Graph – Software Installation Problem

by · March 18, 2018

Objective: Given a list of software’s which you need to install in a computer. Few software’s has dependency on other software’s in the list, means these software can be installed only when all of its dependent software’s are installed. Write an algorithm to print the sequence in which all the software’s in the list can be installed.

Example:

Software’s: A B C D E F
E depends on B, D
D depends on B, C
C depends on A
B depends on A
F no dependency
Output: F A B C D E

Approach: Topological Sorting

  • This problem is the classic example of “topological sorting”.
  • Let’s consider each software as Vertex and dependency between two software’s as Edge between two vertices. So for example B depends on A can be seen as A->B, A has a directed edge towards B. OR it can be read as B can be installed only when the A is installed.
  • Now we can draw a graph and do the topological sort which we have discussed here. So graph for example above is

See the code below for more understanding.

Time Complexity: O(N) n – Number of software’s.

Code:



Output:

Software Sequence:
F A B C D E

Asked in: Amazon

__________________________________________________
Top Companies Interview Questions..-

Google Microsoft Amazon Facebook more..

If you find anything incorrect or you feel that there is any better approach to solve the above problem, please write comment.
__________________________________________________

Tags:

  • lipsa patel

    If A is dependent on B and C then B and C should be installed before A is installed.
    So the order of the output should be E, D, C, B, A, F instead of F, A, B, C, D, E

    • tutorialhorizon

      Thanks lipsa. We have corrected the example description. The image and output is correct.

      • lipsa patel

        The output is correct but the image still doesn’t look correct.

        • tutorialhorizon

          Thanks Lipsa. I have updated the description in approach. Image is correct.

          So for example B depends on A can be seen as A->B, A has a directed edge towards B. OR it can be read as B can be installed only when the A is installed.

  • Archit

    @tutorialhorizon:disqus
    This is a Generic implementation of DirectedGraph . let me know what you think ..


    import java.util.*;
    import java.util.stream.Collectors;

    public class DirectedGraphGeneric {
    K[] vertices; // vertices
    List[] adjacentList; //adjacent lists
    Map positions = new HashMap();

    public DirectedGraphGeneric(K[] vertices) {
    this.vertices = vertices;
    adjacentList = new LinkedList[vertices.length];
    for (int i = 0; i < vertices.length; i++) {
    adjacentList[i] = new LinkedList();
    positions.put(vertices[i],i);
    }
    }

    public void addEdge(K from, K to) {
    int fromIndex = positions.get(from);
    adjacentList[fromIndex].add(to);
    }

    public void dfs(K start) {
    boolean[] seen = new boolean[vertices.length];
    Stack stack = new Stack();
    stack.push(start);
    seen[positions.get(start)] = true;

    while (!stack.isEmpty()) {
    K top = stack.pop();
    System.out.print(top + "->");
    int indexOfTop = positions.get(top);
    if(!adjacentList[indexOfTop].isEmpty()) {

    List unseen = adjacentList[indexOfTop].stream()
    .filter(i -> !seen[positions.get(i)])
    .collect(Collectors.toList());

    unseen.forEach(i -> {
    stack.push(i);
    int index = positions.get(i);
    seen[index] = true;
    });
    }
    }
    }

    public void topologicalSort(K[] nodes) {
    boolean[] seen = new boolean[nodes.length];
    Stack stack = new Stack();
    Arrays.stream(nodes).forEach(node -> {
    int index = positions.get(node);
    if(!seen[index]) {
    topologicalSortUtil(stack,seen,node);
    }
    });
    while (!stack.isEmpty()) {
    System.out.println(stack.pop() + "->");
    }
    }

    private void topologicalSortUtil(Stack stack, boolean[] seen, K node) {
    int index = positions.get(node);
    seen[index] = true;
    List list = adjacentList[index];

    for (int i = 0; i < list.size(); i++) {
    K curr = list.get(i);
    int position = positions.get(curr);

    if(!seen[position]) {
    topologicalSortUtil(stack,seen,curr);
    }
    }
    stack.push(node);
    }

    public void bfs(K start) {
    boolean[] seen = new boolean[vertices.length];

    Queue queue = new LinkedList();
    queue.add(start);
    seen[positions.get(start)] = true;

    while (!queue.isEmpty()) {
    K top = queue.poll();
    System.out.print(top + "->");
    int indexOfTop = positions.get(top);
    if(!adjacentList[indexOfTop].isEmpty()) {

    List unseen = adjacentList[indexOfTop].stream()
    .filter(i -> !seen[positions.get(i)])
    .collect(Collectors.toList());

    unseen.forEach(i -> {
    queue.add(i);
    int index = positions.get(i);
    seen[index] = true;
    });
    }
    }
    }

    public static void main(String[] args) {
    String[] vertices = {"A","B","C","D","E","F"};
    Integer[] v = {0,1,2,3,4,5};
    DirectedGraphGeneric g = new DirectedGraphGeneric(v);
    DirectedGraphGeneric graph = new DirectedGraphGeneric(vertices);
    graph.addEdge("A","B");
    graph.addEdge("A","C");
    graph.addEdge("B","D");
    graph.addEdge("B","E");
    graph.addEdge("C","D");
    graph.addEdge("D","E");
    g.addEdge(5, 2);
    g.addEdge(5, 0);
    g.addEdge(4, 0);
    g.addEdge(4, 1);
    g.addEdge(2, 3);
    g.addEdge(3, 1);
    g.topologicalSort(v);
    graph.topologicalSort(vertices);
    }
    }

Follow:

Select level

Beginner
Intermediate
Expert

Follow me on Twitter

%d bloggers like this: