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Minimum edges to reverse to make path from a source to a destination

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  • Difficulty Level : Hard
  • Last Updated : 04 Mar, 2022
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Given a directed graph and a source node and destination node, we need to find how many edges we need to reverse in order to make at least 1 path from the source node to the destination node.

Examples:  

Minimum edges to reverse to make path from a source to a destination

In above graph there were two paths from node 0 to node 6,
0 -> 1 -> 2 -> 3 -> 6
0 -> 1 -> 5 -> 4 -> 6
But for first path only two edges need to be reversed, so answer will be 2 only.

This problem can be solved assuming a different version of the given graph. In this version we make a reverse edge corresponding to every edge and we assign that a weight 1 and assign a weight 0 to original edge. After this modification above graph looks something like below, 
 

modified graph

Now we can see that we have modified the graph in such a way that, if we move towards original edge, no cost is incurred, but if we move toward reverse edge 1 cost is added. So if we apply Dijkstra’s shortest path on this modified graph from given source, then that will give us minimum cost to reach from source to destination i.e. minimum edge reversal from source to destination. 

Below is the code based on above concept. 

C++




// C++ Program to find minimum edge reversal to get
// atleast one path from source to destination
#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
 
// This class represents a directed graph using
// adjacency list representation
class Graph
{
    int V;
    list<pair<int, int>> *graph;
 
public:
    // Constructor:
    Graph(int V)
    {
        this->V = V;
        graph = new list<pair<int, int>>[V];
    }
 
    // Adding edges into the graph:
    void addEdge(int u, int v, int w)
    {
        graph[u].push_back(make_pair(v, w));
    }
 
    // Returns shortest path from source to all other vertices.
    vector<int> shortestPath(int source)
    {
        // Create a set to store vertices that are being preprocessed
        set<pair<int, int>> setds;
 
        // Create a vector for distances and initialize all
        // distances as infinite (INF)
        vector<int> distance(V, INF);
 
        // Insert source itself in Set and initialize its distance as 0.
        setds.insert(make_pair(0, source));
        distance = 0;
 
        /* Looping till all shortest distance are finalized
           then setds will become empty */
        while (!setds.empty())
        {
            // The first vertex in Set is the minimum distance
            // vertex, extract it from set.
            pair<int, int> tmp = *(setds.begin());
            setds.erase(setds.begin());
 
            // vertex label is stored in second of pair (it
            // has to be done this way to keep the vertices
            // sorted distance (distance must be first item
            // in pair)
            int u = tmp.second;
 
            list<pair<int, int>>::iterator i;
            for (i = graph[u].begin(); i != graph[u].end(); ++i)
            {
                // Get vertex label and weight of current adjacent
                // of u.
                int v = (*i).first;
                int weight = (*i).second;
 
                //  If there is shorter path to v through u.
                if (distance[v] > distance[u] + weight)
                {
                    /*  If distance of v is not INF then it must be in
                        our set, so removing it and inserting again
                        with updated less distance.
                        Note : We extract only those vertices from Set
                        for which distance is finalized. So for them,
                        we would never reach here.  */
                    if (distance[v] != INF)
                        setds.erase(setds.find(make_pair(distance[v], v)));
 
                    // Updating distance of v
                    distance[v] = distance[u] + weight;
                    setds.insert(make_pair(distance[v], v));
                }
            }
        }
        return distance;
    }
 
    Graph modelGraphWithEdgeWeight(int edge[][2], int E, int V)
    {
        Graph g(V);
        for (int i = 0; i < E; i++)
        {
            // original edge : weight 0
            g.addEdge(edge[i][0], edge[i][1], 0);
 
            // reverse edge : weight 1
            g.addEdge(edge[i][1], edge[i][0], 1);
        }
        return g;
    }
 
    int getMinEdgeReversal(int edge[][2], int E, int V, int source, int destination)
    {
        // get modified graph with edge weight.
        Graph g = modelGraphWithEdgeWeight(edge, E, V);
 
        // distance vector stores shortest path.
        vector<int> dist = g.shortestPath(source);
 
