This is a course where you can practice solving various problems for effective algorithm problem-solving. The problem we will tackle today is based on the theme ‘I Don’t Want to Be a Liar’. Through this problem, you will learn statistical thinking, conditional statements, and list processing techniques.<\/p>\n<\/header>\n \n Recently, a participant attended a wedding where he was elected as a slacker. However, there were 100 friends at the wedding whom he did not want to invite. These friends are planning to introduce others as their friends by lying.\n <\/p>\n \n Each friend can claim, ‘I know A and B’, about specific individuals. Your task is to verify all the lies and determine whether two friends know the same person.\n <\/p>\n \n The given input includes the number of friends and the lying information of each friend. Based on the input information, you need to determine which friends are lying.\n <\/p>\n<\/section>\n \n The first line contains the number of friends N, and the following N lines provide the numbers of the friends each friend knows.\n <\/p>\n<\/section>\n \n If the two friends know each other, print ‘Yes’; otherwise, print ‘No’.\n <\/p>\n<\/section>\n \n To solve this problem, we can use graph search algorithms. We represent each friend as a vertex and their relationships as edges to form a graph. We can use search methods such as DFS or BFS to check if two friends know each other.\n <\/p>\n \n The basic approach is to store friend relationships in a list and conduct a graph search for the given two friends.\n <\/p>\n<\/section>\n \n Below is the implementation of the code in Swift.\n <\/p>\n \n The above code operates as follows:\n <\/p>\n \n The method for solving this problem is quite intuitive as it involves basic DFS\/BFS searching. However, if the scope of the problem increases, we may need to explore more efficient methods. For example, if the number of friends becomes large, applying the Union-Find algorithm instead of DFS may be performance advantageous.\n <\/p>\n \n If more complex relationships or random configurations of data are provided, various techniques (e.g., Dynamic Programming options) can be utilized to address them. It is crucial to consider worst-case time and space complexity when selecting the algorithm design.\n <\/p>\n<\/section>\nProblem Description<\/h2>\n
Input Example<\/h2>\n
\n 5\n 1 2\n 2 3\n 3 4\n 4 5\n 1 5\n <\/pre>\n
Output Example<\/h2>\n
\n Yes\n <\/pre>\n
Approach to the Problem<\/h2>\n
Code Implementation<\/h2>\n
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\n import Foundation\n\n func canKnowEachOther(N: Int, friendships: [(Int, Int)], friendA: Int, friendB: Int) -> String {\n \/\/ Initialize graph\n var graph = Array(repeating: [Int](), count: N + 1)\n\n \/\/ Store friendships\n for (a, b) in friendships {\n graph[a].append(b)\n graph[b].append(a)\n }\n\n \/\/ Initialize BFS\n var queue = [friendA]\n var visited = Array(repeating: false, count: N + 1)\n visited[friendA] = true\n\n while !queue.isEmpty {\n let current = queue.removeFirst()\n\n \/\/ Did we find friend B?\n if current == friendB {\n return \"Yes\"\n }\n\n for neighbor in graph[current] {\n if !visited[neighbor] {\n visited[neighbor] = true\n queue.append(neighbor)\n }\n }\n }\n\n return \"No\"\n }\n\n \/\/ Input example\n let N = 5\n let friendships = [(1, 2), (2, 3), (3, 4), (4, 5), (1, 5)]\n let result = canKnowEachOther(N: N, friendships: friendships, friendA: 1, friendB: 5)\n print(result) \/\/ \"Yes\"\n <\/code>\n <\/pre>\n<\/section>\nCode Explanation<\/h2>\n
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Additional Considerations<\/h2>\n