← Databricks Interview Insights
Pretty much Dijkstra with a pre-filter step.
Model the problem as a shortest path on a directed weighted graph, filtering edges by allowed modes. Use Dijkstra's algorithm with a priority queue to efficiently find the minimum-time path. Analyze complexity in terms of vertices, edges, and allowed modes, and discuss the follow-up by noting that filtering becomes unnecessary when all modes are allowed.
Pro tip: Mention that if the graph is large, you can pre-filter edges or use a mode-aware adjacency list to avoid checking modes during traversal. Also, clarify that Dijkstra's algorithm requires non-negative weights, which travel times typically are.
Confirm that edge weights (travel times) are non-negative, the graph is directed, and modes are edge attributes. Ask if multiple edges between same nodes with different modes exist.
Select Dijkstra's algorithm because it efficiently finds shortest paths in graphs with non-negative weights. Explain why BFS is not suitable due to weighted edges.
During relaxation, only consider edges whose mode is in the allowed set. This can be done by checking the mode before relaxing or by pre-filtering the adjacency list.
With a binary heap, time complexity is O((V + E) log V) or O(E log V) depending on implementation. Space complexity is O(V + E) for the graph and O(V) for distances.
If all modes are allowed, the mode filter is removed, so the algorithm remains the same but with no mode checks. Complexity is unchanged, but constant factors may improve.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Recognize this as a variation of the House Robber problem with a circular constraint. Break the circle by considering two linear cases: robbing houses 0 to n-2 (excluding the last house) and robbing houses 1 to n-1 (excluding the first house). Compute the maximum for each using dynamic programming and return the larger result.
Pro tip: Explicitly discuss the trade-off between time and space complexity, and mention that you can optimize space to O(1) by keeping only the last two DP values. This shows you think about efficiency beyond the basic solution.
Restate the problem to ensure understanding: houses are in a circle, adjacent houses cannot both be robbed, and the first and last are adjacent. Confirm that the input is an array of non-negative integers and that you need to return the maximum sum.
Since the first and last houses are adjacent, they cannot both be robbed. So consider two scenarios: (a) rob houses from index 0 to n-2 (exclude last), and (b) rob houses from index 1 to n-1 (exclude first). The answer is the maximum of these two cases.
For a linear arrangement, use dynamic programming: dp[i] = max(dp[i-1], dp[i-2] + nums[i]). This ensures no two adjacent houses are robbed. Handle edge cases like empty array or single house.
Instead of an array, keep only two variables to represent dp[i-1] and dp[i-2], reducing space complexity from O(n) to O(1). Update them iteratively while traversing the houses.
Return the maximum of the two cases. For n=1, return the single house's value. For n=2, return the max of the two houses. Ensure the solution works for all edge cases.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
First, clarify the interval semantics (half-open, sorted, non-overlapping) and walk through the deletion logic with examples, focusing on the three cases: no overlap, full overlap, and partial overlap (splitting). Then, for the follow-up, discuss how to handle high-throughput deletes by using a balanced BST or interval tree for O(log n) updates, and address bounded memory via streaming or compaction strategies.
Pro tip: Explicitly state that you're treating intervals as half-open to avoid off-by-one errors, and mention that you'd write unit tests for edge cases like deleting an interval that exactly matches an existing one or spans multiple intervals.
Confirm that intervals are half-open [start, end), sorted, and non-overlapping. Ask about input size, whether the delete interval can be empty, and expected output format.
Iterate through intervals, and for each, determine if it overlaps with the delete interval. If no overlap, keep it; if fully covered, remove it; if partially overlapped, split into up to two intervals.
For a single delete, O(n) is optimal if the list is an array. For multiple deletes, consider using a balanced BST or interval tree to achieve O(log n + k) per delete, where k is the number of intervals removed.
Discuss using an interval tree or skip list for efficient updates, and bounded memory via periodic compaction or a log-structured approach with tombstones and background merging.
Highlight trade-offs between simplicity (array) and performance (tree). Mention testing edge cases: delete before all, after all, exact match, spanning multiple, and empty delete interval.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.