        // If distance of destination is still INF then we cannot reach destination. Hence, not possible.
        if (dist[destination] == INF)
            return -1;
        else
            return dist[destination];
    }
};
 
int main()
{
    int V = 7;
    Graph g(V);
 
    int edge[][2] = {{0, 1}, {2, 1}, {2, 3}, {5, 1}, {4, 5}, {6, 4}, {6, 3}};
    int E = sizeof(edge) / sizeof(edge[0]);
 
    int minEdgeToReverse = g.getMinEdgeReversal(edge, E, V, 0, 6);
 
    if (minEdgeToReverse != -1)
        cout << minEdgeToReverse << endl;
    else
        cout << "Not Possible." << endl;
 
    return 0;
}

Java




// Java program to find minimum edge reversal to get
// atleast one path from source to destination
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.Iterator;
import java.util.List;
import java.util.Set;
 
class Pair
{
    int first, second;
 
    public Pair(int first, int second)
    {
        this.first = first;
        this.second = second;
    }
}
 
// This class represents a directed graph using
// adjacency list representation
class Graph{
     
final int INF = (int)0x3f3f3f3f;
 
// No. of vertices
int V;
 
// In a weighted graph, we need to store vertex
// and weight pair for every edge
List<Pair>[] adj;
 
// Allocates memory for adjacency list
@SuppressWarnings("unchecked")
public Graph(int V)
{
    this.V = V;
    adj = new ArrayList[V];
 
    for(int i = 0; i < V; i++)
    {
        adj[i] = new ArrayList();
    }
}
 
// Function adds a directed edge from
// u to v with weight w
void addEdge(int u, int v, int w)
{
    adj[u].add(new Pair(v, w));
}
 
// Prints shortest paths from
// src to all other vertices
int[] shortestPath(int src)
{
     
    // Create a set to store vertices
    // that are being preprocessed
    Set<Pair> setds = new HashSet<Pair>();
 
    // Create a vector for distances and
    // initialize all distances as infinite(INF)
    int[] dist = new int[V];
    Arrays.fill(dist, INF);
 
    // Insert source itself in Set and initialize
    // its distance as 0.
    setds.add(new Pair(0, src));
    dist[src] = 0;
 
    // Looping till all shortest distance are
    // finalized then setds will become empty
    while (!setds.isEmpty())
    {
         
        // The first vertex in Set is the minimum
        // distance vertex, extract it from set.
        Iterator<Pair> itr = setds.iterator();
        Pair tmp = itr.next();
        itr.remove();
 
        // Vertex label is stored in second of pair (it
        // has to be done this way to keep the vertices
        // sorted distance (distance must be first item
        // in pair)
        int u = tmp.second;
 
        // 'i' is used to get all adjacent
        // vertices of a vertex
        for(Pair p : adj[u])
        {
             
            // Get vertex label and weight of
            // current adjacent of u.
            int v = p.first;
            int weight = p.second;
 
            // If there is shorter path to v through u.
            if (dist[v] > dist[u] + weight)
            {
                 
                // If distance of v is not INF then it
                // must be in our set, so removing it
                // and inserting again with updated
                // less distance. Note : We extract
                // only those vertices from Set for
                // which distance is finalized. So
                // for them, we would never reach here.
                if (dist[v] != INF)
                {
                    setds.remove(new Pair(dist[v], v));
                }
                 
                // setds.erase(setds.find(new Pair(dist[v], v)));
 
                // Updating distance of v
                dist[v] = dist[u] + weight;
                setds.add(new Pair(dist[v], v));
            }
        }
    }
    return dist;
}
}
 
class GFG{
static final int INF = (int)0x3f3f3f3f;
 
// Function adds reverse edge of each original
// edge in the graph. It gives reverse edge
// a weight = 1 and all original edges a
// weight of 0. Now, the length of the
// shortest path will give us the answer.
// If shortest path is p: it means we
// used p reverse edges in the shortest path.
static Graph modelGraphWithEdgeWeight(int edge[][],
                                      int E, int V)
{
    Graph g = new Graph(V);
    for(int i = 0; i < E; i++)
    {
         
        // Original edge : weight 0
        g.addEdge(edge[i][0], edge[i][1], 0);
 
        // Reverse edge : weight 1
        g.addEdge(edge[i][1], edge[i][0], 1);
    }
    return g;
}
 
