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question:Marge wants to create balanced teams from her class of students for a series of group projects. Each student has an assigned skill level represented by an integer between 1 and 100. Marge believes that teams are well balanced if the difference in skill levels between the most skilled student and the least skilled student in any team is no greater than 10. Marge must form teams of exactly three students. Given the skill levels of the students, help Marge determine if it's possible to form one or more well-balanced teams. ------ Input Format ------ The first line contains an integer N (3 ≤ N ≤ 100), the number of students in the class. The second line contains N space-separated integers, each representing the skill level of a student. ------ Output Format ------ Print “YES” if at least one well-balanced team can be formed, otherwise print “NO”. ------ Constraints ------ 1 ≤ Skill level of each student ≤ 100 ------ Sample Input 1 ------ 5 12 15 10 22 25 ------ Sample Output 1 ------ YES ------ Explanation 1 ------ One possible team with skill levels [10, 12, 15] can be formed, which has a maximum skill difference of 5. ------ Sample Input 2 ------ 4 10 25 40 55 ------ Sample Output 2 ------ NO ------ Explanation 2 ------ No team of three students can be formed where the difference in skill levels is 10 or less.

answer:def can_form_balanced_teams(N, skill_levels): Determine if at least one well-balanced team can be formed. :param N: Number of students :param skill_levels: List of skill levels of the students :return: "YES" if at least one well-balanced team can be formed, otherwise "NO" skill_levels.sort() for i in range(N - 2): if skill_levels[i+2] - skill_levels[i] <= 10: return "YES" return "NO"

question:BuySellStockCompany is a small startup working on stock market analysis tools. They want to create a new feature that helps users identify the best days to buy and sell stocks to maximize their profit. Given an array of stock prices for N consecutive days, you need to determine the maximum profit that can be achieved. However, you are allowed to complete at most one transaction – buying and then later selling one share of the stock. If there is no possibility to achieve any profit, return 0. Input Format The first line contains an integer N, the number of days. The next line contains N integers, the prices of the stock on those days. Output Format Print the maximum profit that can be achieved. If no profit can be achieved, print 0. Constraints 1 ≤ N ≤ 10^5 0 ≤ prices[i] ≤ 10^9 Sample Input 6 7 1 5 3 6 4 Sample Output 5 Explanation The maximum profit can be achieved by buying on day 2 (price = 1) and selling on day 5 (price = 6), for a profit of 6 - 1 = 5.

answer:def max_profit(stock_prices): Returns the maximum profit that can be achieved from buying and selling one share of stock. If no profit can be achieved, return 0. if not stock_prices: return 0 min_price = float('inf') max_profit = 0 for price in stock_prices: if price < min_price: min_price = price elif price - min_price > max_profit: max_profit = price - min_price return max_profit

question:You've been tasked with analyzing an undirected graph consisting of `n` nodes and `m` edges. Each node is identified by a unique integer from `1` to `n`. To assist with your analysis, you've been provided data on various roads connecting these nodes. The objective is to find the shortest path between any two nodes using Breadth First Search (BFS). Your task is to implement an algorithm that will calculate the shortest path (in terms of number of edges) from a specified starting node to all other nodes in the graph. # Input The input consists of a single test case. - The first line contains two integers `n` and `m` (1 ≤ n ≤ 1000, 0 ≤ m ≤ 100000) — the number of nodes and the number of edges. - The next `m` lines describe the edges. Each edge is represented by two integers `u` and `v` (1 ≤ u, v ≤ n), indicating that there is an undirected edge between nodes `u` and `v`. - The last line contains a single integer `s` (1 ≤ s ≤ n) — the starting node for BFS. # Output Output `n` integers on a single line, where the `i-th` integer represents the shortest path from the starting node `s` to the node `i` (1-based index). If a node `i` is not reachable from `s`, the output should contain `-1` for that node. # Example Input ``` 6 7 1 2 1 3 2 4 2 5 3 5 4 6 5 6 1 ``` Output ``` 0 1 1 2 2 3 ``` # Explanation In the example, the shortest path from node `1`: - To node `1` is `0` (the starting node itself); - To node `2` is `1` (direct edge 1-2); - To node `3` is `1` (direct edge 1-3); - To node `4` is `2` (path 1-2-4); - To node `5` is `2` (path 1-2-5); - To node `6` is `3` (path 1-2-4-6 or 1-3-5-6).

answer:from collections import deque, defaultdict def shortest_path_bfs(n, m, edges, s): graph = defaultdict(list) for u, v in edges: graph[u].append(v) graph[v].append(u) distances = [-1] * (n + 1) # Initialize distances as -1 for all nodes distances[s] = 0 # Distance to start node is 0 queue = deque([s]) while queue: current = queue.popleft() for neighbor in graph[current]: if distances[neighbor] == -1: # If not visited distances[neighbor] = distances[current] + 1 queue.append(neighbor) return distances[1:] # Return the distances starting from node 1

question:Your task is to implement a function `rainwater_trap` that calculates the total amount of water that can be trapped after a rainstorm, given an elevation map represented by a list of non-negative integers. The integers in the list represent the height of the terrain at that point. Water can only be trapped in the valleys formed between the peaks of the terrain. **Input:** - A list of non-negative integers representing the heights of the terrain. **Output:** - An integer representing the total amount of trapped water. **Constraints:** - The length of the list will be between 1 and 1000. - The height of each terrain point will be between 0 and 1000. **Example:** ```python def rainwater_trap(heights): # Your implementation here pass # Input heights = [0, 1, 0, 2, 1, 0, 1, 3, 2, 1, 2, 1] # Output print(rainwater_trap(heights)) # should output 6 heights = [4, 2, 0, 3, 2, 5] # Output print(rainwater_trap(heights)) # should output 9 ``` **Explanation:** * For the first example: - At height 0 index, no trapping. - Between 1 index and 3 index, 1 unit is trapped. - Between 3 index and 7 index, 2 units are trapped. - Between 7 index and 11 index, 3 units are trapped. * For the second example: - At height 4 index, no trapping. - Between 4 index and 6 index, 5 units are trapped. - Between 6 index and 8 index, 2 units are trapped. - Between 8 index and 11 index, 2 units are trapped.

answer:def rainwater_trap(heights): if not heights: return 0 n = len(heights) left_max = [0] * n right_max = [0] * n # Fill left_max array left_max[0] = heights[0] for i in range(1, n): left_max[i] = max(left_max[i-1], heights[i]) # Fill right_max array right_max[n-1] = heights[n-1] for i in range(n-2, -1, -1): right_max[i] = max(right_max[i+1], heights[i]) # Calculate the trapped water trapped_water = 0 for i in range(n): trapped_water += min(left_max[i], right_max[i]) - heights[i] return trapped_water