Question
Given a binary tree as a level-order array `arr` (null for missing), return its bottom view: for each horizontal distance (root at 0, left child -1, right child +1), the value of the node that would be seen from directly below. When several nodes share a horizontal distance, the one visited LAST in a left-to-right level-order (BFS) traversal wins. Return values ordered by horizontal distance, left to right. `0 <= node count <= 10000`.
bottom_view(arr: list[int]) → list[int][[20,8,22,5,3,null,25]]out[5,8,3,22,25]State your approach and its time/space complexity out loud before you optimize. Handle the edge cases (empty input, duplicates, overflow), and say why you chose this over the brute force. Green tests are the floor, not the grade.
Vibe coding: describe the solution in plain language (or narrate it) and the coach grades your approach. Generating runnable code from your description is coming next.