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question:<|Analysis Begin|> # Core Identification - **Algorithm/Data Structure**: Linked List-based Addition of Numbers - **Complexity**: - **Time Complexity**: O(max(n, m)), where n and m are the number of nodes in the two input linked lists. Each node in both lists is visited once. - **Space Complexity**: O(max(n, m)) for holding the sum linked list. - **Principles**: - Iterate through both linked lists concurrently, adding corresponding digits along with carrying any overflow from the previous digit's sum. - Store the result as a new linked list in reverse order, representing the sum digit-by-digit. # Characteristics & Applications - **Properties**: - Each node in the linked list contains a single digit. - The linked list represents the number in reverse order. - The solution handles carrying over with inline addition. - **Common Use Cases**: - Summing large numbers that cannot be comfortably handled using traditional data types due to overflow concerns. - Operations involving numbers stored digit-by-digit, typically in scenarios where digitwise manipulation is necessary (e.g., cryptographic applications). - **Strengths/Limitations**: - **Strengths**: Suitable for adding very large numbers without needing arbitrary-precision arithmetic support. - **Limitations**: Not suitable for non-reversed number representations or applications needing floating-point arithmetic. # Implementation Challenges - **Edge Cases**: - Input linked lists of different lengths. - Input nodes containing zero, especially leading or trailing zeros. - **Performance Bottlenecks**: Limited by the length of the linked lists; excessive digits will impact both time and space requirements proportionally. - **Error Scenarios**: - Handling when one of the input lists is empty. - Ensuring correctness when there's a carry over that extends the final result. - **Optimization Points**: - Optimize memory allocation to avoid unnecessary node creation. - Efficient handling of input lists of unequal lengths with reduced redundant operations. <|Analysis End|> <|Question Begin|> # Problem Statement You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order, and each of their nodes contains a single digit. Add the two numbers and return the sum as a linked list. The resulting linked list should also represent the number in reverse order. # Function Signature ```python class Node: def __init__(self, x): self.val = x self.next = None def add_two_numbers(left: Node, right: Node) -> Node: # Write your implementation here ``` # Input - Two non-empty linked lists `left` and `right` each representing a non-negative integer in reverse order. # Output - A linked list representing the sum of the two input numbers, also in reverse order. # Constraints - The input strings will not contain any leading zero, except the number 0 itself. - The algorithm must handle the carry-over efficiently. # Example Input: - `left`: (2 -> 4 -> 3) represents the number 342. - `right`: (5 -> 6 -> 4) represents the number 465. Output: - 7 -> 0 -> 8 represents the number 807 (342 + 465). # Example Input: - `left`: (9 -> 9) represents the number 99. - `right`: (1) represents the number 1. Output: - 0 -> 0 -> 1 represents the number 100 (99 + 1). # Detailed Requirements - Ensure edge cases are handled correctly, particularly when the linked lists have different lengths. - Consider carrying over when the sum of two digits exceeds 9. - Optimize the solution for both time and space to ensure it performs efficiently with the given constraints. # Hints - You may create a dummy node to simplify edge cases handling and iterate through the nodes using a pointer. - While traversing the input lists simultaneously, use a variable to manage the carry-over from the sum of two digits. # Testing **Example Test Cases**: 1. `(2 -> 4 -> 3)` + `(5 -> 6 -> 4)` should return `(7 -> 0 -> 8)`. 2. `(9 -> 9)` + `(1)` should return `(0 -> 0 -> 1)`. 3. `(0)` + `(0)` should return `(0)`. Implement comprehensive unit tests to validate the solution against various scenarios, and ensure edge cases are covered.

answer:class Node: def __init__(self, x): self.val = x self.next = None def add_two_numbers(left: Node, right: Node) -> Node: dummy = Node(0) current = dummy carry = 0 while left is not None or right is not None: l_val = left.val if left else 0 r_val = right.val if right else 0 total = l_val + r_val + carry carry = total // 10 current.next = Node(total % 10) current = current.next if left: left = left.next if right: right = right.next if carry > 0: current.next = Node(carry) return dummy.next

question:# Question: Balanced Binary Search Tree (BST) Insertion and Height Calculation You are given a partially implemented class `BST` representing a Binary Search Tree, with methods to insert elements, search for an element, and get the size of the tree. Your task is to extend this class to ensure the tree remains balanced using the AVL Tree mechanism. Additionally, write a function to calculate the height of the tree. **Objective**: 1. Ensure that the tree remains balanced after every insertion using AVL Tree rotations. 2. Implement a method to calculate the height of the BST. # Specifications 1. **Insert**: Implement the AVL Logic to ensure the tree remains balanced. This involves rotation mechanisms (single and double rotations). The insertion should continue to operate in O(log N) time. 2. **Height Calculation**: Implement a method to calculate the height of the tree. # Constraints: - Values to be inserted are unique integers. - Your height() function should return the height of the entire tree. - Follow AVL balancing rules to manage tree balance after each insert. # Input and Output: - **Input**: A series of integer values to be inserted into the BST. - **Output**: - After each insertion, check the balance of the tree. - The final height of the tree after all insertions. # Example: ```python bst = BST() values_to_insert = [10, 6, 15, 4, 9, 12, 24, 7, 20, 30, 18] for value in values_to_insert: bst.insert(value) print("Height of the tree:", bst.height()) ``` # Expected Output: ``` Height of the tree: 4 ``` Notes: - Implement the necessary rotations (single left, single right, double left-right, double right-left) to keep the tree balanced. - Ensure that each node keeps track of its height. - Implement utility functions to update node heights and balance factors.

