Provide a sequence of 15 integers that maximizes the number ofrotations required by the AVL tree during the insertion process. What is theworst-case time complexity of insertion in an AVL tree? Argue why your sequence meets this bound.

The sequence of 15 integers that maximizes the number of rotations required by the AVL tree during the insertion process is a sequence in descending order. For example, the sequence could be: 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1.

The worst-case time complexity of insertion in an AVL tree is O(log n), where n is the number of nodes in the tree. This is because an AVL tree is a self-balancing binary search tree, and the height of the tree is always log(n). Therefore, the time complexity of searching for the correct place to insert a new node, as well as the time complexity of rebalancing the tree, is proportional to the height of the tree, which is log(n).

The sequence of 15 integers in descending order meets this bound because each insertion requires a traversal down the height of the tree to find the correct place to insert the new node, and then potentially a traversal back up the tree to rebalance it. Therefore, the time complexity of each insertion is O(log n), and since there are 15 insertions, the total time complexity is O(15 log n), which simplifies to O(log n) because the constant factor of 15 can be ignored in big O notation.