This one looked like a bit manipulation problem but it's really just careful state tracking.
Clarify the encoding rules (e.g., UTF-8) and then validate the byte sequence by scanning left to right, checking the leading bits of each byte to determine the character length and ensuring continuation bytes follow. Track the expected number of continuation bytes and verify the sequence ends exactly at a character boundary.
Pro tip: Mention that you can validate in a single pass without extra space, and discuss edge cases like empty input, truncated sequences, and invalid prefixes to demonstrate thoroughness.
Confirm the specific prefix patterns for 1-, 2-, 3-, and 4-byte characters (e.g., UTF-8) and that continuation bytes must be 10xxxxxx.
Iterate through the array, using the first byte to determine the expected number of continuation bytes, then verify each subsequent byte has the 10 prefix.
Check for empty input, sequences ending mid-character, and invalid leading bytes (e.g., 10xxxxxx as first byte).
State that time complexity is O(n) since each byte is visited once, and space complexity is O(1) as only a few variables are used.
Walk through valid and invalid sequences to demonstrate correctness and catch off-by-one errors.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Use BFS to find the shortest path in an unweighted grid with 8-directional movement. Initialize a queue with the start cell, track visited cells, and count steps including both start and end. If the end is reached, return the step count; otherwise, return -1.
Pro tip: Clarify that BFS is optimal for unweighted graphs and mention that 8-directional movement can be handled by iterating over 8 offsets. Also, discuss edge cases like start or end being blocked.
Restate the problem: find shortest path in n×n binary grid with 8-directional moves, count start and end cells, return -1 if no path. Confirm assumptions like start and end are open.
Select BFS because it finds shortest path in unweighted graphs. Explain why DFS or Dijkstra is not optimal here.
Use a queue for BFS, a visited matrix to avoid revisiting, and track distance. For each cell, explore 8 neighbors, enqueue if valid and unvisited.
Check if start or end is blocked; if so, return -1. Also handle n=1 case where start equals end.
Time complexity O(n^2) since each cell visited once, space O(n^2) for queue and visited matrix.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.