← Bytedance Interview Insights
The naive approach is obvious and wrong for any real input size.
Use a monotonic decreasing stack to efficiently compute visible people for each person. Traverse the queue from front to back, maintaining a stack of heights that are strictly decreasing; for each person, pop all shorter or equal heights (they are blocked) and the number of remaining stack elements is the count of visible people. Push the current height onto the stack.
Pro tip: Clarify the visibility condition upfront: a person can see someone ahead if all people between them are strictly shorter than the shorter of the two. This avoids off-by-one errors and demonstrates careful reading.
Restate the problem: person i can see person j (j < i) if for all k with j < k < i, height[k] < min(height[i], height[j]). This means the line of sight is blocked by anyone taller than or equal to the shorter person.
Recognize that a monotonic stack (strictly decreasing) is ideal because it maintains candidates that are not blocked by taller people. The stack size at any point represents the number of visible people for the current person.
Iterate through the queue from left to right. For each person, pop from the stack while the top is less than or equal to the current height. The number of remaining elements in the stack is the count of visible people. Then push the current height.
Consider empty array, single person, all equal heights, and strictly increasing/decreasing heights. Ensure the algorithm correctly returns 0 for the first person and handles duplicates as per the condition.
State that each element is pushed and popped at most once, giving O(n) time and O(n) space. Walk through a small example to verify correctness.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.