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question:# Red-Black Tree Node Retrieval Context You are working on a Red-Black Tree implementation as part of a larger software application requiring efficient order statistics operations. One operation frequently needed is to retrieve the k-th smallest element in the tree. Task Implement a function in the `RBTree` class to retrieve the k-th smallest element in the tree. You may use helper functions and should view the tree as adhering the constraints of the Red-Black Tree as implemented above. Function ```python def kth_smallest(self, k: int) -> int: Retrieve the k-th smallest element in the tree. :param k: The 1-based index of the desired element. :return: The value of the k-th smallest element. ``` Input and Output - **Input**: An integer `k`, representing the 1-based position of the desired element. - **Output**: The value of the k-th smallest element. Raise an exception if `k` is out of bounds. Constraints - The function should handle cases where `k` is less than 1 or greater than the number of elements in the tree. - You may assume the elements are distinct in the tree. - Consider edge cases such as an empty tree or k being larger than the size of the tree. Performance Requirements - The function should efficiently retrieve the k-th smallest element in O(log n) to O(n) time, where n is the number of nodes. Example ```python rb_tree = RBTree() nodes = [RBNode(10, 1), RBNode(20, 1), RBNode(5, 1), RBNode(1, 1), RBNode(15, 1)] for node in nodes: rb_tree.insert(node) assert rb_tree.kth_smallest(1) == 1 assert rb_tree.kth_smallest(3) == 10 assert rb_tree.kth_smallest(5) == 20 ``` Write the implementation for `kth_smallest` function in the given `RBTree` class.

answer:class RBNode: def __init__(self, key, color, left=None, right=None, parent=None): self.key = key self.color = color # 0 for Black, 1 for Red self.left = left self.right = right self.parent = parent self.size = 1 # Initially every node is treated as a single-node tree class RBTree: def __init__(self): self.nil = RBNode(None, 0) self.root = self.nil def insert(self, z): y = self.nil x = self.root while x != self.nil: x.size += 1 # Increment size of subtree y = x if z.key < x.key: x = x.left else: x = x.right z.parent = y if y == self.nil: self.root = z elif z.key < y.key: y.left = z else: y.right = z z.left = self.nil z.right = self.nil z.color = 1 # Red self.fix_insert(z) def fix_insert(self, z): # Fixing logic omitted for brevity pass def kth_smallest(self, k): def inorder_traverse(node): if node == self.nil: return [] return inorder_traverse(node.left) + [node.key] + inorder_traverse(node.right) if k < 1 or k > self.root.size: raise IndexError("k is out of bounds") result = inorder_traverse(self.root) return result[k - 1] # Example setup rb_tree = RBTree() nodes = [RBNode(10, 1), RBNode(20, 1), RBNode(5, 1), RBNode(1, 1), RBNode(15, 1)] for node in nodes: rb_tree.insert(node) print(rb_tree.kth_smallest(1)) # Should print 1 print(rb_tree.kth_smallest(3)) # Should print 10 print(rb_tree.kth_smallest(5)) # Should print 20

question:# Intersection of Two Singly Linked Lists Problem Statement: You are provided with two singly linked lists. Write a function that determines if the two lists intersect and returns the intersecting node. Intersection is defined by reference, not value. Function Signature: ```python def find_intersection(headA: Node, headB: Node) -> Optional[Node]: :param headA: The head node of the first linked list. :param headB: The head node of the second linked list. :return: The node at which the intersection of the two linked lists begins, or None if there is no intersection. ``` Input: * Two singly linked lists defined by their head nodes `headA` and `headB`. Output: * Return the intersecting node or `None` if there is no intersection. Constraints: * The lists should be traversed only once. * Use of extra space should be minimized. Example: ```python # Example 1: # listA: 1 -> 3 -> 5 # # 7 -> 9 -> 11 # / # listB: 2 -> 4 -> 6 # Intersection at node with value 7 # Example 2: # listA: 1 -> 2 -> 3 # listB: 4 -> 5 -> 6 # Intersection: None ``` Note: * The nodes themselves are considered for intersection, not their values.

answer:class Node: def __init__(self, value=0, next=None): self.value = value self.next = next def find_intersection(headA: Node, headB: Node) -> Node: :param headA: The head node of the first linked list. :param headB: The head node of the second linked list. :return: The node at which the intersection of the two linked lists begins, or None if there is no intersection. if not headA or not headB: return None # Two pointers for traversing the two lists ptrA, ptrB = headA, headB # Use two iterations to optimize traversing the lists while ptrA is not ptrB: # Move to the next node in each list, or switch to the start of the other list ptrA = ptrA.next if ptrA else headB ptrB = ptrB.next if ptrB else headA # Either intersection node or None return ptrA

question:--- **Scenario**: You are given an array of integers that you need to sort for a competitive programming contest. The input might not always be sorted, and efficiency is a key consideration to ensure your implemented algorithm runs within allowed time limits even for relatively large arrays. Equipped with the knowledge of Shell Sort, your task is to implement a more flexible version of this sorting algorithm that allows different gap sequences for optimization. # Problem Description: Implement the function `optimized_shell_sort` that sorts an array of integers in ascending order using Shell Sort algorithm. Your implementation should allow the use of three different types of gap sequences: 1. Default sequence (n/2, n/4, ..., 1 [as given in the provided code]). 2. Hibbard Sequence (1, 3, 7, 15, ..., 2^k-1) 3. Sedgewick Sequence (1, 5, 19, 41, 109, ...) # Function Signature: ```python def optimized_shell_sort(arr: List[int], sequence: str) -> List[int]: ''' :param arr: A list of integers to be sorted. :param sequence: A string indicating the gap sequence to use: 'default', 'hibbard', or 'sedgewick'. :return: A list of integers sorted in ascending order. ''' ``` # Input: - `arr`: List of integers (length `n`, 1 ≤ `n` ≤ 10^5). - `sequence`: A string 'default', 'hibbard', or 'sedgewick'. # Output: - A list of integers sorted in ascending order. # Constraints: - Ensure that your implementation performs within O(n^2) time complexity in the worst case. - Regardless of the chosen sequence, the function should be able to sort arrays up to 100,000 elements. # Example Usage: ```python print(optimized_shell_sort([23, 12, 1, 8, 34, 54, 2, 3], 'default')) # Output: [1, 2, 3, 8, 12, 23, 34, 54] print(optimized_shell_sort([4, 2, 7, 1, 3], 'hibbard')) # Output: [1, 2, 3, 4, 7] print(optimized_shell_sort([5, 3, 8, 6, 2], 'sedgewick')) # Output: [2, 3, 5, 6, 8] ``` # Notes: - Carefully design gap sequences for 'hibbard' and 'sedgewick'. - Provide sufficient test cases to ensure the correctness of your implementation. - Consider performance implications for large arrays.

answer:def optimized_shell_sort(arr, sequence): Sorts an array of integers using the Shell Sort algorithm with a specified gap sequence. :param arr: List of integers to be sorted. :param sequence: A string indicating the gap sequence to use: 'default', 'hibbard', or 'sedgewick'. :return: A list of integers sorted in ascending order. def default_sequence(n): seq = [] gap = n // 2 while gap > 0: seq.append(gap) gap //= 2 return seq def hibbard_sequence(n): seq = [] k = 1 while (2**k - 1) < n: seq.append(2**k - 1) k += 1 return list(reversed(seq)) def sedgewick_sequence(n): seq = [] k = 0 while True: if k % 2 == 0: gap = 9 * (2**k - 2**(k//2)) + 1 else: gap = 8 * 2**k - 6 * 2**((k+1)//2) + 1 if gap < n: seq.append(gap) else: break k += 1 return list(reversed(seq)) if sequence == 'default': gaps = default_sequence(len(arr)) elif sequence == 'hibbard': gaps = hibbard_sequence(len(arr)) elif sequence == 'sedgewick': gaps = sedgewick_sequence(len(arr)) else: raise ValueError("Invalid sequence type. Choose from 'default', 'hibbard', or 'sedgewick'.") for gap in gaps: for i in range(gap, len(arr)): temp = arr[i] j = i while j >= gap and arr[j - gap] > temp: arr[j] = arr[j - gap] j -= gap arr[j] = temp return arr

question:In many programming challenges, determining specific digits at certain positions within a large sequence of concatenated numbers can be crucial. Your task is to implement a function that efficiently finds the nth digit in the infinite integer sequence "123456789101112...". **Function Signature**: ```python def find_nth_digit(n: int) -> int: pass ``` # Input - `n`: An integer representing the position of the digit in the sequence (1-indexed). # Output - Returns an integer representing the digit at the nth position. # Examples Example 1 ```python n = 3 print(find_nth_digit(n)) # Output: 3 ``` *Explanation*: The 3rd digit in the sequence "123456789..." is `3`. Example 2 ```python n = 11 print(find_nth_digit(n)) # Output: 0 ``` *Explanation*: The sequence begins as "12345678910...", so the 11th digit is `0`. Example 3 ```python n = 15 print(find_nth_digit(n)) # Output: 2 ``` *Explanation*: The sequence follows "123456789101112131415...", so the 15th digit is `2`. # Constraints - (1 leq n leq 2 times 10^9) # Requirements - The function should be efficient, ideally with logarithmic time complexity. - Handle large values of `n` gracefully without significant performance degradation. # Notes - The sequence is continuous and does not contain separation characters. - Consider using integer division (`//`) to avoid floating-point arithmetic issues where applicable.

answer:def find_nth_digit(n: int) -> int: # Define variables for the current length of numbers and the current count of digits length = 1 count = 9 start = 1 # Find the range where the nth digit lies while n > length * count: n -= length * count length += 1 count *= 10 start *= 10 # Find the exact number that contains the nth digit start += (n - 1) // length str_start = str(start) # Find the exact digit within the number result = str_start[(n - 1) % length] return int(result)