Level-order traversal, take the last node at each level.
Use BFS level-order traversal, recording the last node at each level. Alternatively, use DFS prioritizing the right child and track the maximum depth seen so far. Clearly state the time and space complexity.
Pro tip: Mention that the right-side view is the last node at each level in BFS, and that DFS with right-first traversal can also work by tracking depth. This shows you understand multiple approaches and can choose based on constraints.
Confirm that 'right side view' means the rightmost node at each depth, and that the tree may be empty or skewed.
Decide between BFS (level-order) and DFS (right-first with depth tracking), explaining trade-offs.
Trace the chosen algorithm on a sample tree to verify correctness and edge cases.
State time and space complexity: O(N) time and O(N) space for BFS (queue) or O(H) for DFS (recursion stack).
Mention handling of empty tree, single node, and skewed trees (left or right).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Start by clarifying the problem constraints (e.g., array size, duplicates, expected time complexity). Then propose the optimal O(n^2) solution: sort the array and use a two-pointer technique for each fixed element, skipping duplicates to ensure uniqueness. Discuss trade-offs with brute force and hash-based approaches.
Pro tip: Mention that sorting enables efficient duplicate skipping and two-pointer search, and that the O(n^2) time complexity is optimal for this problem since the output can be O(n^2) in the worst case. Also, handle edge cases like arrays with fewer than 3 elements.
Ask about input size, duplicate handling, and expected time/space complexity. Confirm that triplets must be unique and indices cannot be reused.
Briefly describe brute force O(n^3), hash map O(n^2) with extra space, and the optimal sort + two-pointer O(n^2) with O(1) extra space (excluding output).
Explain: sort the array; for each index i, skip duplicates; use two pointers left=i+1 and right=n-1 to find pairs summing to -nums[i]; skip duplicates for left and right.
State time complexity O(n^2) due to nested loops, and space complexity O(1) extra (or O(n) if counting sorting space).
Walk through a small example like [-1,0,1,2,-1,-4] to demonstrate correctness and duplicate handling.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.