Greedy with a sort felt right to me: sort by points descending, then pick the highest-value item per category until you hit k.
Clarify the problem constraints and edge cases, then propose a greedy approach: for each category, keep only the highest point item, sort these top items in descending order, and sum the top k. If there are fewer than k categories, return an error or -1. Discuss time and space complexity, and consider alternative approaches like dynamic programming if constraints differ.
Pro tip: Mention that this is a variation of the 'maximum sum of k items from distinct categories' problem, and that the greedy choice is optimal because selecting the highest point item from each category never hurts. Also, proactively discuss how to handle ties or if k exceeds the number of categories.
Restate the problem in your own words and ask clarifying questions about input format, constraints, and expected output for edge cases.
Recognize that to maximize sum with distinct categories, you should pick the highest point item from each category, then choose the top k among those.
Describe steps: group items by category, find max per category, collect these maxes, sort descending, and sum the first k. If fewer than k categories, handle appropriately.
State time complexity O(n + m log m) where n is number of items and m is number of categories, and space complexity O(m).
Mention cases like k=0, k > number of categories, negative points, and briefly note that if categories were not distinct, a different approach (e.g., heap) might be needed.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Basically a hash map from book ID to a location tuple.
Start by clarifying requirements and constraints, then propose a hash map-based design that maps book IDs to location objects, with auxiliary indexes for efficient queries by location. Discuss trade-offs between different data structures and how to handle updates and deletions.
Pro tip: Mention that you would use a composite key (branch, aisle, shelf, position) for the location index to enable efficient range queries and that you would consider concurrency control for multi-user access.
Ask about expected operations, frequency, data size, and whether location queries need to be efficient (e.g., find all books in a branch).
Suggest a primary hash map from book ID to location, and a secondary index (e.g., another hash map or sorted structure) from location to book ID(s) for reverse lookups.
Explain how add, move, remove, and get_location work with the proposed structures, ensuring O(1) average time for key operations.
Compare alternatives (e.g., B-tree for range queries) and mention potential optimizations like caching or sharding for scale.
Cover handling of duplicate locations, missing books, and thread-safety if needed.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
This one tripped me up more than I want to admit.
Use a hash map to track the current holder and last timestamp for each book, and iterate through the events once, validating each event against the rules. Return false with the index of the first invalid event, or true if all events are valid.
Pro tip: Clarify edge cases upfront, such as whether a book can be checked out immediately after return, and mention that the solution runs in O(n) time with O(b) space where b is the number of books.
Confirm assumptions: e.g., can a book be renewed multiple times? Is a return allowed if the book was never checked out? What about timestamps equal to previous?
Use a hash map keyed by book ID to store the current holder (member ID or null) and the last event timestamp for that book.
For each event, check timestamp non-decreasing, then apply action-specific rules: checkout requires no current holder; return requires current holder matches; renew requires current holder matches.
If any rule is violated, immediately return false and the current index. Otherwise, update the state for that book.
After processing all events, return true if no violations were found.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Standard anagram check with a frequency array of size 26.
First, clarify that anagrams have the same character counts. Then, choose a fixed-size array (e.g., size 26) to count character frequencies, incrementing for the first string and decrementing for the second. Finally, verify all counts are zero.
Pro tip: Mention that a fixed-size array is O(1) space and faster than a hash map for lowercase English letters, and always confirm the character set with the interviewer.
Confirm that the strings contain only lowercase English letters and that anagrams must have the same length. Ask if the character set is fixed or could include other characters.
Select a fixed-size array of 26 integers (or a dictionary if the character set is unknown) to count character frequencies. Explain why this is efficient.
Iterate through the first string, incrementing the count for each character. Then iterate through the second string, decrementing the count for each character.
After processing both strings, check that all counts are zero. If any count is non-zero, the strings are not anagrams.
State that the time complexity is O(n) where n is the length of the strings, and space complexity is O(1) for a fixed-size array (or O(k) for a dictionary with k distinct characters).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.