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question:# Question: Implement Huffman Coding for Data Compression Given an input file, implement the Huffman Coding algorithm to encode the file's content and then decode the file back to its original content to verify that the lossless compression works correctly. The main tasks are to: 1. Calculate the frequency of each character. 2. Build the Huffman tree. 3. Generate Huffman codes. 4. Encode the input file. 5. Decode the encoded file and verify the original content. Your task is to implement the following two functions: **Function 1: encode_file** ```python def encode_file(input_filename: str, encoded_filename: str) -> None: Encodes the content of the input file using Huffman Coding and stores the encoded result in a binary file along with the Huffman tree structure. :param input_filename: The name of the file to be encoded :param encoded_filename: The name of the file to store the encoded content pass ``` **Function 2: decode_file** ```python def decode_file(encoded_filename: str, decoded_filename: str) -> None: Decodes the content of the encoded file using Huffman Coding and stores the decoded result in a new file. :param encoded_filename: The name of the file with the encoded content :param decoded_filename: The name of the file to store the decoded content pass ``` # Input and Output Formats 1. **Input (encode_file)**: - `input_filename`: Path to the input file containing the text to be encoded. - `encoded_filename`: Path to the output file where the encoded binary data and Huffman tree will be stored. 2. **Output (encode_file)**: - The function should compress the content of the input file using Huffman Coding and save the binary encoded data and tree to the specified output file. 3. **Input (decode_file)**: - `encoded_filename`: Path to the encoded file containing the binary Huffman coded data and tree. - `decoded_filename`: Path to the output file where the decoded text will be stored. 4. **Output (decode_file)**: - The function should decode the binary Huffman coded data and save the original text into the specified output file. # Constraints 1. Assume all input files contain only ASCII characters (0-127). 2. The size of the input file will not exceed 1MB. # Performance Requirements 1. Ensure that the encoding and decoding processes are efficient and complete in a reasonable time for the input file size limit. 2. Minimize the memory usage while storing the Huffman tree and encoded data. # Example Suppose `input.txt` contains: `hello huffman` - Encode the file: ```python encode_file('input.txt', 'encoded.bin') ``` - Decode the file and verify the content: ```python decode_file('encoded.bin', 'decoded.txt') ``` - After decoding, `decoded.txt` should contain: `hello huffman` Implement the `encode_file` and `decode_file` functions to achieve the tasks as described. Make sure to handle edge cases and optimize your code for performance.
answer:import heapq import os from collections import defaultdict, Counter import pickle class Node: def __init__(self, char=None, freq=None, left=None, right=None): self.char = char self.freq = freq self.left = left self.right = right def __lt__(self, other): return self.freq < other.freq def build_huffman_tree(text): frequency = Counter(text) priority_queue = [Node(char, freq) for char, freq in frequency.items()] heapq.heapify(priority_queue) while len(priority_queue) > 1: node1 = heapq.heappop(priority_queue) node2 = heapq.heappop(priority_queue) merged = Node(left=node1, right=node2, freq=node1.freq + node2.freq) heapq.heappush(priority_queue, merged) return priority_queue[0] if priority_queue else None def build_codes(node, prefix='', codebook={}): if node: if node.char is not None: codebook[node.char] = prefix build_codes(node.left, prefix + '0', codebook) build_codes(node.right, prefix + '1', codebook) return codebook def encode_text(text, codebook): return ''.join(codebook[char] for char in text) def pad_encoded_text(encoded_text): extra_padding = 8 - len(encoded_text) % 