointer to include new characters, and when the distinct count exceeds K, shrink the window from the left until it's valid again. Keep track of the maximum window length seen.
Pro tip: Clarify edge cases upfront (e.g., K=0, empty string, K >= unique characters) and discuss the time/space complexity trade-offs. Mention that the sliding window approach is optimal for this problem and can be extended to variations like 'at most K distinct' vs 'exactly K distinct'.
Restate the problem in your own words and ask clarifying questions about input size, character set, and edge cases. Confirm that the substring must be contiguous and that K is a non-negative integer.
Use a hash map (dictionary) to count the frequency of characters in the current window. This allows O(1) updates and quick checks of the number of distinct characters.
Initialize left and right pointers at 0. Expand right, adding characters to the map. While the map size exceeds K, remove characters from the left and increment left. Update the maximum length at each step.
State that the time complexity is O(n) because each character is processed at most twice (once by right, once by left), and space is O(K) for the map. Walk through a small example to verify correctness.
Mention that if the character set is small (e.g., ASCII), an array can replace the hash map for faster access. Also, note how to adapt the solution for 'exactly K distinct' by maintaining a window with at most K and at most K-1 distinct characters.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.