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Shortest path in a complement graph

  • Difficulty Level : Hard
  • Last Updated : 28 Jan, 2021

Given an undirected non-weighted graph G. For a given node start return the shortest path that is the number of edges from start to all the nodes in the complement graph of G.

Complement Graph is a graph such that it contains only those edges which are not present in the original graph.

Examples:

Input: Undirected Edges = (1, 2), (1, 3), (3, 4), (3, 5), Start = 1 
Output: 0 2 3 1 1 
Explanation: 
Original Graph:



Complement Graph: 
 

The distance from 1 to every node in the complement graph are: 
1 to 1 = 0, 
1 to 2 = 2, 
1 to 3 = 3, 
1 to 4 = 1, 
1 to 5 = 1

Naive Approach: A Simple solution will be to create the complement graph and use Breadth-First Search on this graph to find the distance to all the nodes.
Time complexity: O(n2) for creating the complement graph and O(n + m) for breadth first search.
Efficient Approach: The idea is to use Modified Breadth-First Search to calculate the answer and then there is no need to construct the complement graph.

  • For each vertex or node, reduce the distance of a vertex which is a complement to the current vertex and has not been discovered yet.
  • For the problem, we have to observe that if the Graph is Sparse then the undiscovered nodes will be visited very fast.

Below is the implementation of the above approach:

C++




// C++ implementation to find the
// shortest path in a complement graph
 
#include <bits/stdc++.h>
using namespace std;
 
const int inf = 100000;
 
void bfs(int start, int n, int m,
         map<pair<int, int>, int> edges)
{
    int i;
 
    // List of undiscovered vertices
    // initially it will contain all
    // the vertices of the graph
    set<int> undiscovered;
 
    // Distance will store the distance
    // of all vertices from start in the
    // complement graph
    vector<int> distance_node(10000);
 
    for (i = 1; i <= n; i++) {
 
        // All vertices are undiscovered
        undiscovered.insert(i);
 
        // Let initial distance be infinity
        distance_node[i] = inf;
    }
 
    undiscovered.erase(start);
    distance_node[start] = 0;
    queue<int> q;
 
    q.push(start);
 
    // Check if queue is not empty and the
    // size of undiscovered vertices
    // is greater than 0
    while (undiscovered.size() && !q.empty()) {
        int cur = q.front();
        q.pop();
 
        // Vector to store all the complement
        // vertex to the current vertex
        // which has not been
        // discovered or visited yet.
        vector<int> complement_vertex;
 
        for (int x : undiscovered) {
           
            if (edges.count({ cur, x }) == 0 &&
                edges.count({ x, cur })==0)
                complement_vertex.push_back(x);
        }
        for (int x : complement_vertex) {
 
            // Check if optimal change
            // the distance of this
            // complement vertex
            if (distance_node[x]
                > distance_node[cur] + 1) {
                distance_node[x]
                    = distance_node[cur] + 1;
                q.push(x);
            }
 
            // Finally this vertex has been
            // discovered so erase it from
            // undiscovered vertices list
            undiscovered.erase(x);
        }
    }
    // Print the result
    for (int i = 1; i <= n; i++)
        cout << distance_node[i] << " ";
}
 
// Driver code
int main()
{
    // n is the number of vertex
    // m is the number of edges
    // start - starting vertex is 1
    int n = 5, m = 4;
 
    // Using edge hashing makes the
    // algorithm faster and we can
    // avoid the use of adjacency
    // list representation
    map<pair<int, int>, int> edges;
 
    // Initial edges for
    // the original graph
    edges[{ 1, 3 }] = 1,
                 edges[{ 3, 1 }] = 1;
    edges[{ 1, 2 }] = 1,
                 edges[{ 2, 1 }] = 1;
    edges[{ 3, 4 }] = 1,
                 edges[{ 4, 3 }] = 1;
    edges[{ 3, 5 }] = 1,
                 edges[{ 5, 3 }] = 1;
 
    bfs(1, n, m, edges);
 
    return 0;
}

Java




// Java implementation to find the
// shortest path in a complement graph
import java.io.*;
import java.util.*;
 
class GFG{
     
// Pair class is made so as to
// store the edges between nodes
static class Pair
{
    int left;
    int right;
     
    public Pair(int left, int right)
    {
        this.left = left;
        this.right = right;
    }
     
