I knew Quickselect going in but fumbled the pivot discussion more than I expected.
Start by comparing brute force, sorting, and heap approaches, then focus on Quickselect as the optimal average-case O(n) solution. Explain pivot selection strategies (random, median-of-medians) and their impact on worst-case complexity, then implement with careful partitioning and handling of duplicates.
Pro tip: Mention that random pivot selection gives expected O(n) time and that median-of-medians guarantees O(n) worst-case, but is rarely used in practice due to high constants. Also, clarify how duplicates are handled in partitioning to avoid infinite loops.
Confirm that k is 1-indexed, duplicates are allowed, and the array can be modified. Discuss edge cases like k out of bounds or empty array.
Mention sorting (O(n log n)), min-heap of size k (O(n log k)), and Quickselect (average O(n), worst O(n^2)). Highlight Quickselect as optimal for average case.
Describe the algorithm: partition around a pivot, recurse only on the side containing the k-th element. Discuss pivot strategies: random (expected O(n)), median-of-medians (worst-case O(n)), and first/last (bad for sorted input).
Code a partition function (e.g., Lomuto or Hoare) that handles duplicates. Implement Quickselect iteratively or recursively, adjusting k based on pivot index.
State time complexity: average O(n), worst O(n^2) with random pivot, O(n) with median-of-medians. Space: O(1) iterative, O(log n) recursive. Walk through an example and edge cases.
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Clarify the problem constraints and edge cases, then propose an efficient solution using BFS/DFS from one node to find the other, or preprocess the tree for LCA if multiple queries are expected. Discuss how to validate input and handle invalid cases like duplicate edges, missing nodes, or cycles.
Pro tip: Mention that you'd first build an adjacency list and validate the tree structure (e.g., check for cycles or duplicate edges) before answering queries, showing you think about robustness and real-world data issues.
Ask about input format, whether the tree is guaranteed valid, and if multiple queries will be made. Confirm return values for same node, missing nodes, and invalid edges.
For a single query, BFS/DFS from u to v is O(N). For multiple queries, preprocess with LCA (binary lifting or Euler tour + RMQ) for O(log N) per query.
Validate that u and v exist in the node set. Detect duplicate edges by checking if an edge already exists in the adjacency list; if duplicates are present, decide whether to ignore or treat as invalid.
Write clean code with helper functions. Test with cases: same node, adjacent nodes, distant nodes, non-existent nodes, duplicate edges, and disconnected components (if tree not guaranteed).
Discuss time/space complexity of chosen approach. Compare BFS/DFS vs LCA preprocessing, and explain when each is preferable based on query frequency.
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First, verify that the string s starts with the target prefix; if not, return 0 (and -1 for the follow-up). If it does, the count of substrings starting with target is simply the number of possible ending positions, which is n - m + 1, where n is the length of s and m is the length of target. For the shortest substring, the answer is m itself (the prefix itself), so return m if the prefix matches, else -1.
Pro tip: Clarify whether overlapping substrings are counted (they are) and mention that the count can be large, so use a 64-bit integer to avoid overflow. Also, explicitly state that the shortest substring is always the prefix itself if it exists, demonstrating you understand the problem's simplicity.
Confirm that substrings are contiguous and that we count all occurrences, including overlapping ones. Ask if the prefix must match exactly at the start of the substring.
Compare the first m characters of s with target. If they don't match, return 0 for the count and -1 for the shortest length.
If the prefix matches, the number of substrings starting with target is n - m + 1. Use a 64-bit integer to store the result.
The shortest substring starting with target is the prefix itself, so its length is m. Return m if the prefix matches, else -1.
The solution runs in O(m) time for the prefix check and O(1) additional time for the count and shortest length, with O(1) space.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.