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question:def longest_common_subsequence(seq1, seq2): Computes the length of the longest common subsequence between two sequences. Args: seq1 (List[int]): The first sequence of integers. seq2 (List[int]): The second sequence of integers. Returns: int: The length of the longest common subsequence. >>> longest_common_subsequence([1, 3, 4, 1, 2, 5], [1, 4, 1, 2, 5]) 5 >>> longest_common_subsequence([9, 8, 7], [5, 6, 7, 8]) 1

answer:def longest_common_subsequence(seq1, seq2): Computes the length of the longest common subsequence between two sequences. n1, n2 = len(seq1), len(seq2) # Create a 2D dp array with (n1+1)x(n2+1) dimensions dp = [[0] * (n2 + 1) for _ in range(n1 + 1)] # Fill the dp array for i in range(1, n1 + 1): for j in range(1, n2 + 1): if seq1[i - 1] == seq2[j - 1]: dp[i][j] = dp[i - 1][j - 1] + 1 else: dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]) # The value in dp[n1][n2] will be the length of the longest common subsequence return dp[n1][n2]

question:def first_non_repeating_character(s: str) -> str: Returns the first non-repeating character in a given string s. If there is no non-repeating character, returns 'None'. >>> first_non_repeating_character("a") 'a' >>> first_non_repeating_character("abbcdde") 'a' >>> first_non_repeating_character("aabbccefd") 'e' >>> first_non_repeating_character("aabbccd") 'd' >>> first_non_repeating_character("aabbcc") 'None'

answer:def first_non_repeating_character(s): Returns the first non-repeating character in a given string s. If there is no non-repeating character, returns 'None'. char_count = {} # Count occurrences of each character in the string for char in s: if char in char_count: char_count[char] += 1 else: char_count[char] = 1 # Find the first character in the string with a count of 1 for char in s: if char_count[char] == 1: return char return 'None'

question:def min_operations_to_equal_subsequence(A, N, K): Find the minimum number of operations required to make the array have at least one subsequence of length K such that all elements in the subsequence are equal. >>> min_operations_to_equal_subsequence([1, 2, 2, 2, 3], 5, 3) 0 >>> min_operations_to_equal_subsequence([1, 2, 3, 4, 5, 6, 7], 7, 5) 4 >>> min_operations_to_equal_subsequence([4, 4, 4, 4], 4, 2) 0 pass def solve(test_cases): Given multiple test cases, find and return the minimum number of operations required for each test case to achieve a subsequence of specified length K with equal elements. >>> test_cases = [(5, 3, [1, 2, 2, 2, 3]), (7, 5, [1, 2, 3, 4, 5, 6, 7]), (4, 2, [4, 4, 4, 4])] >>> solve(test_cases) [0, 4, 0] pass

answer:def min_operations_to_equal_subsequence(A, N, K): from collections import Counter count = Counter(A) max_frequency = max(count.values()) if max_frequency >= K: return 0 return K - max_frequency def solve(test_cases): results = [] for N, K, A in test_cases: result = min_operations_to_equal_subsequence(A, N, K) results.append(result) return results

question:def smallest_length_after_removals(s: str) -> int: Determines the smallest possible length of the string after removing palindrome substrings any number of times. >>> smallest_length_after_removals("a") 0 >>> smallest_length_after_removals("abba") 0 >>> smallest_length_after_removals("racecar") 0 >>> smallest_length_after_removals("abacdc") 1 >>> smallest_length_after_removals("abcde") 1 >>> smallest_length_after_removals("ababab") 1 >>> smallest_length_after_removals("aabbccddeeffgg") 1 pass

answer:def smallest_length_after_removals(s): Determines the smallest possible length of the string after removing palindrome substrings any number of times. # If the whole string is a palindrome, we can remove it entirely to get an empty string if s == s[::-1]: return 0 # If the entire string is not a palindrome, the smallest possible length after all removals is 1 return 1