answer:class Node: def __init__(self, key): self.key = key self.left = None self.right = None self.height = 1 class BST: def __init__(self): self.root = None def insert(self, key): self.root = self._insert(self.root, key) def _insert(self, node, key): if not node: return Node(key) if key < node.key: node.left = self._insert(node.left, key) else: node.right = self._insert(node.right, key) node.height = 1 + max(self._get_height(node.left), self._get_height(node.right)) balance = self._get_balance(node) # Left Left Case if balance > 1 and key < node.left.key: return self._right_rotate(node) # Right Right Case if balance < -1 and key > node.right.key: return self._left_rotate(node) # Left Right Case if balance > 1 and key > node.left.key: node.left = self._left_rotate(node.left) return self._right_rotate(node) # Right Left Case if balance < -1 and key < node.right.key: node.right = self._right_rotate(node.right) return self._left_rotate(node) return node def height(self): return self._get_height(self.root) def _get_height(self, node): if not node: return 0 return node.height def _get_balance(self, node): if not node: return 0 return self._get_height(node.left) - self._get_height(node.right) def _left_rotate(self, z): y = z.right T2 = y.left y.left = z z.right = T2 z.height = 1 + max(self._get_height(z.left), self._get_height(z.right)) y.height = 1 + max(self._get_height(y.left), self._get_height(y.right)) return y def _right_rotate(self, z): y = z.left T3 = y.right y.right = z z.left = T3 z.height = 1 + max(self._get_height(z.left), self._get_height(z.right)) y.height = 1 + max(self._get_height(y.left), self._get_height(y.right)) return y

question:# Description You are working on a networked application that requires sending a list of strings over a network connection. The requirement is to encode the list into a single string format and decode it back to the original list of strings on the other end. Your task is to implement the functions `encode` and `decode`. # Task 1. **Function 1 - `encode`**: - **Purpose**: Encodes a list of strings into a single string. - **Input**: A list of strings, `strs`. For example, `["hello", "world"]`. - **Output**: A single encoded string. For example, `"5:hello5:world"`. 2. **Function 2 - `decode`**: - **Purpose**: Decodes a single string back into a list of strings. - **Input**: An encoded string, `s`. For example, `"5:hello5:world"`. - **Output**: The original list of strings. For example, `["hello", "world"]`. # Requirements - Implement the functions `encode` and `decode` such that they can handle the edge cases mentioned in the analysis. - Ensure the encoded string format is maintainable and correctly decodable. # Examples ```python def encode(strs): Encodes a list of strings to a single string. :type strs: List[str] :rtype: str # Your code goes here def decode(s): Decodes a single string to a list of strings. :type s: str :rtype: List[str] # Your code goes here # Example 1 encoded = encode(["hello", "world"]) print(encoded) # Expected output: "5:hello5:world" decoded = decode(encoded) print(decoded) # Expected output: ["hello", "world"] # Example 2 encoded = encode([""]) print(encoded) # Expected output: "0:" decoded = decode(encoded) print(decoded) # Expected output: [""] ``` # Constraints - The provided input list `strs` will have at most (10^5) strings. - Each string in `strs` will have a maximum length of (10^4).

answer:def encode(strs): Encodes a list of strings to a single string. :type strs: List[str] :rtype: str encoded_str = '' for s in strs: encoded_str += f'{len(s)}:{s}' return encoded_str def decode(s): Decodes a single string to a list of strings. :type s: str :rtype: List[str] decoded_list = [] i = 0 while i < len(s): j = s.find(':', i) length = int(s[i:j]) decoded_list.append(s[j + 1:j + 1 + length]) i = j + 1 + length return decoded_list

question:# Palindromic Partition Finder Context You are a software developer responsible for implementing a feature in a text processing application aimed at revealing all possible ways to partition input text into palindromic substrings. A palindrome is a string that reads the same forward and backward (such as "level" or "radar"). Objective Write a Python function `palindromic_partitions(s: str) -> List[List[str]]` that takes a string as an input and returns all possible ways to partition that string into non-empty palindromic substrings. Each partition should be represented as a list of substrings. Input and Output Formats - **Input**: A single string `s` where `0 <= len(s) <= 16`. The string contains lowercase English letters. - **Output**: A list of lists where each inner list represents a valid partition of the input string into palindromic substrings. Constraints 1. The input string `s` will only contain lowercase alphabetical characters. 2. You need to capture all possible partitions where every substring in the partition is a palindrome. 3. Complexity may reach exponential due to the combinatorics involved. Function Signature ```python def palindromic_partitions(s: str) -> List[List[str]]: pass ``` Example ```python >>> palindromic_partitions("aab") [['a', 'a', 'b'], ['aa', 'b']] >>> palindromic_partitions("abcbab") [['a', 'b', 'c', 'b', 'a', 'b'], ['a', 'b', 'c', 'bab'], ['a', 'bcb', 'a', 'b'], ['abcba', 'b']] ``` Note - An empty input should return an empty list of lists: `[[]]`.

answer:from typing import List def is_palindrome(s: str) -> bool: Helper function to check if the input string is a palindrome. return s == s[::-1] def palindromic_partitions(s: str) -> List[List[str]]: Function to find all palindromic partitions of the input string. def backtrack(start: int, path: List[str], result: List[List[str]]): if start == len(s): result.append(path[:]) return for end in range(start + 1, len(s) + 1): substring = s[start:end] if is_palindrome(substring): path.append(substring) backtrack(end, path, result) path.pop() result = [] backtrack(0, [], result) return result

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