This felt like two questions stitched together.
Start by clarifying the problem and edge cases, then present a two-pointer merge for the sorted lists, followed by binary search for lower and upper bounds. Discuss time and space complexity, and explain how duplicates and empty inputs are handled.
Pro tip: Mention that the merge step can be skipped if the lists are already sorted and you only need bounds, but since the problem requires a merged array, do it efficiently. Also, note that Python's bisect module provides lower_bound and upper_bound, but be prepared to implement them manually.
Ask about input sizes, whether lists can be empty, if duplicates are allowed, and if the merged array should be returned or just the bounds. Confirm that x can be any integer.
Use two pointers to merge the lists into a new sorted array in O(m+n) time. Handle empty lists by returning the other list.
Implement binary search to find the first index where element >= x (lower bound) and first index where element > x (upper bound). Return -1 or len(arr) if not found, as appropriate.
State that merge takes O(m+n) time and O(m+n) space for the new array. Binary search takes O(log(m+n)) time and O(1) space. Overall O(m+n) time and space.
Explain that duplicates are naturally handled by binary search: lower bound returns first occurrence, upper bound returns index after last occurrence. Empty inputs: if both empty, return empty array and bounds 0; if one empty, merge returns the other.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Start by clarifying the problem constraints (sorted vs. unsorted, array size, data types) and then discuss the trade-offs between linear scan and binary search. For unsorted arrays, a linear scan is optimal; for sorted arrays, binary search or simply taking the first element (if ascending) is better. Also cover edge cases like empty arrays, single elements, duplicates, and negative numbers.
Pro tip: Mention that in a sorted array, the minimum is at one of the ends depending on sort order, and that binary search can find it in O(log n) if the array is rotated. This shows you think beyond the obvious and consider variations.
Ask about input size, whether the array is sorted, if it can be empty, data types, and if there are duplicates. This ensures you handle all cases correctly.
Explain that a linear scan (O(n)) is optimal because you must examine each element at least once. Mention that sorting first would be O(n log n) and is unnecessary.
For a sorted array (ascending), the minimum is the first element (O(1)). For a rotated sorted array, use binary search to find the minimum in O(log n).
Cover empty array (return null or throw exception), single element (return it), all duplicates, negative numbers, and large arrays (consider memory and time).
Write clean code for the chosen approach, then walk through test cases including edge cases to verify correctness.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
I went with BFS immediately since it finds shortest path and said so upfront.
Clarify the problem constraints (grid size, movement allowed, whether diagonal moves are permitted) and then propose BFS as the optimal solution for finding the shortest path to any boundary cell. Explain the algorithm step-by-step, including how to reconstruct the path using parent pointers, and analyze time and space complexity. Finally, compare BFS with DFS, highlighting trade-offs in terms of optimality and memory usage.
Pro tip: Mention that BFS guarantees the shortest path in an unweighted grid, which is often crucial in real-world applications like robotics or game AI. Also, note that if the grid is very large, bidirectional BFS or A* with a heuristic could be more efficient, showing awareness of advanced techniques.
Ask about grid size, movement directions (4 or 8), whether the start cell can be on the boundary, and if multiple paths exist. Confirm that the goal is any boundary cell and that the path should be returned as a list of coordinates.
Explain that BFS explores level by level, guaranteeing the shortest path in an unweighted grid. Contrast with DFS, which may find a path but not necessarily the shortest and can get stuck in deep branches.
Describe initializing a queue with the start cell, a visited set, and a parent map. While the queue is not empty, dequeue a cell, check if it's on the boundary, and if so, reconstruct the path using the parent map. Otherwise, enqueue all valid unvisited neighbors.
State that time complexity is O(R*C) where R and C are grid dimensions, as each cell is visited at most once. Space complexity is also O(R*C) for the queue, visited set, and parent map in the worst case.
Compare BFS vs DFS: BFS uses more memory but finds shortest path; DFS uses less memory but may not find shortest and can be slower in practice. Mention edge cases: start on boundary, no path exists, and large grids where bidirectional BFS or A* might be better.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Circular buffer was the obvious move here and I said it right away.
Start by clarifying the requirements: define the behavior when fewer than k elements have been seen (e.g., return the average of all elements seen so far). Then, design a class using a queue (or circular buffer) to maintain the last k elements and a running sum to compute the average in O(1) time per operation. Explain how you achieve amortized O(1) by ensuring each element is added and removed at most once.
Pro tip: Mention that you would use a fixed-size circular buffer to avoid the overhead of dynamic resizing and to keep memory usage constant, which is crucial for high-throughput ML systems. Also, discuss how you would handle edge cases like k=0 or negative k gracefully.
Ask about the expected behavior when fewer than k elements have been seen (e.g., return average of all seen elements or 0). Also clarify constraints: k > 0, stream size, and whether the stream can be infinite.
Select a queue (or circular buffer) to store the last k elements and maintain a running sum. This allows O(1) insertion and removal, and O(1) average computation.
Define methods: add(value) to insert a new number and update the sum and queue; getAverage() to return the current average. Ensure getAverage() is O(1).
In add(value): if queue size == k, remove the oldest element and subtract it from sum. Then add the new value to queue and sum. In getAverage(): if queue is empty, return 0 (or handle as per requirement); else return sum / queue.size().
Explain that each element is added and removed at most once, so amortized O(1) per operation. Discuss potential optimizations like using a fixed-size array for the circular buffer to avoid dynamic memory allocation.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.