8 for _ in range(extra_padding): encoded_text += '0' padded_info = "{0:08b}".format(extra_padding) encoded_text = padded_info + encoded_text return encoded_text def to_bytearray(padded_encoded_code): if len(padded_encoded_code) % 8 != 0: raise ValueError("Encoded text length is not a multiple of 8.") byte_array = bytearray() for i in range(0, len(padded_encoded_code), 8): byte_byte = padded_encoded_code[i:i+8] byte_array.append(int(byte_byte, 2)) return byte_array def encode_file(input_filename: str, encoded_filename: str) -> None: with open(input_filename, 'r') as file: text = file.read() huffman_tree = build_huffman_tree(text) codebook = build_codes(huffman_tree) encoded_text = encode_text(text, codebook) padded_encoded_text = pad_encoded_text(encoded_text) byte_array = to_bytearray(padded_encoded_text) with open(encoded_filename, 'wb') as file: pickle.dump((huffman_tree, byte_array), file) def remove_padding(padded_encoded_text): padded_info = padded_encoded_text[:8] extra_padding = int(padded_info, 2) encoded_text = padded_encoded_text[8:] encoded_text = encoded_text[:-extra_padding] return encoded_text def decode_text(encoded_text, tree): decoded_text = [] current_node = tree for bit in encoded_text: if bit == '0': current_node = current_node.left else: current_node = current_node.right if current_node.char is not None: decoded_text.append(current_node.char) current_node = tree return ''.join(decoded_text) def decode_file(encoded_filename: str, decoded_filename: str) -> None: with open(encoded_filename, 'rb') as file: huffman_tree, byte_array = pickle.load(file) bit_string = ''.join(format(byte, '08b') for byte in byte_array) encoded_text = remove_padding(bit_string) decoded_text = decode_text(encoded_text, huffman_tree) with open(decoded_filename, 'w') as file: file.write(decoded_text)
question:You are given the implementation of a BST Iterator that allows for in-order traversal of a binary search tree (BST). Your task is to extend this functionality by implementing the following additional methods: 1. **peek()**: This method should return the next element in the in-order traversal without advancing the iterator. 2. **is_empty()**: This method should return true if there are no more elements to be iterated, otherwise false. 3. **reset()**: This method should reset the iterator to the initial state as if it hasn't been used to traverse the tree yet. # Function Specifications 1. `peek() -> int`: * **Description**: Returns the next element in the in-order traversal without moving the iterator. * **Constraints**: Assumes that `has_next()` would return true before `peek()` is called. * **Output**: An integer representing the next element in in-order traversal. 2. `is_empty() -> bool`: * **Description**: Checks if there are no more elements left in the iterator. * **Output**: A boolean value indicating if the iterator is empty. 3. `reset() -> None`: * **Description**: Resets the state of the iterator to the beginning of the traversal. * **Output**: None. # Example Usage ```python Assume class BSTIterator is already defined as above. # Example to construct a binary search tree class TreeNode: def __init__(self, val=0, left=None, right=None): self.val = val self.left = left self.right = right # Example tree root = TreeNode(4) root.left = TreeNode(2) root.right = TreeNode(6) root.left.left = TreeNode(1) root.left.right = TreeNode(3) root.right.left = TreeNode(5) root.right.right = TreeNode(7) iterator = BSTIterator(root) # Existing methods print(iterator.has_next()) # should return True print(iterator.next()) # should return 1 # New methods to implement print(iterator.peek()) # should return 2 print(iterator.is_empty()) # should return False iterator.reset() print(iterator.next()) # should return 1 ``` # Requirements * Your implementation must have efficient time complexity for all the new methods. * Ensure that your iterator still adheres to the constraints and characteristics of in-order traversal. * Handle edge cases such as an empty tree or single-node trees correctly.
