Pretty straightforward join between Books and Loans, filter on condition and a NULL check on returned_at.
Start by clarifying the schema and the meaning of 'good condition' and 'not yet returned'. Then write a query that filters on condition and null return date, and counts the rows, using appropriate joins if needed.
Pro tip: Mention that you would confirm the definition of 'good condition' (e.g., condition = 'good' vs. condition IN ('good', 'excellent')) and whether 'not yet returned' means return_date IS NULL or return_date > CURRENT_DATE. This shows attention to detail and avoids incorrect assumptions.
Ask about the table structure (e.g., loans, books) and the exact meaning of 'good condition' and 'not yet returned'. Confirm if there are multiple conditions or if 'good' is a specific value.
Determine if condition is stored in the loans table or a separate books table. If separate, plan an INNER JOIN on book_id to filter by condition.
Use WHERE clauses to filter for condition = 'good' (or appropriate set) and return_date IS NULL (or return_date > CURRENT_DATE).
Use COUNT(*) or COUNT(loan_id) to get the number of loans. Consider if DISTINCT is needed if joins cause duplicates.
Check for edge cases (e.g., NULL conditions, future return dates) and suggest indexes on condition and return_date for performance.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
First, aggregate the data to compute lifetime value per book, then filter for books with more than 10 copies. Finally, sort by lifetime value descending and book_id ascending, and limit to the top 3 results.
Pro tip: Clarify the definition of 'lifetime value' and 'copies' upfront, as these may be ambiguous. Also, consider whether ties should be handled with a window function like RANK or DENSE_RANK if the top 3 should include all tied books.
Identify the relevant tables and columns, and clarify what 'lifetime value' and 'copies' mean in this context.
Use GROUP BY on book_id and compute the sum of revenue or appropriate metric to get lifetime value.
Apply a HAVING clause to include only books where the total copies exceed 10.
Order by lifetime value descending and book_id ascending, then limit to 3 rows.
If ties in lifetime value should not be arbitrarily cut off, use a window function like RANK or DENSE_RANK to include all tied books.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Use a self-join on the Members table to pair each inviter with their invitee, then compute the absolute difference in reserved_copies for each pair. Finally, select the maximum difference from those computed values.
Pro tip: Clarify whether the table stores inviter_id or invitee_id, and handle NULLs appropriately; also mention that using ABS() simplifies the query and ensures correctness.
Identify the self-referential relationship (e.g., inviter_id references member_id) and the reserved_copies column.
Join Members as inviter and invitee on the inviter's member_id matching the invitee's inviter_id (or vice versa).
Use ABS(inviter.reserved_copies - invitee.reserved_copies) to get the absolute difference for each pair.
Apply MAX() to the computed differences to get the maximum absolute difference.
Combine the join, ABS, and MAX into a single SQL statement, ensuring proper aliasing and filtering if needed.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Clarify the problem and edge cases, then propose a dynamic programming solution that tracks the maximum score up to each index by considering whether to include the current element or skip it. Derive the recurrence relation and optimize space to O(1) by keeping only the last two values.
Pro tip: After presenting the DP solution, mention that this is the classic House Robber problem and that the same pattern applies to many 'pick or skip' scenarios; also note that for very large inputs, a greedy approach fails, so DP is necessary.
Restate the problem to ensure understanding: given a list of non-negative integers, select a subset with no two adjacent elements to maximize the sum. Ask about edge cases like empty list, single element, or all zeros.
Define dp[i] as the maximum score achievable from the first i elements. Explain that at each index i, you either skip the element (dp[i-1]) or take it plus the best from i-2 (nums[i] + dp[i-2]).
Write the recurrence: dp[i] = max(dp[i-1], nums[i] + dp[i-2]). Base cases: dp[0] = 0 (empty), dp[1] = nums[0] (first element).
Observe that only the last two dp values are needed, so use two variables (prev2 and prev1) to achieve O(1) space. Update them iteratively.
State time complexity O(n) and space O(1). Walk through a small example (e.g., [2,7,9,3,1]) to verify correctness and handle edge cases.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Used a set to track currently open resources.
Use a stack to track open resources: push on OPEN, pop on CLOSE, and validate that the stack is not empty before popping and is empty at the end. Clearly state the invariants (no double-open, no close-before-open, no unclosed at end) and map each to a stack condition. Discuss time and space complexity, then consider edge cases and potential optimizations.
Pro tip: Clarify whether the sequence is per-resource or global; if per-resource, a single stack won't suffice and you'll need a map of resource IDs to states. Mentioning this shows you think about real-world logging scenarios and avoids a common pitfall.
Ask whether entries are for a single resource or multiple resources, and whether the input is a stream or a static list. Confirm the exact validation rules and expected return type (e.g., boolean or error details).
For a single resource, a counter suffices; for multiple resources, use a stack or a map of resource IDs to states. Explain why the chosen structure fits the problem.
Iterate through entries: on OPEN, check for double-open and push/increment; on CLOSE, check for close-before-open and pop/decrement. After the loop, ensure no unclosed resources remain.
State O(n) time and O(n) space (or O(1) for a single resource counter). Discuss edge cases like empty input, all opens, all closes, and interleaved resources.
Walk through examples to verify correctness. If needed, propose optimizations such as early termination or streaming validation.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Dict comprehension, skip if id is in the closed set, otherwise len of the employees list.
Clarify the input structure and edge cases, then iterate through the list of office objects, skipping any whose ID is in the closed set, and build a dictionary mapping open office IDs to their employee counts. Consider using a set for O(1) lookups and handle potential duplicates or missing fields gracefully.
Pro tip: Mention that you'd use a set for closed IDs to achieve O(1) membership checks, and discuss how you'd handle edge cases like duplicate office IDs or missing employee counts. This shows attention to efficiency and robustness.
Ask about the structure of office objects (e.g., attributes like id, employees), whether employee count is always present, and if there can be duplicate IDs. Confirm the expected output format.
Use a set for closed office IDs to enable O(1) lookups. Use a dictionary to accumulate results, mapping open office IDs to employee counts.
Loop through each office object. If its ID is not in the closed set, add its ID and employee count to the result dictionary. Handle duplicates by deciding whether to sum, overwrite, or skip.
Consider empty input lists, all offices closed, missing employee counts (default to 0 or skip), and duplicate IDs. Ensure the function returns an empty dictionary when appropriate.
State time complexity O(n) and space O(m) where m is number of open offices. 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.
zip_longest from itertools makes this trivial but I wasn't sure if they wanted that or a manual index approach.
Start by clarifying edge cases and constraints, then propose a two-pointer approach that iterates through both strings simultaneously, appending characters alternately. After one string is exhausted, append the remaining characters from the longer string. Analyze time and space complexity, and discuss potential optimizations or alternative implementations.
Pro tip: Meta interviewers value clean, efficient code and clear communication. Before coding, walk through a few examples to confirm understanding, and after coding, test with edge cases like empty strings and very different lengths.
Ask about input types, constraints (e.g., string lengths, character sets), and expected behavior for edge cases like empty strings or one string being much longer.
Explain the two-pointer technique: iterate up to the length of the shorter string, alternating characters, then append the remainder of the longer string. State that time complexity is O(n+m) and space complexity is O(n+m) for the output.
Write clean code, using a loop to alternate characters and then appending the rest. Use a StringBuilder or list for efficiency in languages like Java/Python.
Walk through test cases: equal lengths, different lengths, one empty string, both empty, and strings with special characters. Verify output matches expectations.
Mention that the solution is optimal in time, but if in-place modification were required, it would be more complex. Also, note that the output string length is fixed, so pre-allocating space can be beneficial.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.