{"id":33468,"date":"2024-11-01T09:16:54","date_gmt":"2024-11-01T09:16:54","guid":{"rendered":"http:\/\/atmokpo.com\/w\/?p=33468"},"modified":"2024-11-01T11:38:34","modified_gmt":"2024-11-01T11:38:34","slug":"java-coding-test-course-finding-the-critical-path","status":"publish","type":"post","link":"https:\/\/atmokpo.com\/w\/33468\/","title":{"rendered":"Java Coding Test Course, Finding the Critical Path"},"content":{"rendered":"<div class=\"post\">\n<h2>Introduction<\/h2>\n<p>\n        In many companies related to software development, coding tests have become a mandatory part of the hiring process.<br \/>\n        In particular, Java is one of the preferred programming languages in many companies.<br \/>\n        In this post, we will take a closer look at how to solve the &#8216;Finding the Critical Path&#8217; problem using Java.<br \/>\n        The critical path refers to finding the longest path in a directed graph and plays an important role in scheduling in project management tools.\n    <\/p>\n<h2>Problem Description<\/h2>\n<p>\n        In the given directed graph, each node represents a task, and each edge indicates the dependencies between tasks.<br \/>\n        Our goal is to find the longest path from the starting node to the final node in the directed graph.<br \/>\n        For example, when the execution time of each task is given, the essence of the problem is to determine the maximum time it takes to complete the project.\n    <\/p>\n<h3>Input Format<\/h3>\n<ul>\n<li>Integer N: Number of tasks (1 \u2264 N \u2264 10,000)<\/li>\n<li>Integer M: Number of dependencies (1 \u2264 M \u2264 100,000)<\/li>\n<li>Execution time of each task: Given in the form of an array<\/li>\n<li>Dependency information: Given in the form of an array (a \u2192 b: Task a can only be performed after task b is completed)<\/li>\n<\/ul>\n<h3>Output Format<\/h3>\n<p>\n        Output the maximum time required to complete the final task.\n    <\/p>\n<h2>Example<\/h2>\n<pre>\n        **Input**:\n        5 4\n        [2, 3, 4, 1, 5]\n        [\n            (0, 1),\n            (1, 2),\n            (1, 3),\n            (3, 4)\n        ]\n        \n        **Output**:\n        10\n    <\/pre>\n<p>\n        In this example, the maximum path is based on the continuity of None: 2 (from 0 to 1) + 3 (from 1 to 2) + 5 (from 3 to 4), requiring a total time of 10 to complete the final task.\n    <\/p>\n<h2>Approach to the Problem<\/h2>\n<p>\n        To solve the &#8216;Finding the Critical Path&#8217; problem, we can utilize graph traversal algorithms.<br \/>\n        Generally, we will use Topological Sort to determine the order of tasks and then calculate the cumulative time for each node to determine the final time.<br \/>\n        The overall solution proceeds through the following steps:\n    <\/p>\n<ol>\n<li>Create the graph: Construct a directed graph connecting each task to its dependent tasks.<\/li>\n<li>Topological Sorting: Sort the dependent tasks to allow for sequential execution of tasks.<\/li>\n<li>Calculate time: Accumulate the time taken for each task to compute the time for the final target task.<\/li>\n<\/ol>\n<h3>Step 1: Create the Graph<\/h3>\n<p>\n        We create the graph using the inputted dependency information.<br \/>\n        Each task is viewed as a vertex, and edges are established according to their dependencies.<br \/>\n        In Java, we can represent the graph using an ArrayList.\n    <\/p>\n<pre>\n        List<List<Integer>> graph = new ArrayList<>();\n        int[] indegree = new int[N];\n        int[] time = new int[N];\n\n        for (int i = 0; i < N; i++) {\n            graph.add(new ArrayList<>());\n        }\n\n        \/\/ Construct graph based on dependency information\n        \/\/ Add edge from a to b through (a, b)\n        for (int[] dep : dependencies) {\n            graph.get(dep[0]).add(dep[1]);\n            indegree[dep[1]]++;\n            time[dep[0]] = taskTimes[dep[0]];\n        }\n    <\/List<Integer><\/pre>\n<h3>Step 2: Topological Sorting<\/h3>\n<p>\n        Once the graph is created, we start exploring the vertices using topological sorting, beginning with nodes that have an indegree of 0.<br \/>\n        This can typically be implemented using a queue.\n    <\/p>\n<pre>\n        Queue<Integer> queue = new LinkedList<>();\n        for (int i = 0; i < N; i++) {\n            if (indegree[i] == 0) {\n                queue.offer(i);\n            }\n        }\n\n        int[] dp = new int[N]; \/\/ Store completion time for each task\n        while (!queue.isEmpty()) {\n            int u = queue.poll();\n            dp[u] = Math.max(dp[u], time[u]);\n\n            for (int v : graph.get(u)) {\n                indegree[v]--;\n                dp[v] = Math.max(dp[v], dp[u] + time[v]);\n                if (indegree[v] == 0) {\n                    queue.offer(v);\n                }\n            }\n        }\n    <\/Integer><\/pre>\n<h3>Step 3: Calculate Time<\/h3>\n<p>\n        When we reach the final task, the maximum value in the dp array represents the maximum time required to complete the project.