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S1 : An insertion in an AVL with n nodes requires O(n) rotations.

answer is false in answer key,but is guess for 1 insetion its O(1).so for n it will be O(n).

tell me if i am wrong and correct me please.
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You are correct that the answer "An insertion in an AVL with n nodes requires O(n) rotations" is false. The correct answer is that an insertion in an AVL tree takes O(log n) time in the worst case, and requires a small number of rotations, typically less than O(log n).

Each insertion operation in an AVL tree involves traversing down the tree from the root to the appropriate leaf node where the new element should be inserted. This operation takes O(log n) time since the height of an AVL tree is at most log(n) for n elements.

After insertion, the AVL tree may become unbalanced, and so rotations are performed to rebalance the tree. However, the number of rotations required to rebalance the tree after an insertion is typically small, and in the worst case, it is still O(log n). In fact, most insertions in an AVL tree require only one or two rotations to rebalance the tree.

Therefore, the statement that "An insertion in an AVL with n nodes requires O(n) rotations" is false, and the correct time complexity of an insertion operation in an AVL tree is O(log n).

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