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question:**Objective**: Demonstrate your understanding of dynamic programming and integer partitioning. # Problem Statement You are given a positive integer `n`. Your task is to write a Python function `count_partitions(n)` that returns the number of distinct ways to partition `n` into a sum of positive integers. Each partition should be a non-increasing sequence of integers. # Function Signature ```python def count_partitions(n: int) -> int: ``` # Input * An integer `n` where (1 leq n leq 100). # Output * An integer representing the number of distinct ways to partition `n`. # Constraints * Time complexity should be (O(n^2)). * Space complexity should be (O(n^2)). # Example ```python assert count_partitions(4) == 5 assert count_partitions(7) == 15 ``` # Explanation For `n = 4`, the possible partitions are: ``` 4 3 + 1 2 + 2 2 + 1 + 1 1 + 1 + 1 + 1 ``` For `n = 7`, the possible partitions are: ``` 7 6 + 1 5 + 2 5 + 1 + 1 4 + 3 4 + 2 + 1 4 + 1 + 1 + 1 3 + 3 + 1 3 + 2 + 2 3 + 2 + 1 + 1 3 + 1 + 1 + 1 + 1 2 + 2 + 2 + 1 2 + 2 + 1 + 1 + 1 2 + 1 + 1 + 1 + 1 + 1 1 + 1 + 1 + 1 + 1 + 1 + 1 ``` # Implementation Note You may refer to the dynamic programming table approach for the integer partition problem. Make sure to follow proper initialization and controlled looping for accurate results. **Hint**: Use a 2-dimensional dynamic programming table where `dp[i][j]` represents the number of ways to partition `i` using integers up to `j`.

answer:def count_partitions(n: int) -> int: # Initialize the dp table dp = [[0] * (n + 1) for _ in range(n + 1)] # Base case: There is one way to partition the number 0 (using no numbers) for i in range(n + 1): dp[0][i] = 1 # Fill the dp table for i in range(1, n + 1): for j in range(1, n + 1): if j > i: dp[i][j] = dp[i][i] else: dp[i][j] = dp[i][j - 1] + dp[i - j][j] # The number of ways to partition n using any numbers up to n return dp[n][n]

question:# Separate Chaining Hash Table Extension As a software developer, you are responsible for maintaining and upgrading a library that implements a separate chaining hash table. The current implementation supports basic insertion, deletion, and retrieval of key-value pairs. To assess your understanding and skills, implement the following additional functionalities to the `SeparateChainingHashTable`: 1. **Resize**: Implement a method to resize the hash table when the load factor exceeds a specified threshold to maintain efficient operations. 2. **Keys**: Provide a method to return all keys in the hash table. 3. **Values**: Provide a method to return all values in the hash table. # Task Extend the `SeparateChainingHashTable` class with the following methods: Methods to Implement: 1. `resize(new_size: int) -> None`: Resizes the hash table to a new bucket size and rehashes all existing entries. 2. `keys() -> list`: Returns a list of all keys currently stored in the hash table. 3. `values() -> list`: Returns a list of all values currently stored in the hash table. # Constraints & Requirements: * For the sake of this exercise, assume the initial implementation is provided and the following constraints hold: - Keys are unique and hashable. - The size of the hash table should double when resized and rehashed. * The load factor (number of elements/number of buckets) threshold for resizing should be specified as a parameter or class variable (e.g., 0.75). # Example Usage: ```python # Initialize hash table with default size table = SeparateChainingHashTable() # Insert elements table.put('key1', 'value1') table.put('key2', 'value2') table.put('key3', 'value3') # Retrieve keys and values print(table.keys()) # Output: ['key1', 'key2', 'key3'] print(table.values()) # Output: ['value1', 'value2', 'value3'] # Induce resize by putting more elements table.put('key4', 'value4') table.put('key5', 'value5') # Resize should be triggered internally if the load factor exceeds the threshold # Verify if all elements are still accessible print(table.get('key1')) # Output: 'value1' print(table.get('key5')) # Output: 'value5' ``` # Implementation Hints: - Ensure `resize` adjusts the size of `_table` and rehashes all existing nodes to their new positions. - `keys` and `values` should traverse all buckets and linked lists to collect the required information.

answer:class SeparateChainingHashTable: def __init__(self, size=10, load_factor_threshold=0.75): self.initial_size = size self._size = size self._load_factor_threshold = load_factor_threshold self._table = [[] for _ in range(size)] self._num_elements = 0 def _hash(self, key): return hash(key) % self._size def put(self, key, value): hash_index = self._hash(key) for item in self._table[hash_index]: if item[0] == key: item[1] = value return self._table[hash_index].append([key, value]) self._num_elements += 1 if self._num_elements / self._size > self._load_factor_threshold: self.resize(self._size * 2) def get(self, key): hash_index = self._hash(key) for item in self._table[hash_index]: if item[0] == key: return item[1] return None def delete(self, key): hash_index = self._hash(key) for i, item in enumerate(self._table[hash_index]): if item[0] == key: del self._table[hash_index][i] self._num_elements -= 1 return True return False def resize(self, new_size): old_table = self._table self._table = [[] for _ in range(new_size)] self._size = new_size old_num_elements = self._num_elements self._num_elements = 0 for bucket in old_table: for key, value in bucket: self.put(key, value) self._num_elements = old_num_elements def keys(self): keys = [] for bucket in self._table: for key, _ in bucket: keys.append(key) return keys def values(self): values = [] for bucket in self._table: for _, value in bucket: values.append(value) return values

