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question:# Question: You are given an array of integers `nums` which contains exactly two unique elements and every other element appears exactly twice. Write a function `find_unique_elements(nums)` to find the two unique elements. Your solution must run in O(N) time and use only O(1) extra space. Input * An array of integers, `nums` (0 ≤ len(nums) ≤ 10^5), with exactly two unique numbers and all other numbers appearing exactly twice. Output * A list of the two unique integers. Function Signature ```python def find_unique_elements(nums: List[int]) -> List[int]: # your code here ``` Example ```python assert find_unique_elements([1, 2, 1, 3, 2, 5]) == [3, 5] or find_unique_elements([1, 2, 1, 3, 2, 5]) == [5, 3] assert find_unique_elements([4, 4, 6, 6, 9, 7]) == [9, 7] or find_unique_elements([4, 4, 6, 6, 9, 7]) == [7, 9] ``` Constraints * You must achieve a time complexity of O(N) and a space complexity of O(1). * The order of the result does not matter. Guideline 1. Use XOR bitwise operation to find the combined result of the two unique numbers. 2. Isolate one bit that is set in the combined result. 3. Divide the array into two groups based on this isolated bit and use XOR again to find the unique numbers for each group.
answer:from typing import List def find_unique_elements(nums: List[int]) -> List[int]: # Step 1: XOR all numbers to get XOR of the two unique numbers xor_result = 0 for num in nums: xor_result ^= num # Step 2: Find a set bit in the xor_result (that differs in the two unique numbers) diff_bit = xor_result & -xor_result # Step 3: Divide numbers into two groups based on the diff_bit and XOR individually unique1 = 0 unique2 = 0 for num in nums: if num & diff_bit: unique1 ^= num else: unique2 ^= num return [unique1, unique2]
question:Scenario You are working on a text segmentation feature for an application that processes sentences to identify words from a given dictionary. Your task is to write a function that determines the number of valid segmentations of a given string into dictionary words. Each word in the string must be contained within the given dictionary. Function Specification Implement the following function: ```python def count_sentences(s: str, dictionary: set) -> int: Given a string s and a set of valid words (dictionary), this function returns the number of ways to segment s such that every segment is a valid word in the dictionary. :param s: str - Input string to be segmented. :param dictionary: set - A set of valid words. :return: int - Number of ways to segment the string s using words in dictionary. pass ``` Input and Output Formats * **Input**: * `s` (1 <= `s` <= 1000): A string worth splitting. * `dictionary` (size <= 1000): A set containing valid dictionary words. * **Output**: * Return an integer denoting the number of possible segmentations. Constraints * The characters in the string `s` are all lowercase English letters. * The dictionary contains lowercase English words. * Performance is crucial; therefore, consider optimizing your solution to avoid time complexity bottlenecks discussed. Example ```python assert count_sentences("appletablet", {"apple", "tablet", "app", "let", "able", "t", "applet"}) == 4 assert count_sentences("thing", {"thing"}) == 1 assert count_sentences("thing", {"th", "ing", "thing"}) == 2 assert count_sentences("helloworld", {"hello", "world"}) == 1 ``` Hints * Consider using a dynamic programming approach to store results of already computed segmentations. * An iterative approach with a queue might also be beneficial to avoid deep recursion.
