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question:# Problem Statement: Circular Array Duplicate Detection You are given an array of integers that represent a circular list. Your task is to implement a function that detects if there are any duplicates in the array considering its circular nature. A circular list means that the last element of the array loops back to the first element (i.e., next element after the last element is the first element of the list). # Function Specifications detect_circular_duplicates **Parameters**: - `arr`: A list of integers representing the circular array. **Returns**: - A boolean indicating whether there are duplicates in the circular array. Example ```python def detect_circular_duplicates(arr: list) -> bool: # Your implementation here ``` Input: ```python arr = [1, 2, 3, 4, 5, 1] ``` Output: ```python True ``` Input: ```python arr = [1, 2, 3, 4, 5] ``` Output: ```python False ``` Constraints - ( 0 leq |arr| leq 100 ) - Elements in the array can range from -10^6 to 10^6 # Notes - Consider the circular nature when looking for duplicates. - Ensure your function handles edge cases such as empty arrays efficiently. - Aim for an optimized solution that gracefully handles potentially large number ranges efficiently. - Test your solution with varied input to ensure correctness across different scenarios.
answer:def detect_circular_duplicates(arr: list) -> bool: Detects if there are any duplicates in the circular array. Parameters: arr (list): A list of integers representing the circular array. Returns: bool: True if there are duplicates, False otherwise. return len(arr) != len(set(arr))
question:# Coding Assessment Question **Scenario:** A logistics company manages a fleet of delivery trucks each able to carry a maximum load. The company aims to optimize the loading process, ensuring each truck is filled as much as possible without exceeding its weight limit. They have an existing algorithm that calculates the maximum load for each truck using a form of the knapsack problem but need to enhance it to return the combination of items loaded. **Task:** Implement the 0/1 Knapsack algorithm to determine the maximum weight that can be carried by a truck and to identify which items are included. # Requirements: 1. **Function 1:** `knapsack(capacity: int, weights: list[int], values: list[int]) -> int` - Takes the maximum capacity of the truck (`capacity`), a list of weights `weights` and a list of corresponding values `values`. - Returns the maximum value that can be carried within the given capacity. 2. **Function 2:** `find_items(capacity: int, weights: list[int], values: list[int]) -> list[int]` - Takes the same inputs as `knapsack`. - Returns a list representing the indices of the items included in the optimal combination. # Input: - **`capacity`**: An integer representing the maximum capacity of the truck (e.g., 50). - **`weights`**: A list of integers representing the weights of the items (e.g., [10, 20, 30]). - **`values`**: A list of integers representing the values of the items (e.g., [60, 100, 120]). # Output: - **From `knapsack`**: An integer representing the maximum value within the given capacity. - **From `find_items`**: An integer list representing the indices of items included in the optimal solution. # Constraints: - Weights and values lists are non-empty and of the same length. - Elements in weights and values lists are positive integers. - Capacity is a non-negative integer. # Example: ```python capacity = 50 weights = [10, 20, 30] values = [60, 100, 120] max_value = knapsack(capacity, weights, values) print(max_value) # Output: 220 items = find_items(capacity, weights, values) print(items) # Output: [1, 2] ``` **Edge Cases to Consider:** 1. Capacity is zero. 2. All items weigh more than the capacity. 3. Large lists requiring optimized memory or processing time.
answer:def knapsack(capacity, weights, values): Returns the maximum value that can be carried within the given capacity. n = len(weights) K = [[0 for _ in range(capacity + 1)] for _ in range(n + 1)] for i in range(n + 1): for w in range(capacity + 1): if i == 0 or w == 0: K[i][w] = 0 elif weights[i-1] <= w: K[i][w] = max(values[i-1] + K[i-1][w-weights[i-1]], K[i-1][w]) else: K[i][w] = K[i-1][w] return K[n][capacity] def find_items(capacity, weights, values): Returns the list of indices representing the items included in the optimal solution. n = len(weights) K = [[0 for _ in range(capacity + 1)] for _ in range(n + 1)] for i in range(n + 1): for w in range(capacity + 1): if i == 0 or w == 0: K[i][w] = 0 elif weights[i-1] <= w: K[i][w] = max(values[i-1] + K[i-1][w-weights[i-1]], K[i-1][w]) else: K[i][w] = K[i-1][w] # Traceback to find the items to include selected_items = [] w = capacity for i in range(n, 0, -1): if K[i][w] != K[i-1][w]: selected_items.append(i-1) w -= weights[i-1] selected_items.reverse() return selected_items
