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question:In the bustling town of Algorithmia, there is a popular marketplace where merchants from all over the land come to sell their goods. Each merchant has a unique stall where they sell a certain kind of fruit. The marketplace is lined up in a straight row of n stalls, and each stall has a sign indicating the type of fruit sold there, such as apples, oranges, bananas, etc. As part of a festival, the mayor of Algorithmia has decided to organize a raffle where the townspeople can win prizes. To participate, each person must choose two different stalls (X and Y) such that the total number of distinct types of fruits between those stalls, including the fruits at stalls X and Y, is maximized. However, due to the high number of stalls, you need to help the mayor by writing a program that determines the maximum number of distinct fruit types between any two different stalls. Note that if X is before Y, you must consider all stalls from X to Y (inclusive) and count the number of distinct types of fruit in those stalls. Input The first line contains an integer n (2 ≤ n ≤ 10^5) — the number of stalls in the marketplace. The second line contains n integers f_1, f_2, ..., f_n (1 ≤ f_i ≤ 10^5) — the types of fruit sold at each stall. Output Print a single integer — the maximum number of distinct fruit types between any two different stalls in the marketplace. Examples Input 5 1 2 1 3 2 Output 3 Input 6 5 5 5 5 1 1 Output 2 Note In the first example, if we choose stalls 1 and 4, the total number of distinct types of fruits from stall 1 to stall 4 is 3 (types 1, 2, and 3). In the second example, no matter which pairs of stalls we choose, the maximum number of distinct types of fruits between any two stalls is 2 (type 5 and type 1).
answer:def max_distinct_fruits(n, fruits): Returns the maximum number of distinct fruit types between any two different stalls. max_distinct = 0 for i in range(n): seen = set() for j in range(i, n): seen.add(fruits[j]) max_distinct = max(max_distinct, len(seen)) return max_distinct # Example usage: # n = 5 # fruits = [1, 2, 1, 3, 2] # print(max_distinct_fruits(n, fruits)) # Output: 3
question:You are given an array of n integers where each element represents the cost of producing a single item. Your goal is to determine the maximum profit you can obtain by selling exactly k items from this set, subject to the condition that you can choose the selling price of each item as long as the selling price is not lower than the production cost of the item. To achieve the maximum profit, you can increase each item's selling price. The profit for each item is defined as the selling price minus the production cost. Your task is to find out the maximum total profit you can achieve by strategically choosing k items to sell at their respective optimal selling prices. Input The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. The first line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 100) — the number of items and the number of items to be sold. The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the array of production costs. Output For each test case, print a single integer — the maximum total profit you can achieve by selling exactly k items. Example Input 3 5 3 4 3 2 5 1 4 2 10 10 10 10 6 5 1 1 1 1 1 1 Output 9 0 0 Explanation For the first test case: - If you select the items with production costs 1, 2, and 3, and increase their selling prices to 5, 6, and 7 respectively, the achievable profits would be: - (5 - 1) = 4 - (6 - 2) = 4 - (7 - 3) = 4 However, the actual maximum profit for choosing 3 items would be obtained by: - Selecting items 1, 2, and 5 and setting their selling prices to 11, 10, and 9, where profit would be: - (11 - 4) = 7 - (10 - 3) = 7 - (9 - 1) = 8 But since you select the numbers 1, 2, and 3 and their increment is very close to each other to maximize the collective profit. For the second test case, all production costs are equal, so there is no additional profit to be gained apart from the production cost. For the third case, all items have the same cost, selling at cost hence no additional profit can be itemized through incremented price.
