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question:Zara the novice coder has an array b of length n consisting of distinct positive integers. Zara loves sorting and permutation problems. She decided to challenge herself by trying to rearrange the numbers in the array b into a specific ordering criterion. She wants to create an array c by rearranging the elements of b such that the following conditions are met: 1. c is a permutation of b (contains the same elements as b). 2. For every i and j such that 1 ≤ i < j ≤ n, the value of c_i + c_j is not divisible by 2 (in other words, c_i + c_j should be odd). If it is possible to create such an array c, output "YES" and print the array c. Otherwise, output "NO". -----Input----- The first line of input contains an integer n (1 ≤ n ≤ 10^5) — the length of the array b. The second line contains n distinct positive integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^9). -----Output----- If it is possible to reorder the array to meet the specified conditions, print "YES" on the first line and the rearranged array c on the second line. If it is not possible, print "NO". -----Examples----- Input 5 1 3 2 4 5 Output YES 1 2 3 4 5 Input 2 2 4 Output NO -----Note----- In the first example, the reordered array c can be [1, 2, 3, 4, 5]. We can check: 1 + 2 = 3 (odd) 1 + 3 = 4 (even) 1 + 4 = 5 (odd) 1 + 5 = 6 (even) 2 + 3 = 5 (odd) 2 + 4 = 6 (even) 2 + 5 = 7 (odd) 3 + 4 = 7 (odd) 3 + 5 = 8 (even) 4 + 5 = 9 (odd) Thus, the condition 2 is met. In the second example, we cannot reorder the array to meet the condition.

answer:def reorder_array(n, b): This function reorders the array such that for every i < j, c_i + c_j is not divisible by 2. If it's possible, it returns "YES" and the reordered array, otherwise "NO". odd = [x for x in b if x % 2 == 1] even = [x for x in b if x % 2 == 0] # If both even and odd elements are present, returning them separated fulfills the requirement. if odd and even: return "YES", odd + even else: return "NO", [] # Example usage # n = 5 # b = [1, 3, 2, 4, 5] # print(reorder_array(n, b)) # Output: ("YES", [1, 3, 5, 2, 4])

question:You are given two strings, `S` and `T`, of length `N` and `M` respectively, both consisting of lowercase English letters. You need to find the length of the longest common subsequence between the given two strings. A subsequence is a sequence that appears in the same relative order, but not necessarily consecutively. For example, given `S = "abcde"` and `T = "ace"`, the longest common subsequence is "ace" with length 3. ------ Input Format ------ - The first line of input contains an integer `Q`, denoting the number of test cases. The `Q` test cases then follow: - Each test case consists of two lines: The first line contains the string `S`, and the second line contains the string `T`. ------ Output Format ------ For each test case, output in a single line the length of the longest common subsequence. ------ Constraints ------ 1 leq Q leq 100 1 leq N, M leq 1000 Strings `S` and `T` only contain lowercase English letters. ------ Sample Input 2 ------ 3 abcde ace xyz yz abcdef ghijkl ------ Sample Output 2 ------ 3 2 0 ------ Explanation 2 ------ Test case 1: The strings are `S = "abcde"` and `T = "ace"`. The longest common subsequence is "ace" which has a length of 3. Test case 2: The strings are `S = "xyz"` and `T = "yz"`. The longest common subsequence is "yz" which has a length of 2. Test case 3: The strings are `S = "abcdef"` and `T = "ghijkl"`. There are no common characters, hence the length of common subsequence is 0.

answer:def longest_common_subsequence(S, T): Returns the length of the longest common subsequence between strings S and T. N = len(S) M = len(T) dp = [[0] * (M + 1) for _ in range(N + 1)] for i in range(1, N + 1): for j in range(1, M + 1): if S[i - 1] == T[j - 1]: dp[i][j] = dp[i - 1][j - 1] + 1 else: dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]) return dp[N][M] def process_test_cases(Q, cases): results = [] for case in cases: S, T = case results.append(longest_common_subsequence(S, T)) return results

question:In a faraway land, there is a unique language consisting of lowercase Latin letters. A peculiar feature of this language is that certain pairs of letters are considered "friends" and each pair of friends can replace each other without changing the overall meaning of the message. You are given a string `s` and a list of friend pairs. Your task is to determine if it's possible to rearrange the string such that it becomes a palindrome by considering that the friends can be swapped with each other. A palindrome is a word that reads the same backward as forward. For example, "abba" or "racecar" are palindromes. Input The input consists of multiple lines: - The first line contains a single string `s` (1 ≤ |s| ≤ 200,000) consisting of lowercase Latin letters. - The second line contains an integer `m` (0 ≤ m ≤ 100,000), the number of friend pairs. - Each of the next `m` lines contains two different lowercase Latin letters `a` and `b` indicating that `a` and `b` are friends. Output Print "YES" if it is possible to rearrange the string to form a palindrome considering the friend pairs; otherwise, print "NO". Examples Input abac 2 a b c a Output YES Input abcfbc 2 a b c d Output NO Note In the first example, the string "abac" can be rearranged to form "acca" which is a palindrome by considering `a` and `b` as friends and `a` and `c` as friends. In the second example, regardless of how the friends `a` and `b` or `c` and `d` are swapped, a palindrome cannot be formed.

answer:def possible_to_rearrange_palindrome(s, friends): from collections import defaultdict, Counter def find(x): if parent[x] == x: return x parent[x] = find(parent[x]) return parent[x] def union(x, y): rootX = find(x) rootY = find(y) if rootX != rootY: parent[rootY] = rootX # Initialize each letter's parent as itself parent = {chr(i): chr(i) for i in range(ord('a'), ord('z') + 1)} # union sets based on friend relationships for a, b in friends: union(a, b) # Create a count for each character, considering their root representative char_count = Counter(find(c) for c in s) # To form a palindrome, at most 1 character's count can be odd odd_count = sum(1 for count in char_count.values() if count % 2 != 0) return "YES" if odd_count <= 1 else "NO"

question:Alice and Bob are playing a game with a sequence of integers. Initially, there are N integers in the sequence, and the i-th integer is A_i. They take turns alternately starting from Alice, and on each turn, a player can perform one of the following operations: - Choose an integer in the sequence and decrease it by 1. - Choose two different integers in the sequence and remove both of them. The game ends when all integers are zero or one integer remains. The player who makes the last move wins. Assuming both players play optimally, determine the winner of the game. -----Constraints----- - 1 leq N leq 200 - 1 leq A_i leq 10^6 - All input values are integers. -----Input----- Input is given from Standard Input in the following format: N A_1 A_2 ... A_N -----Output----- If Alice will win, print Alice; if Bob will win, print Bob. -----Sample Input----- 3 2 3 4 -----Sample Output----- Alice Initially, the sequence is [2, 3, 4]. - If Alice takes the 2 and the 3 and removes both, the sequence becomes [4]. - Then Bob will decrease the remaining 4 by 1, making it 3. - Then Alice will decrease the 3 by 1, making it 2. - Then Bob will decrease the 2 by 1, making it 1. - Then Alice will decrease the remaining 1 by 1, ending the game. Hence, Alice wins.

answer:def determine_winner(N, sequence): Determine the winner of the game given the initial sequence of integers. # Calculate the nim-sum (XOR of all elements) nim_sum = 0 for num in sequence: nim_sum ^= num # If nim-sum is zero, Bob wins, otherwise Alice wins if nim_sum == 0: return "Bob" else: return "Alice"