Binary search on a peak-finding problem feels unintuitive at first because you're not searching for a specific value.
Use a binary search approach by comparing the middle element with its neighbors to determine which half contains a peak. If the middle element is greater than both neighbors, it's a peak; otherwise, move towards the side with the larger neighbor. This guarantees O(log n) time because we halve the search space each iteration.
Pro tip: Clarify that the algorithm works even with duplicates by treating equal neighbors as not strictly greater, and emphasize that the peak is guaranteed to exist due to the boundary conditions.
Confirm that the array is non-empty, elements can be negative, and out-of-bounds are treated as negative infinity. Ask if multiple peaks are acceptable and if any peak is fine.
Describe how to use binary search: compute mid, compare with neighbors, and decide which half to search based on the slope. Emphasize that this ensures O(log n) time.
Trace the algorithm on a sample array (e.g., [1,2,3,1]) to demonstrate how it finds a peak. Show the steps and the decision-making process.
Discuss edge cases: single element array, peak at boundaries, and arrays with duplicates. Explain how the algorithm handles them.
State that time complexity is O(log n) and space is O(1). Summarize why the approach is optimal and mention potential pitfalls.
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Start by clarifying the abbreviation rules and constraints, then propose a solution that groups words by their abbreviation patterns to efficiently find the shortest unique abbreviation for each. Discuss trade-offs between time and space complexity, and consider edge cases like words that are already short or have no valid abbreviation.
Pro tip: Mention that you can use a trie or a hash map to group words by their prefix and suffix, and that the problem can be solved in O(n * L^2) time where L is the maximum word length, but you can optimize by binary searching the abbreviation length for each word.
Ask about input size, character set, and whether abbreviations must be unique across all words. Confirm that if the abbreviation is not shorter, the original word is kept.
Explain that an abbreviation is formed as prefix + count + last character, and that it must be unique among all words. For example, 'apple' with 1 skipped character becomes 'a3e'.
Propose grouping words by their first and last characters, then for each group, find the shortest prefix length that makes the abbreviation unique. Use a trie or sorting to compare prefixes efficiently.
Discuss time and space complexity. A naive approach checks all possible abbreviation lengths for each word, but you can optimize by binary searching the length or using a trie to share prefix computations.
Consider words of length 1 or 2, words that are already unique, and cases where no abbreviation is shorter. 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.