answer:def count_sentences(s: str, dictionary: set) -> int: Given a string s and a set of valid words (dictionary), this function returns the number of ways to segment s such that every segment is a valid word in the dictionary. :param s: str - Input string to be segmented. :param dictionary: set - A set of valid words. :return: int - Number of ways to segment the string s using words in dictionary. # Create a list to store the number of ways to segment s up to each index dp = [0] * (len(s) + 1) dp[0] = 1 # There's one way to segment an empty string for i in range(1, len(s) + 1): for j in range(i): if s[j:i] in dictionary: dp[i] += dp[j] return dp[len(s)]
question:# Comb Sort Implementation and Optimization Challenge You are provided with a basic implementation of the Comb Sort algorithm. This algorithm is known to sort an array in a manner somewhat similar to Bubble Sort by reducing the gap between compared elements over iterations. Task Your task is to implement the function `optimized_comb_sort` which will improve upon the given Comb Sort algorithm by using a more efficient gap reduction sequence. Specifically, you should implement the "Hibbard sequence" (1, 3, 7, ... up to the largest increment less than the length of the array). # Function Signature ```python def optimized_comb_sort(arr: List[int]) -> List[int]: pass ``` Input * An array `arr` of integers, with a maximum length of 10^5 and values ranging between -10^9 and 10^9. Output * A sorted list of integers in ascending order. Constraints * The algorithm should run efficiently with an average-case time complexity better than O(N^2). Example ```python >>> optimized_comb_sort([34, 8, 64, 51, 32, 21]) [8, 21, 32, 34, 51, 64] >>> optimized_comb_sort([3, 1, 2]) [1, 2, 3] >>> optimized_comb_sort([10, 9, 8, 7, 6, 5, 4, 3, 2, 1]) [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] ``` Notes * Ensure that your implementation handles edge cases such as empty arrays, single-element arrays, and arrays with all identical elements correctly. * Consider performance optimizations in your implementation to handle large arrays efficiently.
answer:from typing import List def optimized_comb_sort(arr: List[int]) -> List[int]: def next_gap(gap): # Using the Hibbard sequence `1, 3, 7, 15, ...` return (gap + 1) // 2 if gap > 1 else 0 n = len(arr) if n <= 1: return arr gap = n swapped = True while gap > 1 or swapped: gap = next_gap(gap) swapped = False for i in range(n - gap): if arr[i] > arr[i + gap]: arr[i], arr[i + gap] = arr[i + gap], arr[i] swapped = True return arr
question:# Context: You are working on a text generator that uses a Markov chain to predict the next word based on the current word. Given a text, you build a Markov chain to learn the transition probabilities between words. Your task is to implement a function that predicts the next sequence of words given an initial word and a Markov chain. # Input and Output Formats: - **Input**: - A Markov chain defined as a dictionary where each key is a word and the value is another dictionary containing possible next words and their transition probabilities. - An initial word as a string. - An integer `num_words` indicating the sequence length to generate. - **Output**: - A list of words representing the sequence generated from the initial word using the Markov chain. # Constraints & Limitations: - Ensure all probabilities in the chain dictionary sum to 1 for each word. - Handle the edge case where the initial word has no corresponding transitions. - Avoid infinite loops and ensure the function terminates practically. # Performance Requirements: - The function should run efficiently even for large Markov chains with thousands of states. # Function Signature: ```python def generate_sequence(chain: dict, initial_word: str, num_words: int) -> List[str]: pass ``` # Example: ```python my_chain = { 'A': {'A': 0.6, 'E': 0.4}, 'E': {'A': 0.7, 'E': 0.3} } sequence = generate_sequence(my_chain, 'A', 5) print(sequence) # Expected output might be: ['E', 'A', 'E', 'A', 'A'] # Note: Due to randomness, actual output may vary. ``` # Additional Notes: - You are not required to handle training the Markov chain from the text, only using the provided chain for generation. - Make sure to seed the random number generator for consistency when testing.
answer:import random def generate_sequence(chain: dict, initial_word: str, num_words: int) -> list: Generate a sequence of words from a Markov chain. Parameters: - chain: A dictionary representing the Markov chain with transition probabilities. - initial_word: The word to start the sequence from. - num_words: The number of words to generate for the sequence. Returns: - A list of words representing the generated sequence. if initial_word not in chain: return [] sequence = [initial_word] current_word = initial_word for _ in range(num_words - 1): if current_word not in chain or not chain[current_word]: break next_word = random.choices( population=list(chain[current_word].keys()), weights=list(chain[current_word].values()) )[0] sequence.append(next_word) current_word = next_word return sequence