Find the starting node in a directed graph which covers the maximum number of nodes

#tutorial #datastructure #graph #beginners

Given a directed graph with N number of nodes and exactly N number of edges. Each node has exactly one outgoing edge from it. Find a path such that the maximum number of nodes can be covered from the starting point, and return the starting point.

NOTE: A node can have multiple edges which are pointing towards him, but only one outgoing edge



Input: N = 5
1->2, 2->1, 3->1, 4->2, 5->3

Graph Structure:



Output: 5
Explanation: 
If we start from node 1 as beginning then the path is: 1 -> 2
If we start from node 2 as beginning then the path is: 2 -> 1
If we start from node 3 as beginning then the path is: 3 -> 1 -> 2
If we start from node 4 as beginning then the path is: 4 -> 2 -> 1
If we start from node 5 as beginning then path is: 5 -> 3 -> 1 -> 2

Hence, we can clearly see that if we start for 5, we cover the 
maximum number of nodes in the
graph i.e. 4.

Approach:

To find the number of nodes reachable from a particular node, one thing to observe is that we can reach from a node X to a node Y only when they are in the same connected component.

Since the graph is directional, any node in a connected component to any other node in the same connected components. Hence for a particular node X number of nodes that will be reachable will be the number of nodes in that particular component. Use a depth-first search to find the answer.

CODE
Following is a Java implementation of the code:



import java.util.HashSet;
import java.util.Set;

public class Main{

    static int[] graph;

    // Driver Function
    public static void main(String[] args) {

        // Taking the number of nodes from the user
        int n = 5;

        // Array to store the nodes direction
        graph = new int[n];

        // Initializing graph
        // 1->2, 2->1, 3->1, 4->2, 5->3
        graph[0] = 1;
        graph[1] = 0;
        graph[2] = 0;
        graph[3] = 1;
        graph[4] = 2;

        System.out.println(find(n));
    }

    static int find(int n) {

        int max = 0; // Holds the maximum count of nodes visited
        int node = 0; // Starting index of the node with maximum max count

        // Considering each node one-by-one as beginning node
        // Using DFS to fully explore that node
        for (int i = 0; i < n; i++) {

            // Finding the total number of node that can be covered by
            // considering the ith node as beginning
            int visits = canVisit(i);

            if (visits > max) { // If ith node covers more node then
                max = visits; // Store the number of nodes covered
                node = i; // Store the node index
            }
        }

        // As we are considering the indices from 0 just add 1 into the index
        return node + 1; // and return it
    }

    // Function to perform the DFS calculating the
    // count of the elements in a connected component
    static int canVisit(int n) {
        // This set contains the indices of the visited elements
        // This will help use to make sure that we are not running
        // inside a cycle in the graph
        Set<Integer> set = new HashSet<>();

        set.add(n); // Add the current node into the graph as it is visited

        // Go to the next node in the graph towards with the current directs
        int visit = graph[n];

        // Hold the total number of nodes visited
        // Since we at least visit the beginning node hence assign count to 1
        int count = 1;

        // Explore until the node repeats or we reach at the dead-end
        while (!set.contains(visit)) {
            set.add(visit); // Add the next visit into the set
            visit = graph[visit]; // Jump to next directed node
            count++; // Increment the count as one more has been explored
        }

        return count; // Return the total number of nodes visited
    }
}

Complexity:

We will have the worst case if the graph is linearly connected with no internal cycles, which will give us O(N²). With an Auxiliary space of O(N).