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To solve the problem efficiently in O(log(m×n)), we can treat the matrix as a single sorted 1D array. Even though the matrix is 2D, its structure allows us to apply binary search principles. The matrix's rows are sorted, and the first element of each row is greater than the last element of the previous row. This guarantees that the entire matrix if flattened, would behave like a sorted 1D array. Instead of flattening the matrix explicitly, we can compute the index of any element in this virtual 1D array by calculating its corresponding row and column indices.

We perform a binary search by initializing two pointers: l at the start (index 0) and r at the end (m×n−1). At each step, we calculate the midpoint mid, which corresponds to an index in the virtual 1D array. We convert this index back to the matrix by using the formulas row = mid / n and col = mid % n to access the corresponding matrix element. We then compare the element at matrix[row][col] with the target, adjusting the search space accordingly. If the target is found, we return true; otherwise, the search continues until the pointers overlap. If the target is not in the matrix, the function returns false.

To implement the solution, the first step is to sort the words by length. Sorting helps us ensure that when we attempt to create a chain, we are only comparing shorter words with longer ones. This is crucial because a shorter word can only be a predecessor of a longer word. After sorting, we use a dynamic programming approach where we maintain an array lis (Longest Increasing Subsequence) to store the length of the longest word chain that ends at each word. Initially, every word is its own chain, so we initialize the lis array to all ones.

Next, we loop over each word in the sorted list and compare it with all the words that come before it (because they are shorter). For each pair of words, we use the helper function isChainable to check if the shorter word can be transformed into the longer word by adding exactly one character in any position. If the shorter word is a valid predecessor, we update the current word's chain length by setting lis[i] = max(lis[i], lis[j] + 1) where i is the index of the current word, and j is the index of the shorter word. The helper function isChainable works by iterating through both words and confirming that we can insert one extra character into the shorter word to match the longer one. Finally, the answer is the maximum value in the lis array, which represents the length of the longest possible word chain.

To begin solving this problem, let’s first discuss a brute-force approach. Since the array is rotated at an unknown pivot, one way to find the target is to perform a simple linear search. In this approach, we iterate over each element in the array from start to end, comparing each one to the target. If we find the target, we return true; otherwise, we return false after checking all elements. While this approach is simple, it takes O(n) time, which is not optimal, especially for larger arrays. Given that the array is partially sorted, we can improve this by applying a binary search, which reduces the time complexity significantly.

To optimize, we can use binary search by taking advantage of the fact that one-half of the rotated array is always sorted. We maintain two pointers, l and r, to represent the current search bounds. First, we compute the middle index mid. If the element at mid equals the target, we return true immediately. If not, we check whether the right or left half of the array is sorted. If the right half (nums[mid] to nums[r]) is sorted and the target lies within this range, we search in the right half by setting l = mid + 1; otherwise, we search in the left half by setting r = mid - 1. Similarly, if the left half is sorted, we check whether the target lies within this range. If it does, we narrow our search to the left half, otherwise, to the right.

To find the maximum depth of a binary tree, the recursive approach is ideal. We start by defining a helper function depth that will calculate the depth of the tree for each node. In the main function maxDepth, the base case checks if the root is NULL, meaning the tree is empty, and returns a depth of 0. If the tree has nodes, we call the depth function, which will recursively traverse the tree starting from the root node, exploring both the left and right subtrees. For each node, the depth function is responsible for returning the maximum depth between the left and right subtrees, and it adds 1 to account for the current node.

In the recursive depth function, for each node, we recursively call the function on its left and right child nodes. We compute the depth of the left subtree and the right subtree. The function then returns the larger of the two depths (i.e., max(left, right) + 1). Meanwhile, the variable ans keeps track of the overall maximum depth encountered during the recursion. After processing all nodes, the final maximum depth is stored in ans, which is returned by the maxDepth function. This approach ensures that we check all nodes and find the longest path from the root to any leaf node in the binary tree.

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