// Function returns minimum number of edges to be
// reversed to reach from src to dest
static int getMinEdgeReversal(int edge[][], int E,
                              int V, int src, int dest)
{
     
    // Get modified graph with edge weight
    Graph g = modelGraphWithEdgeWeight(edge, E, V);
 
    // Get shortes path vector
    int[] dist = g.shortestPath(src);
 
    // If distance of destination is still INF,
    // not possible
    if (dist[dest] == INF)
        return -1;
    else
        return dist[dest];
}
 
// Driver code
public static void main(String[] args)
{
    int V = 7;
    int edge[][] = { { 0, 1 }, { 2, 1 },
                     { 2, 3 }, { 5, 1 },
                     { 4, 5 }, { 6, 4 },
                     { 6, 3 } };
    int E = edge.length;
 
    int minEdgeToReverse = getMinEdgeReversal(
        edge, E, V, 0, 6);
     
    if (minEdgeToReverse != -1)
        System.out.println(minEdgeToReverse);
    else
        System.out.println("Not possible");
}
}
 
// This code is contributed by sanjeev2552

Python3




# Python3 Program to find minimum edge reversal to get
# atleast one path from source to destination
 
# method adds a directed edge from u to v with weight w
def addEdge(u, v, w):
    global adj
    adj[u].append((v, w))
 
# Prints shortest paths from src to all other vertices
def shortestPath(src):
   
    # Create a set to store vertices that are being
    # preprocessed
    setds = {}
 
    # Create a vector for distances and initialize all
    # distances as infinite (INF)
    dist = [10**18 for i in range(V)]
 
    # Insert source itself in Set and initialize its
    global adj
    setds[(0, src)] = 1
    dist[src] = 0
 
    # /* Looping till all shortest distance are finalized
 
    # then setds will become empty */
    while (len(setds) > 0):
       
        # The first vertex in Set is the minimum distance
        # vertex, extract it from set.
        tmp = list(setds.keys())[0]
        del setds[tmp]
 
        # vertex label is stored in second of pair (it
        # has to be done this way to keep the vertices
        # sorted distance (distance must be first item
        # in pair)
        u = tmp[1]
 
        # 'i' is used to get all adjacent vertices of a vertex
        # list< pair<int, int> >::iterator i;
        for i in adj[u]:
           
            # Get vertex label and weight of current adjacent
            # of u.
            v = i[0];
            weight = i[1]
 
            # If there is shorter path to v through u.
            if (dist[v] > dist[u] + weight):
               
                # /* If distance of v is not INF then it must be in
                #     our set, so removing it and inserting again
                #     with updated less distance.
                #     Note : We extract only those vertices from Set
                #     for which distance is finalized. So for them,
                #     we would never reach here. */
                if (dist[v] != 10**18):
                    del setds[(dist[v], v)]
 
                # Updating distance of v
                dist[v] = dist[u] + weight
                setds[(dist[v], v)] = 1
 
    return dist
 
# /* method adds reverse edge of each original edge
# in the graph. It gives reverse edge a weight = 1
# and all original edges a weight of 0. Now, the
# length of the shortest path will give us the answer.
# If shortest path is p: it means we used p reverse
# edges in the shortest path. */
def modelGraphWithEdgeWeight(edge, E, V):
    global adj
    for i in range(E):
       
        # original edge : weight 0
        addEdge(edge[i][0], edge[i][1], 0)
 
        # reverse edge : weight 1
        addEdge(edge[i][1], edge[i][0], 1)
 
# Method returns minimum number of edges to be
# reversed to reach from src to dest
def getMinEdgeReversal(edge, E, V,src, dest):
   
    # get modified graph with edge weight
    modelGraphWithEdgeWeight(edge, E, V)
 
    # get shortes path vector
    dist = shortestPath(src)
 
    # If distance of destination is still INF,
    # not possible
    if (dist[dest] == 10**18):
        return -1
    else:
        return dist[dest]
 
# Driver code
if __name__ == '__main__':
    V = 7
    edge = [[0, 1], [2, 1], [2, 3], [5, 1],[4, 5], [6, 4], [6, 3]]
    E, adj = len(edge), [[] for i in range(V + 1)]
    minEdgeToReverse = getMinEdgeReversal(edge, E, V, 0, 6)
    if (minEdgeToReverse != -1):
        print(minEdgeToReverse)
    else:
        print("Not possible")
 