answer:def max_profit(t, test_cases): results = [] for test in test_cases: n, k, costs = test costs.sort() max_possible_profit = sum(costs[-k:]) - sum(costs[:k]) results.append(max_possible_profit) return results # Helper function to parse input and handle multiple test cases def handle_inputs(input_data): input_lines = input_data.strip().split("n") t = int(input_lines[0]) test_cases = [] line_idx = 1 for _ in range(t): n, k = map(int, input_lines[line_idx].split()) costs = list(map(int, input_lines[line_idx + 1].split())) test_cases.append((n, k, costs)) line_idx += 2 return t, test_cases # Example usage input_data = 3 5 3 4 3 2 5 1 4 2 10 10 10 10 6 5 1 1 1 1 1 1 t, test_cases = handle_inputs(input_data) print(max_profit(t, test_cases)) # Output: [6, 0, 0]
question:Given an array of integers and a sequence of rotate operations, you are to determine the state of the array after applying all the operations in order. There are three types of rotate operations: 1. "L k" - Rotate the array to the left by k positions. 2. "R k" - Rotate the array to the right by k positions. 3. "Q" - Output the current state of the array. Each operation in the sequence should be executed exactly once and in the given order. For the rotate operations, an element that is rotated off the end of the array reappears at the opposite end. For example, rotating the array [1, 2, 3, 4, 5] to the left by 2 positions results in [3, 4, 5, 1, 2]. Input The first line contains an integer n (1 ≤ n ≤ 10^5) - the number of elements in the array. The second line contains n integers separated by spaces, representing the initial state of the array. The third line contains an integer m (1 ≤ m ≤ 10^5) - the number of operations. The fourth line contains m operations, each separated by a space. Each operation is of the form "L k", "R k", or "Q". Output For each "Q" operation in the sequence, output the current state of the array as a space-separated list of integers on a new line. Example Input 5 1 2 3 4 5 6 L 2 Q R 1 Q L 3 Q Output 3 4 5 1 2 2 3 4 5 1 5 1 2 3 4 Note Initially, the array is [1, 2, 3, 4, 5]. After "L 2", the array becomes [3, 4, 5, 1, 2]. The first "Q" operation outputs this array as "3 4 5 1 2". After "R 1", the array becomes [2, 3, 4, 5, 1]. The second "Q" operation outputs this array as "2 3 4 5 1". After "L 3", the array becomes [5, 1, 2, 3, 4]. The third "Q" operation outputs this array as "5 1 2 3 4".
answer:def execute_operations(n, array, m, operations): result = [] for operation in operations: if operation[0] == 'L': k = int(operation[1]) k = k % n array = array[k:] + array[:k] elif operation[0] == 'R': k = int(operation[1]) k = k % n array = array[-k:] + array[:-k] elif operation[0] == 'Q': result.append(" ".join(map(str, array))) return result
question:Implement a function to compute the minimum number of operations required to convert a given string S1 into another string S2. The allowed operations are: 1. Insert a character 2. Delete a character 3. Replace a character This is a classic problem in computer science known as the Edit Distance or Levenshtein Distance. Input The input consists of two lines. - The first line contains the string S1. - The second line contains the string S2. Both strings consist of lowercase English letters and have lengths between 1 and 2000. Output Return the minimum number of operations required to convert S1 into S2. Examples Input kitten sitting Output 3 Input flaw lawn Output 2 Input intention execution Output 5 Note: In the first example, the sequence of operations could be: 1. Replace 'k' with 's' 2. Replace 'e' with 'i' 3. Insert 'g' at the end In the second example, the sequence of operations could be: 1. Replace 'f' with 'l' 2. Insert 'n' at the end
answer:def min_edit_distance(S1, S2): Returns the minimum number of operations required to convert string S1 into string S2. len1, len2 = len(S1), len(S2) # Create a 2D array to store results of subproblems dp = [[0 for j in range(len2 + 1)] for i in range(len1 + 1)] # Fill dp array for i in range(len1 + 1): for j in range(len2 + 1): if i == 0: dp[i][j] = j # Minimum operations = j (all insertions) elif j == 0: dp[i][j] = i # Minimum operations = i (all deletions) elif S1[i - 1] == S2[j - 1]: dp[i][j] = dp[i - 1][j - 1] else: dp[i][j] = 1 + min(dp[i - 1][j], # Delete dp[i][j - 1], # Insert dp[i - 1][j - 1]) # Replace return dp[len1][len2]