Pretty transparent binary search setup once you see it.
Recognize this as a classic binary search problem where the API acts as a monotonic predicate. Use binary search to find the largest floor where the API returns 'safe', which is the threshold T, ensuring O(log N) API calls.
Pro tip: Clarify edge cases upfront (e.g., if all floors are unsafe or all safe) and discuss how to handle them within the binary search to avoid off-by-one errors.
Restate the problem: floors 1 to N, API returns safe/unsafe, find T such that floors ≤ T are safe and > T are unsafe. Emphasize the O(log N) API call constraint.
Explain that the safety predicate is monotonic: if a floor is safe, all lower floors are safe; if unsafe, all higher floors are unsafe. This enables binary search.
Initialize low=1, high=N. While low <= high, compute mid, call API. If safe, record mid as candidate T and search higher (low=mid+1); else search lower (high=mid-1).
After loop, return the last safe floor found. If no safe floor, T=0 (or indicate none). If all safe, T=N. Discuss how the algorithm naturally handles these.
State that each iteration halves the search space, so O(log N) API calls. Walk through a small example (e.g., N=10, T=6) to verify correctness.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Model the problem as a shortest path in a graph where each state is a 4-digit combination and edges represent single-wheel moves. Use BFS from the start state, skipping blocked states, and return the distance when the target is reached or -1 if exhausted.
Pro tip: Mention that BFS is optimal for unweighted graphs and that bidirectional BFS can significantly reduce the search space, especially when the target is far. Also, pre-check if start or target is blocked to return -1 immediately.
Confirm input format, edge cases (start equals target, blocked start/target), and that moves are bidirectional with wraparound. Ask if the lock can have repeated states or if blocked states are guaranteed unique.
Represent each combination as a node (e.g., integer 0000-9999). Edges connect states differing by one wheel ±1 mod 10. Blocked states are removed from the graph.
Use BFS to find the shortest path because all edges have unit weight. Initialize a queue with the start state, a visited set, and a distance counter.
While queue is not empty, dequeue a state. If it's the target, return distance. Otherwise, generate all 8 neighbors (each wheel +1/-1 mod 10), skip blocked or visited states, mark visited, and enqueue with distance+1.
If BFS exhausts without reaching the target, return -1. Also, early return -1 if start or target is blocked.
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 open cell is a node and edges connect adjacent open cells. Use BFS from the starting position to compute the shortest distance to any boundary open cell (excluding the start), returning the minimum distance or -1 if unreachable.
Pro tip: Clarify edge cases upfront: if the start is already on the boundary, you still need to find another boundary cell; also confirm whether diagonal moves are allowed. This shows attention to detail and avoids incorrect assumptions.
Ask about grid size, movement directions (4 or 8), and whether the starting cell can be a boundary cell. Confirm that the exit must be a different open boundary cell.
Explain that BFS guarantees the shortest path in an unweighted graph. Initialize a queue with the start position and a visited set or distance matrix.
Process cells level by level, and when dequeuing a cell, check if it's on the boundary and not the start. If so, return the current distance.
If the queue empties without finding a valid boundary cell, return -1. Also consider early termination if the start is surrounded by walls.
State that time complexity is O(m*n) since each cell is visited at most once, and space complexity is O(m*n) for the queue and visited set.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
The random placement part is where people overthink it and I was no exception.
Start by clarifying the problem constraints and edge cases, then outline a simple algorithm: create a 2D array, randomly place mines using a set or shuffle to ensure distinct cells, and compute neighbor counts by iterating over each cell. Emphasize clean code by avoiding unnecessary data structures and using helper functions for clarity.
Pro tip: Mention that you can avoid a separate mine placement data structure by using a sentinel value (e.g., -1) to mark mines, then compute counts in a single pass. This demonstrates awareness of memory and simplicity trade-offs.
Ask about board dimensions, mine count validity (e.g., N <= rows*cols), and expected output format. Discuss edge cases like zero mines or full board.
Choose a 2D array (list of lists) to represent the board. Use a sentinel value (e.g., -1) for mines to avoid a separate boolean array.
Generate N distinct random positions. Use a set to track placed mines or shuffle a list of all cells and take the first N.
Iterate over each non-mine cell and count mines in its eight neighbors using boundary checks. Update the cell with the count.
Walk through a small example to verify correctness. Discuss time and space complexity (O(rows*cols) time, O(rows*cols) space).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.