Start by clarifying the problem constraints and then present both solutions: for the in-place version, use three pointers starting from the end of the arrays to merge backwards; for the auxiliary buffer version, use two pointers from the start and copy into a new array. After presenting each, analyze time and space complexity, and discuss stability, noting that merging from the end can disrupt stability if equal elements are handled incorrectly.
Pro tip: Mention that merging from the end is optimal for the in-place case because it avoids overwriting unprocessed elements, and explicitly state that stability requires taking from the second array when elements are equal during forward merging, but this is tricky in the in-place backward approach.
Confirm the input format: two sorted arrays, first has extra capacity at the end, and ask about stability requirements and whether the arrays contain comparable elements.
Describe using three pointers: one at the end of the merged array (m+n-1), one at the last element of the first array (m-1), and one at the last element of the second array (n-1). Compare elements and place the larger one at the end, moving pointers backward.
Describe using two pointers starting at the beginning of each array, comparing elements and appending the smaller to a new array, then copying back if needed.
For both solutions, time complexity is O(m+n) because each element is processed once. In-place uses O(1) extra space, while auxiliary buffer uses O(m+n) extra space.
Explain that stability means equal elements retain their relative order. The auxiliary buffer approach can be stable if we take from the first array when elements are equal. The in-place backward approach is not stable by default because it may reverse the order of equal elements; to maintain stability, we would need to take from the second array when equal, but that complicates the in-place merge.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Start by clarifying the problem and constraints, then describe a recursive algorithm that processes the tree in reverse preorder (right, left, root) to flatten it in-place. Explain how each node's right pointer is set to the previously processed node, and left pointers are nulled. Conclude with a correctness argument (e.g., by induction) and complexity analysis (O(n) time, O(h) space).
Pro tip: Mention that the recursive solution uses O(h) stack space, and if asked to optimize to O(1) space, you can use Morris traversal or an iterative approach with parent pointers. This shows awareness of trade-offs and scalability.
Confirm the input/output format, in-place requirement, and that preorder means root-left-right. Ask if recursion is acceptable or if O(1) space is required.
Explain the reverse preorder recursive approach: recursively flatten right subtree, then left subtree, then set current node's right to the previously processed node and left to null. Maintain a 'prev' pointer.
Use induction: assume subtrees are correctly flattened; then linking them in reverse preorder yields the correct preorder sequence. Emphasize that the order of processing (right before left) ensures the 'prev' node is the next in preorder.
State time complexity O(n) since each node is visited once. Space complexity O(h) due to recursion stack, where h is tree height; worst-case O(n) for skewed tree, best-case O(log n) for balanced.
Mention that an iterative solution using a stack can also achieve O(n) time and O(h) space. For O(1) space, describe Morris traversal or a two-pointer approach that rewires pointers without extra space.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.