Three parts in one question, which I did not fully appreciate until I was already mid-explanation on part one.
First, clarify that the O(1) space constraint implies using Morris inorder traversal to check for the subsequence in O(n) time. For the modification part, recognize that the minimum operations correspond to the length of the sequence minus the length of the longest subsequence of the sequence that already appears in the inorder traversal, and explain how to compute that efficiently.
Pro tip: Mention that while Morris traversal achieves O(1) space, it temporarily modifies the tree; discuss the trade-off of mutating the tree during traversal and how to restore it. Also, note that the modification problem is equivalent to finding the longest common subsequence between the given sequence and the inorder traversal, but since the tree can be modified arbitrarily, the answer simplifies to the number of elements in the sequence not already present in the inorder traversal in the correct order.
Restate the problem to ensure understanding: check if the sequence is a subsequence of the inorder traversal in O(n) time and O(1) extra space, then find the minimum operations to modify the tree so that the sequence becomes a subsequence. Ask clarifying questions about the tree structure and operation definitions.
Explain that Morris inorder traversal allows traversing the tree without recursion or stack, using O(1) extra space. During traversal, maintain a pointer to the current element in the sequence and advance it when a match is found; if the pointer reaches the end, the sequence is a subsequence.
To make the sequence a subsequence, we need to ensure that the inorder traversal contains the sequence in order. The minimum operations equal the number of elements in the sequence that are not already present in the inorder traversal in the correct order. This is equivalent to finding the length of the longest subsequence of the given sequence that is already a subsequence of the inorder traversal, then subtracting from the sequence length.
Use a two-pointer approach: traverse the tree inorder (using Morris) and simultaneously scan the sequence. Count how many elements of the sequence can be matched in order; the unmatched count is the minimum operations. Alternatively, if the tree can be modified arbitrarily, the answer is simply the number of elements in the sequence not present in the tree, but order matters, so the two-pointer method is correct.
Address edge cases: empty tree, empty sequence, duplicate values, and the impact of modifying the tree on the O(1) space constraint. Discuss whether the modification should preserve the binary search tree property or not, and how that affects the operations.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.