    // We need to override hashCode so that
    // we can use Set's properties like contains()
    @Override
    public int hashCode()
    {
        final int prime = 31;
        int result = 1;
        result = prime * result + left;
        result = prime * result + right;
        return result;
    }
 
    @Override
    public boolean equals( Object other )
    {
        if (this == other){return true;}
        if (other instanceof Pair)
        {
            Pair m = (Pair)other;
            return this.left == m.left &&
                  this.right == m.right;
        }
        return false;
    }
}
 
public static void bfs(int start, int n, int m,
                       Set<Pair> edges)
{
    int i;
 
    // List of undiscovered vertices
    // initially it will contain all
    // the vertices of the graph
    Set<Integer> undiscovered = new HashSet<>();
 
    // Distance will store the distance
    // of all vertices from start in the
    // complement graph
    int[] distance_node = new int[1000];
 
    for(i = 1; i <= n; i++)
    {
 
        // All vertices are undiscovered initially
        undiscovered.add(i);
 
        // Let initial distance be maximum value
        distance_node[i] = Integer.MAX_VALUE;
    }
     
    // Start is discovered
    undiscovered.remove(start);
     
    // Distance of the node to itself is 0
    distance_node[start] = 0;
     
    // Queue used for BFS
    Queue<Integer> q = new LinkedList<>();
 
    q.add(start);
 
    // Check if queue is not empty and the
    // size of undiscovered vertices
    // is greater than 0
    while (undiscovered.size() > 0 && !q.isEmpty())
    {
         
        // Current node
        int cur = q.peek();
        q.remove();
 
        // Vector to store all the complement
        // vertex to the current vertex
        // which has not been
        // discovered or visited yet.
        List<Integer>complement_vertex = new ArrayList<>();
 
        for(int x : undiscovered)
        {
            Pair temp1 = new Pair(cur, x);
            Pair temp2 = new Pair(x, cur);
             
            // Add the edge if not already present
            if (!edges.contains(temp1) &&
                !edges.contains(temp2))
            {
                complement_vertex.add(x);
            }
        }
         
        for(int x : complement_vertex)
        {
             
            // Check if optimal change
            // the distance of this
            // complement vertex
            if (distance_node[x] >
                distance_node[cur] + 1)
            {
                distance_node[x] =
                distance_node[cur] + 1;
                q.add(x);
            }
 
            // Finally this vertex has been
            // discovered so erase it from
            // undiscovered vertices list
            undiscovered.remove(x);
        }
    }
     
    // Print the result
    for(i = 1; i <= n; i++)
        System.out.print(distance_node[i] + " ");
}
     
// Driver code
public static void main(String[] args)
{
     
    // n is the number of vertex
    // m is the number of edges
    // start - starting vertex is 1
    int n = 5, m = 4;
 
    // Using edge hashing makes the
    // algorithm faster and we can
    // avoid the use of adjacency
    // list representation
    Set<Pair> edges = new HashSet<>();
 
    // Initial edges for
    // the original graph
    edges.add(new Pair(1, 3));
    edges.add(new Pair(3, 1));
    edges.add(new Pair(1, 2));
    edges.add(new Pair(2, 1));
    edges.add(new Pair(3, 4));
    edges.add(new Pair(4, 3));
    edges.add(new Pair(3, 5)) ;
    edges.add(new Pair(5, 3));
    Pair t = new Pair(1, 3);
 
    bfs(1, n, m, edges);
}
}
 
// This code is contributed by kunalsg18elec
Output: 
0 2 3 1 1

Time complexity: O(V+E) 
Auxiliary Space: O(V)




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