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First, clarify the problem constraints and edge cases, then walk through a concrete example to derive the logic for updating the cover. Implement the solution by iterating through intervals, computing cumulative lengths, and handling the three cases: deletion within an interval, at a boundary, or spanning multiple intervals.
Pro tip: Emphasize that the cover intervals are non-overlapping and sorted; this allows a single pass with cumulative length tracking, achieving O(n) time and O(1) extra space beyond the output. Mention that you would write unit tests for edge cases like deleting the first or last character, and intervals of length 1.
Ask about input guarantees: Are intervals sorted and non-overlapping? Can intervals be empty? What should be returned if the index is out of bounds? Confirm the expected output format.
Choose a small cipherbook and cover, compute the ciphertext, and manually delete a character. Observe how the cover changes, especially when an interval shrinks or splits.
Iterate through intervals while maintaining a cumulative length. Locate the interval containing the deletion index, then update that interval and adjust subsequent intervals' indices by subtracting 1.
Consider deletion at the start or end of an interval, intervals of length 1, and deletion at the very beginning or end of the ciphertext. Ensure the cover remains sorted and non-overlapping.
State that the solution runs in O(n) time and O(1) extra space (excluding output). Suggest writing tests for various scenarios to validate correctness.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Classic interval merge underneath, but framed weirdly enough that I had to think for a second.
First, clarify the data structures: intervals are represented as objects with start and end fields, and the cover is a list of such intervals. Then, sort the intervals by start, iterate through them, and merge any overlapping or contiguous intervals into a new list. Finally, return the simplified cover.
Pro tip: Mention that this is essentially the classic merge intervals problem, but emphasize the importance of handling edge cases like empty input, single interval, and intervals that are adjacent (end + 1 == next start). Also, discuss time complexity: O(n log n) due to sorting, and space O(n) for the output.
Ask clarifying questions: Are intervals inclusive? What does 'contiguous' mean exactly (e.g., end + 1 == next start)? Can intervals be unsorted? What is the expected output format?
Sort intervals by start. Then iterate and merge if the current interval's start is <= the last merged interval's end + 1 (to handle contiguity).
Write code, handling edge cases: empty list, single interval, all intervals mergeable, none mergeable. Test with examples.
State time complexity O(n log n) due to sorting, and space O(n) for the result. Mention that if input is already sorted, it's O(n).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
This is where I probably spent too long hedging.
Start by clarifying the problem: we need a data structure that maps an index to an interval, supports deletions, and maintains efficiency under many operations. Then propose a balanced binary search tree (e.g., red-black tree) or a skip list, explaining how it supports O(log n) search and deletion, and discuss the amortized complexity when deletions are frequent.
Pro tip: Mention that if deletions are always of the interval containing the index, a union-find structure with path compression can give near O(α(n)) amortized time, but if arbitrary deletions are needed, a balanced BST is more appropriate. This shows you consider the specific deletion pattern.
Confirm that intervals are non-overlapping and sorted, and that we need to support point queries (given index, find interval) and deletions of intervals.
Suggest a balanced binary search tree (e.g., red-black tree) or a skip list where each node stores an interval and the tree is keyed by interval start. This allows O(log n) search and deletion.
For a given index, traverse the tree to find the interval whose start is the largest start ≤ index, then check if index ≤ end. This is O(log n).
Deleting an interval from a balanced BST takes O(log n) worst-case. If many deletions occur, the amortized cost per deletion remains O(log n) because each deletion is a standard tree operation.
Mention that if deletions are always of the interval containing the queried index, a union-find with path compression can achieve O(α(n)) amortized. Compare with other structures like segment trees or interval trees.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.