answer:# Definition of a simple TreeNode for BST class TreeNode: def __init__(self, val=0, left=None, right=None): self.val = val self.left = left self.right = right # Definition of the BSTIterator class class BSTIterator: def __init__(self, root: TreeNode): self.root = root self.stack = [] self._leftmost_inorder(root) self.initial_stack = list(self.stack) # Save initial state to support reset() def _leftmost_inorder(self, root): while root: self.stack.append(root) root = root.left def has_next(self) -> bool: return len(self.stack) > 0 def next(self) -> int: if not self.has_next(): raise StopIteration("No more elements in the BSTIterator") topmost_node = self.stack.pop() if topmost_node.right: self._leftmost_inorder(topmost_node.right) return topmost_node.val def peek(self) -> int: if not self.has_next(): raise StopIteration("No more elements in the BSTIterator") topmost_node = self.stack[-1] return topmost_node.val def is_empty(self) -> bool: return not self.has_next() def reset(self) -> None: self.stack = list(self.initial_stack)
question:You are tasked with implementing efficient functions to compute the n-th Fibonacci number. This involves understanding and utilizing dynamic programming and iterative techniques for optimization. The scenarios assume all inputs are valid non-negative integers. **Part 1: Dynamic Programming** Write a function `fib_dp(n: int) -> int` that computes the n-th Fibonacci number using dynamic programming. This approach should store intermediate results in a list or array. **Part 2: Iterative Approach** Write a function `fib_iterative(n: int) -> int` that computes the n-th Fibonacci number using an iterative approach. This approach should use constant space for the calculations. # Input Format Each function should take a single integer `n` (0 ≤ n ≤ 10^6). # Output Format Both functions should return a single integer that is the n-th Fibonacci number. # Constraints * `0 ≤ n ≤ 10^6` * Your solution should be optimized for large values of n and run in O(n) time complexity. * Dynamic programming solutions should use O(n) space, whereas iterative solutions should use O(1) space. # Example Usage ```python print(fib_dp(10)) # Should output 55 print(fib_iterative(10)) # Should output 55 print(fib_dp(1000)) # Should output 43466557686937456435688527675040625802564660517371780402481729089536555417949 print(fib_iterative(1000)) # Should output 43466557686937456435688527675040625802564660517371780402481729089536555417949 ``` # Notes * Consider handling large values efficiently. * Ensure proper validation for the input constraints. * Expect solutions to be tested for performance accuracy.
answer:def fib_dp(n: int) -> int: Computes the n-th Fibonacci number using dynamic programming. :param n: An integer index n where 0 <= n <= 10^6 :returns: The n-th Fibonacci number if n == 0: return 0 elif n == 1: return 1 fibs = [0] * (n + 1) fibs[0], fibs[1] = 0, 1 for i in range(2, n + 1): fibs[i] = fibs[i - 1] + fibs[i - 2] return fibs[n] def fib_iterative(n: int) -> int: Computes the n-th Fibonacci number using an iterative approach. :param n: An integer index n where 0 <= n <= 10^6 :returns: The n-th Fibonacci number if n == 0: return 0 elif n == 1: return 1 a, b = 0, 1 for _ in range(2, n + 1): a, b = b, a + b return b
question:# Objective Implement a function to find the index of a target element in a sorted array using binary search. # Problem Statement You are given a sorted (in non-decreasing order) integer array `nums` and an integer `target`. Implement the function `binary_search(nums: List[int], target: int) -> int` to search for `target` in `nums`. If the target exists, return its index. Otherwise, return `-1`. # Function Signature ```python def binary_search(nums: List[int], target: int) -> int: ``` Input - `nums`: A list of integers sorted in non-decreasing order, where `1 <= len(nums) <= 10^4` and `-10^4 <= nums[i] <= 10^4`. - `target`: An integer `-10^4 <= target <= 10^4`. Output - The integer index of the target in the array, or `-1` if the target is not found. # Constraints - You must implement the function using binary search. - The algorithm should run in O(log n) time complexity. # Example ```python # Example 1 nums = [-1, 0, 3, 5, 9, 12] target = 9 print(binary_search(nums, target)) # Output: 4 # Example 2 nums = [-1, 0, 3, 5, 9, 12] target = 2 print(binary_search(nums, target)) # Output: -1 # Example 3 nums = [5] target = 5 print(binary_search(nums, target)) # Output: 0 ``` # Explanation In the first example, the array is `[-1, 0, 3, 5, 9, 12]` and the target is `9`. The algorithm should correctly identify that `9` is at index `4`. In the second example, the algorithm should return `-1` as `2` is not present in the array. In the third example, the array has only one element `5` which is the target, so the index `0` should be returned.
answer:from typing import List def binary_search(nums: List[int], target: int) -> int: Uses binary search to find the index of the target element in a sorted array. Returns -1 if target does not exist in the array. left, right = 0, len(nums) - 1 while left <= right: mid = (left + right) // 2 if nums[mid] == target: return mid elif nums[mid] < target: left = mid + 1 else: right = mid - 1 return -1