question:# AVL Tree Coding Challenge Context: You have been hired to improve the implementation of a database indexing system by incorporating a self-balancing AVL Tree. This will ensure efficient search, insertion, and deletion operations, crucial for the performance of the system. Problem Statement: Implement a function for the AVL Tree that: 1. Inserts elements and maintains tree balance. 2. Removes elements while ensuring the tree remains balanced after deletion. Function Signatures: ```python class TreeNode: def __init__(self, key): self.key = key self.left = None self.right = None ``` ```python class AvlTree: def __init__(self): self.node = None self.height = -1 self.balance = 0 def insert(self, key: int) -> None: pass def delete(self, key: int) -> None: pass def in_order_traverse(self) -> list: pass ``` Requirements: - **Insert Function**: Given an integer, insert it into the AVL Tree and preserve its balance. - **Delete Function**: Given an integer key, delete it from the AVL Tree and maintain balance. - **In-Order Traversal Function**: Implement an in-order traversal that returns the elements in sorted order. Constraints: - Input keys are unique integers. - AVL Tree must remain balanced after each insertion and deletion. - The number of operations (insertions and deletions) will not exceed 10^6 in aggregate. Example: ```python avl = AvlTree() avl.insert(10) avl.insert(20) avl.insert(30) avl.insert(15) print(avl.in_order_traverse()) # Output: [10, 15, 20, 30] avl.delete(20) print(avl.in_order_traverse()) # Output: [10, 15, 30] ``` Performance: - Your solution should efficiently handle large inputs and maintain a balanced tree after multiple operations.

answer:class TreeNode: def __init__(self, key): self.key = key self.left = None self.right = None self.height = 1 class AvlTree: def __init__(self): self.root = None def insert(self, key: int) -> None: self.root = self._insert(self.root, key) def _insert(self, node, key): if not node: return TreeNode(key) elif 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) if balance > 1 and key < node.left.key: return self._right_rotate(node) if balance < -1 and key > node.right.key: return self._left_rotate(node) if balance > 1 and key > node.left.key: node.left = self._left_rotate(node.left) return self._right_rotate(node) if balance < -1 and key < node.right.key: node.right = self._right_rotate(node.right) return self._left_rotate(node) return node def delete(self, key: int) -> None: self.root = self._delete(self.root, key) def _delete(self, node, key): if not node: return node elif key < node.key: node.left = self._delete(node.left, key) elif key > node.key: node.right = self._delete(node.right, key) else: if node.left is None: return node.right elif node.right is None: return node.left temp_val = self._get_min_value_node(node.right) node.key = temp_val.key node.right = self._delete(node.right, temp_val.key) node.height = 1 + max(self._get_height(node.left), self._get_height(node.right)) balance = self._get_balance(node) if balance > 1 and self._get_balance(node.left) >= 0: return self._right_rotate(node) if balance < -1 and self._get_balance(node.right) <= 0: return self._left_rotate(node) if balance > 1 and self._get_balance(node.left) < 0: node.left = self._left_rotate(node.left) return self._right_rotate(node) if balance < -1 and self._get_balance(node.right) > 0: node.right = self._right_rotate(node.right) return self._left_rotate(node) return node def _get_min_value_node(self, node): if node is None or node.left is None: return node return self._get_min_value_node(node.left) 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 def in_order_traverse(self) -> list: result = [] self._in_order_traverse(self.root, result) return result def _in_order_traverse(self, node, result): if not node: return self._in_order_traverse(node.left, result) result.append(node.key) self._in_order_traverse(node.right, result)

question:# Special Digit Powers Context In number theory, there exist certain numbers whose digits, when raised to consecutive powers starting from 1, sum up to the number itself. Task Write a function `special_digit_powers(low, high)` that returns a list of numbers within the range `[low, high]` (inclusive) such that the number equals the sum of its digits raised to powers corresponding to their positions. Input - An integer `low` representing the lower bound of the range (inclusive). - An integer `high` representing the upper bound of the range (inclusive). Output - A list of integers that meet the condition described; the integers should be in ascending order. Constraints - `0 <= low <= high <= 10^4` Examples 1. `special_digit_powers(1, 10)` should return `[1, 2, 3, 4, 5, 6, 7, 8, 9]` because all single-digit numbers trivially satisfy the property. 2. `special_digit_powers(1, 100)` should return `[1, 2, 3, 4, 5, 6, 7, 8, 9, 89]` because 89 = 8^1 + 9^2. Edge Cases - When `low` is greater than `high`, the function should return an empty list. - When `low` or `high` are negative, the function should ignore these and return appropriate results based on positive integers within the range. Performance Requirements - Ensure the function works efficiently within the given constraints. - Minimize redundant computations where possible.

answer:def special_digit_powers(low, high): Returns a list of special digit power numbers within the range [low, high]. def is_special_num(num): Check if a number is a special digit power number. str_num = str(num) return num == sum(int(digit) ** (idx + 1) for idx, digit in enumerate(str_num)) return [num for num in range(low, high + 1) if is_special_num(num)]