\n    <\/p>\n<pre>\n        int result = 0;\n        for (int t : dp) {\n            result = Math.max(result, t);\n        }\n        System.out.println(result);\n    <\/pre>\n<h2>Java Code<\/h2>\n<pre>\n        import java.util.*;\n\n        public class CriticalPath {\n            public static void main(String[] args) {\n                \/\/ Input processing\n                Scanner sc = new Scanner(System.in);\n                int N = sc.nextInt();\n                int M = sc.nextInt();\n                int[] taskTimes = new int[N];\n                for (int i = 0; i < N; i++) {\n                    taskTimes[i] = sc.nextInt();\n                }\n                List<List<Integer>> graph = new ArrayList<>();\n                int[] indegree = new int[N];\n\n                for (int i = 0; i < N; i++) {\n                    graph.add(new ArrayList<>());\n                }\n\n                for (int i = 0; i < M; i++) {\n                    int a = sc.nextInt();\n                    int b = sc.nextInt();\n                    graph.get(a).add(b);\n                    indegree[b]++;\n                }\n\n                \/\/ Topological sorting and time calculation\n                Queue<Integer> queue = new LinkedList<>();\n                for (int i = 0; i < N; i++) {\n                    if (indegree[i] == 0) {\n                        queue.offer(i);\n                    }\n                }\n\n                int[] dp = new int[N];\n                while (!queue.isEmpty()) {\n                    int u = queue.poll();\n                    dp[u] = Math.max(dp[u], taskTimes[u]);\n\n                    for (int v : graph.get(u)) {\n                        indegree[v]--;\n                        dp[v] = Math.max(dp[v], dp[u] + taskTimes[v]);\n                        if (indegree[v] == 0) {\n                            queue.offer(v);\n                        }\n                    }\n                }\n\n                \/\/ Calculate maximum time\n                int result = Arrays.stream(dp).max().getAsInt();\n                System.out.println(result);\n            }\n        }\n    <\/Integer><\/List<Integer><\/pre>\n<h2>Conclusion<\/h2>\n<p>\n        In this lecture, we covered the process of solving the critical path finding problem using Java.<br \/>\n        We learned how to represent the dependencies of each task using graph theory and to calculate the completion time of the final task.<br \/>\n        As such, a step-by-step approach is crucial in solving algorithmic problems, and this can lead to good results in actual coding tests.<br \/>\n        We will continue to write articles covering various algorithms, so please stay tuned for more.\n    <\/p>\n<h2>References<\/h2>\n<ul>\n<li>\n<a href=\"https:\/\/en.wikipedia.org\/wiki\/Longest_path_problem\">Longest Path Problem (Wikipedia)<\/a>\n<\/li>\n<li>\n<a href=\"https:\/\/www.geeksforgeeks.org\/find-longest-path-directed-acyclic-graph\/\">Find Longest Path in a Directed Acyclic Graph (GeeksforGeeks)<\/a>\n<\/li>\n<li>\n<a href=\"https:\/\/www.coursera.org\/learn\/algorithms-part1\">Coursera Algorithms Course<\/a>\n<\/li>\n<\/ul>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Introduction In many companies related to software development, coding tests have become a mandatory part of the hiring process. In particular, Java is one of the preferred programming languages in many companies. In this post, we will take a closer look at how to solve the &#8216;Finding the Critical Path&#8217; problem using Java. The critical &hellip; <a href=\"https:\/\/atmokpo.com\/w\/33468\/\" class=\"more-link\">\ub354 \ubcf4\uae30<span class=\"screen-reader-text\"> &#8220;Java Coding Test Course, Finding the Critical Path&#8221;<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[139],"tags":[],"class_list":["post-33468","post","type-post","status-publish","format-standard","hentry","category-java-coding-test"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.2 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Java Coding Test Course, Finding the Critical Path - \ub77c\uc774\ube0c\uc2a4\ub9c8\ud2b8<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/atmokpo.com\/w\/33468\/\" \/>\n<meta property=\"og:locale\" content=\"ko_KR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Java Coding Test Course, Finding the Critical Path - \ub77c\uc774\ube0c\uc2a4\ub9c8\ud2b8\" \/>\n<meta property=\"og:description\" content=\"Introduction In many companies related to software development, coding tests have become a mandatory part of the hiring process. 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