/* 4.7 Build Order: You are given a list of projects and a list of dependencies (which is a list of pairs of
* projects, where the second project is dependent on the first project). All of a project's dependencies
* must be built before the project is. Find a build order that will allow the projects to be built. If there
* is no valid build order, return an error.
* EXAMPLE
* Input:
* projects: a, b, c, d, e, f
* dependencies: (a, d), (f, b), (b, d), (f, a), (d, c)
* Output: f, e, a, b, d, c
*
* 1- Build a graph with dependencies list for each project
* 2 - For each project in the graph we will perfom DFS.
* - We mark the current node as State.Partial and go deep;
* - When we hit the leaf, we mark the leaf as completed (avoid do again in next steps) and add it to the stack created to build as order as go
* add all the previus call to that
* PSEUDO LOGIC:
* if (project.getState() == Project.State.PARTIAL)
* {
* return false; // Cycle
* }
*
* if (project.getState() == Project.State.BLANK)
* {
* project.setState(Project.State.PARTIAL);
* foreach (var child in Project.children)
* {
* do DFS(child)
* }
* project.setState(Project.State.COMPLETE);
* stack.Push(project);
*/
public static CCIChallenges.BinaryTree.Graph BuildGraph(String[] projects, String[][] dependencies)
{
var graph = new CCIChallenges.BinaryTree.Graph();
foreach (var dependency in dependencies)
{
string first = dependency[0];
string second = dependency[1];
graph.addEdge(first, second);
}
return(graph);
}
public static Stack <Project> findBuildOrder(String[] projects, String[][] dependencies)
{
CCIChallenges.BinaryTree.Graph graph = BuildGraph(projects, dependencies);
return(OrderProjects(graph.getNodes()));
}
