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The base problem is just two binary searches, fine.
Start by explaining the binary search approach to find the first and last occurrences in O(log n) time, then discuss how to adapt the data structure for the follow-up scenarios. For the growing array, consider a dynamic array with binary search or a balanced BST; for arbitrary insertions, propose a balanced BST or skip list with order statistics to support efficient range queries.
Pro tip: Emphasize the trade-offs between different data structures and always mention time/space complexity; Bloomberg values practical, efficient solutions and clear communication of engineering decisions.
Restate the problem to ensure understanding, ask about array size, data types, and whether the array can contain duplicates. Confirm that the array is sorted and that we need both first and last indices.
Explain how to modify binary search to find the leftmost and rightmost occurrences of the target. Describe two separate binary searches: one biased to the left, one to the right, both O(log n).
Discuss that if the array only grows at the end with larger values, the sorted property is maintained. A dynamic array with binary search still works, but insertions at the end are O(1) amortized. For range queries, binary search remains O(log n).
Explain that arbitrary insertions while keeping sorted require a data structure like a balanced BST (e.g., AVL, Red-Black) or a skip list. To support range queries efficiently, augment nodes with subtree sizes to find indices, or use an order-statistic tree.
Compare the approaches: static array with binary search is simplest and fastest for lookups but costly for insertions; dynamic structures offer O(log n) insertions and queries but add complexity. Choose based on expected workload.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Combine a hash map (for O(1) add/remove by ID) with a dynamic array (for O(1) random access). Store each participant's ID and its index in the array within the hash map, and on removal, swap the last element with the removed element to maintain a contiguous array.
Pro tip: Emphasize that the swap-with-last trick ensures O(1) removal even though array deletion is typically O(n), and mention that this approach is used in real systems like random sampling from a stream.
Confirm that IDs are unique, that add/remove are by ID, and that random pick must be uniform. Ask about expected size and whether duplicates are allowed.
Use a hash map to map ID to index in a dynamic array, and a dynamic array to store the IDs. This gives O(1) add, O(1) random pick, and O(1) removal with a swap.
For add: append to array and record index in map. For remove: swap the element with the last, update the moved element's index in the map, then pop the last element and remove the ID from the map. For random: pick a random index in the array and return the ID.
Explain that all operations are expected O(1) due to hash map operations and array indexing. Discuss edge cases: removing the last element, removing a non-existent ID, and handling empty structure.
Mention possible variations like allowing duplicates (use a set of indices per ID) or thread safety. Compare with alternative approaches like balanced BST (O(log n)) and explain why the hash map + array is optimal.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Took me a second to parse what 'depth' meant exactly.
Use a single pass to track the current depth and the maximum depth seen so far, collecting characters at the maximum depth. When a new maximum is found, reset the result list; when the current depth equals the maximum, append the character. This handles balanced parentheses and preserves order in O(n) time.
Pro tip: Clarify whether parentheses are only '(' and ')' or include other bracket types, and confirm that the string is guaranteed balanced. Mention edge cases like empty string or no parentheses, and that you'll return an empty list if no characters are at max depth.
Confirm the definition of nesting depth, the types of parentheses, and whether the string is guaranteed balanced. Ask about edge cases like empty string or no parentheses.
Set current depth = 0, max depth = 0, and an empty list for results. These will track the state during traversal.
Iterate through each character: if '(', increment depth; if ')', decrement depth; otherwise, it's a letter. For letters, compare current depth with max depth and update results accordingly.
When current depth exceeds max depth, update max depth and reset the result list to contain only the current character. When current depth equals max depth, append the character to the result list.
After traversal, return the result list. Discuss time and space complexity: O(n) time, O(k) space where k is the number of characters at max depth.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Clarify the problem constraints and then propose an efficient algorithm using a 2D prefix sum over a rotated coordinate system to count houses within Manhattan distance k for any candidate turret position. Discuss trade-offs between brute force O(m*n*k^2) and the optimized O(m*n) approach, and consider edge cases like houses on the boundary.
Pro tip: Mention that Manhattan distance balls become axis-aligned squares after rotating coordinates 45 degrees, which allows using a 2D prefix sum for O(1) range queries. This shows deep algorithmic insight and practical optimization.
Ask about grid size limits, whether turret can be placed on a house or only empty cells, and if multiple turrets are allowed. Confirm that Manhattan distance is used and that we want to maximize houses covered.
Explain that for each cell, we could check all cells within Manhattan distance k, counting houses. This takes O(m*n*k^2) time, which may be too slow for large grids.
Describe transforming coordinates (u = x+y, v = x-y) so that Manhattan distance becomes Chebyshev distance, making the coverage area an axis-aligned square. Then use a 2D prefix sum on the transformed grid to query the number of houses in O(1) per candidate position.
Discuss mapping transformed coordinates to a bounded grid, handling negative indices, and ensuring the turret is placed only on valid cells (empty or house, as specified). Also consider if k is large enough to cover the entire grid.
State that the optimized solution runs in O(m*n) time and O(m*n) space, which is optimal. Compare with brute force and mention that if k is small, brute force might be acceptable, but the prefix sum approach scales better.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Model the grid as a graph where each cell is a node and edges connect adjacent cells with weight 0 for empty cells and weight 1 for walls. Then use 0-1 BFS or Dijkstra to find the shortest path from (0,0) to (n-1,m-1) with total cost ≤ k. Alternatively, use BFS with state (row, col, walls_broken) to track the minimum steps.
Pro tip: Clarify whether 'shortest' means minimum steps or minimum walls broken; if ambiguous, assume minimum steps and mention that you can adapt. Also, discuss early termination and pruning to optimize.
Confirm grid dimensions, movement directions (4-way or 8-way), and whether k is inclusive. Ask about edge cases like start/end being walls.
Decide between 0-1 BFS (treating walls as cost 1) or BFS with state (r,c,walls). Explain why 0-1 BFS is efficient (O(nm)) and handles the constraint naturally.
For 0-1 BFS, state is just cell; for BFS with state, include walls broken. Describe how to update cost and check if walls broken ≤ k.
Write code with a deque for 0-1 BFS, or a queue for BFS with state. Check if start or end is a wall and if k is sufficient.
State time and space complexity (O(nm) for 0-1 BFS, O(nm*k) for BFS with state). Walk through a small example to verify.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.