question:# Problem: Implement Custom HashMap with Collision Resolution You are required to implement a custom HashMap class in Python. This class should support basic operations like insertion, deletion, and retrieval of key-value pairs, along with handling collisions using chaining (linked list method). Instructions: 1. **Create the HashMap Class**: Define a class `CustomHashMap` that uses an array of linked lists to store key-value pairs. Each array index will represent a bucket that can contain multiple items in case of collisions. 2. **Implement Basic Operations**: - `put(key, value)`: Inserts the key-value pair into the HashMap. If the key already exists, update the value. - `get(key)`: Retrieves the value associated with the key if it exists, else return `None`. - `remove(key)`: Removes the key-value pair from the HashMap if it exists. 3. Ensure that the class efficiently handles collisions using linked lists, and provides average-case constant-time complexity for the operations. Example Usage: ```python # Instantiate CustomHashMap hash_map = CustomHashMap() # Insert key-value pairs hash_map.put(1, "value1") hash_map.put(2, "value2") hash_map.put(3, "value3") # Retrieve values print(hash_map.get(1)) # Output: "value1" print(hash_map.get(2)) # Output: "value2" print(hash_map.get(4)) # Output: None # Remove a key hash_map.remove(2) print(hash_map.get(2)) # Output: None ``` Constraints: - You may assume that keys are integers and values can be any type. - Implement the code in Python. - You are not allowed to use Python’s built-in dictionary for this implementation. - Aim to minimize the time complexity of each operation. Assume a reasonable size for the underlying array of buckets (e.g., initial size of 1000, and resize as needed). Details: Each bucket in your HashMap will be a linked list. Design a simple linked list structure to store key-value pairs where each node contains: - `key`: The key associated with this node. - `value`: The value associated with this node. - `next`: A pointer to the next node in the list (or `None` if it’s the end of the list). You will be graded on: - Correctness: The implementation of the `put`, `get`, and `remove` methods handling all edge cases. - Efficiency: Proper collision handling and time complexity. - Code quality: Clear, readable code with proper documentation and comments. Implement the `CustomHashMap` class and ensure it passes all provided test cases.
answer:class Node: def __init__(self, key, value): self.key = key self.value = value self.next = None class CustomHashMap: def __init__(self, initial_size=1000): self.size = initial_size self.buckets = [None] * self.size def _hash(self, key): return key % self.size def put(self, key, value): index = self._hash(key) if self.buckets[index] is None: self.buckets[index] = Node(key, value) else: current = self.buckets[index] while True: if current.key == key: current.value = value return if current.next is None: break current = current.next current.next = Node(key, value) def get(self, key): index = self._hash(key) current = self.buckets[index] while current is not None: if current.key == key: return current.value current = current.next return None def remove(self, key): index = self._hash(key) current = self.buckets[index] prev = None while current is not None: if current.key == key: if prev is None: # head node self.buckets[index] = current.next else: prev.next = current.next return prev = current current = current.next
question:# Binary Tree Path Sum As a software engineer, you may often encounter problems related to tree data structures. One common problem is finding paths in a binary tree that sum to a given value. This task involves writing a function that finds all root-to-leaf paths where each path's sum equals the given sum. # Problem Statement You need to write a function `path_sum(root: TreeNode, sum: int) -> List[List[int]]` that finds all paths in a binary tree where each path's sum is equal to the given sum. # Constraints * The function should return a list of lists, where each inner list represents a path from the root to a leaf node that sums up to the given value. * If no such path exists, return an empty list. * TreeNode is a class that represents a node in a binary tree, and each node has a value, a left child, and a right child. You may assume the value of each node is an integer. # Input - `root`: The root node of the binary tree. - `sum`: An integer representing the desired path sum. # Output A list of lists of integers, where each inner list represents a path from the root to a leaf node that sums up to the given sum. # Examples ```python class TreeNode: def __init__(self, val=0, left=None, right=None): self.val = val self.left = left self.right = right def path_sum(root: TreeNode, sum: int) -> List[List[int]]: pass if __name__ == "__main__": # Example Test Cases root = TreeNode(5) root.left = TreeNode(4) root.right = TreeNode(8) root.left.left = TreeNode(11) root.right.left = TreeNode(13) root.right.right = TreeNode(4) root.left.left.left = TreeNode(7) root.left.left.right = TreeNode(2) root.right.right.left = TreeNode(5) root.right.right.right = TreeNode(1) print(path_sum(root, 22)) # Expected [[5, 4, 11, 2], [5, 8, 4, 5]] print(path_sum(root, 26)) # Expected [[5, 8, 13]] print(path_sum(root, 18)) # Expected [[5, 8, 4, 1]] print(path_sum(None, 22)) # Expected [] ``` # Additional Information * You may assume that the input tree is a binary tree, and each node contains a single integer value. * Consider using Depth-First Search (DFS) to traverse the tree and track the paths that sum to the given value. # Hint * Think about recursive approaches to traverse the tree. * Keep track of the current path and sum as you traverse each node. If you reach a leaf node and the path sum equals the given sum, record the path.
answer:from typing import List class TreeNode: def __init__(self, val=0, left=None, right=None): self.val = val self.left = left self.right = right def path_sum(root: TreeNode, sum: int) -> List[List[int]]: def dfs(node, current_path, total, res): if not node: return current_path.append(node.val) total += node.val if not node.left and not node.right and total == sum: res.append(list(current_path)) else: dfs(node.left, current_path, total, res) dfs(node.right, current_path, total, res) current_path.pop() result = [] dfs(root, [], 0, result) return result