Classic BFS level-order traversal, just grab the last node at each level.
Use a level-order traversal (BFS) with a queue, and for each level, record the last node's value. Alternatively, use DFS with depth tracking, updating the result array at each depth with the current node's value. Both approaches yield O(n) time and O(n) space, but BFS is more intuitive for this problem.
Pro tip: Clarify with the interviewer whether the tree can be empty or have only one node, and mention that the right-side view includes the rightmost node at each depth even if it's not the right child. Also, discuss trade-offs: BFS uses O(width) space, while DFS uses O(height) space, which can be more efficient for skewed trees.
Confirm that you need to return the rightmost node at each depth, not just nodes that are right children. Clarify edge cases like empty tree or single node.
Decide between BFS and DFS. BFS naturally processes level by level, making it easy to pick the last node. DFS can be more space-efficient for deep trees if you track depth.
For BFS: use a queue, process each level, and add the last node's value to the result. For DFS: traverse right-first, and if the current depth equals the result size, append the node's value.
Walk through a sample tree, including edge cases like a left-skewed tree where the right view includes left children. Verify the output order is top to bottom.
State that both approaches are O(n) time. Discuss space: BFS O(width), DFS O(height). Mention that DFS can be more memory-efficient for skewed trees.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
This one tripped me up more than it should have.
Use a binary search approach: compare the middle element with its neighbors to decide which half contains a local minimum. If the middle is smaller than both neighbors, return it; otherwise, move towards the smaller neighbor. This works because the array has distinct elements and boundaries are treated as positive infinity.
Pro tip: Clarify edge cases upfront (e.g., array of size 1, local minimum at boundaries) and mention that the algorithm finds any local minimum, not necessarily the global one. This shows attention to detail and prevents misunderstandings.
Confirm that the array is unsorted, elements are distinct, and out-of-bounds neighbors are positive infinity. Discuss edge cases like empty array (return -1 or handle as needed) and single-element array (return index 0).
Describe how to use binary search: pick the middle index, compare its value with its left and right neighbors. If it's smaller than both, it's a local minimum. Otherwise, move towards the side with the smaller neighbor.
Trace the algorithm on a small example array to demonstrate correctness. Show how the search space halves each step and why a local minimum must exist in the chosen half.
State that the time complexity is O(log n) due to halving the search space, and space complexity is O(1) for iterative implementation (or O(log n) for recursive due to call stack).
Explain why the algorithm always finds a local minimum: by moving towards the smaller neighbor, we ensure that the boundary of the search space eventually leads to a local minimum. Mention handling of boundaries using positive infinity.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.