        # This code is contributed by mohit kumar 29

Output: 

2

One more efficient approach to this problem would be by using 0-1 BFS concept

Below is the implementation of that algorithm:

Java




//Java code to find minimum edge reversal to get
//atleast one path from source to destination using 0-1 BFS
//Code By: Sparsh_CBS
import java.util.*;
 
class Node{
    private int val;
    private int weight;
    private Integer parent;
    Node(int val, int weight){
        this.val = val;
        this.weight = weight;
        parent = null;
    }
    //We have used the concept of parent to avoid
      //a child revisiting its parent and pushing it in
      //the deque during the 0-1 BFS
    Node(int val, int distance, Integer parent){
        this.val = val;
        this.weight = distance;
        this.parent = parent;
    }
 
    public int getVal(){
        return val;
    }
 
    public int getWeight(){
        return weight;
    }
 
    public Integer getParent(){
        return parent;
    }
}
 
public class Gfg{
    public static void main(String[] args) {
        List<List<Integer>> adj = new ArrayList<>();
        for(int i = 0; i < 7; i++)
            adj.add(new ArrayList<>());
 
        adj.get(0).add(1);
         
        adj.get(2).add(1);
         
        adj.get(5).add(1);
         
        adj.get(2).add(3);
         
        adj.get(6).add(3);
         
        adj.get(6).add(4);
         
        adj.get(4).add(5);
 
        int ans = getMinRevEdges(adj, 0, 6);
       
        if(ans == Integer.MAX_VALUE)
            System.out.println(-1);
        else
            System.out.println(ans);
    }
 
    private static int getMinRevEdges(List<List<Integer>> adj, int src, int dest){
        int n = adj.size();
 
        //Create the given graph into bidirectional graph
        List<List<Node>> newAdj = getBiDirectionalGraph(adj);
 
        //Now, Apply 0-1 BFS using Deque to get the shortest path
 
        //In the implementation, we will only add the
        //encountered node into the deque if and only if
        //the distance at which it was earlier explored was
        //strictly larger than the currently encountered distance
        Deque<Node> dq = new LinkedList<>();
 
        //Here Node is made up of : Node(int node_val, int node_distance, int node_parent)
        dq.offer(new Node(src,0,0));
        int[] dist = new int[n];
        //Set the distance of all nodes to infinity(Integer.MAX_VALUE)
        Arrays.fill(dist, Integer.MAX_VALUE);
        //set distance of source node as 0
        dist[src] = 0;
 
        while(!dq.isEmpty()){
            Node curr = dq.pollFirst();
            int currVal = curr.getVal();
            int currWeight = curr.getWeight();
            int currParent = curr.getParent();
            //If we encounter the destination node, we return
            if(currVal == dest)
                return currWeight;
            //Iterate over the neighbours of the current Node
            for(Node neighbourNode: newAdj.get(currVal)){
                int neighbour = neighbourNode.getVal();
                if(neighbour == currParent)
                    continue;
 
                int wt = neighbourNode.getWeight();
                if(wt == 0 && dist[neighbour] > currWeight){
                    dist[neighbour] = currWeight;
                    dq.offerFirst(new Node(neighbour,currWeight, currVal));
                }
                else if(dist[neighbour] > currWeight+wt){
                    dist[neighbour] = currWeight+wt;
                    dq.offerLast(new Node(neighbour, currWeight+wt, currVal));
                }
            }
        }
        return Integer.MAX_VALUE;
    }
   
      private static List<List<Node>> getBiDirectionalGraph(List<List<Integer>> adj){
        int n = adj.size();
        List<List<Node>> newAdj = new ArrayList<>();
         
        for(int i = 0; i < n; i++)
            newAdj.add(new ArrayList<>());
         
        boolean[] visited = new boolean[n];
        Queue<Integer> queue = new LinkedList<>();
 
        for(int i = 0; i < n; i++){
            if(!visited[i]){
                visited[i] = true;
                queue.offer(i);
                 
                while(!queue.isEmpty()){
                    int curr = queue.poll();
                    for(int neighbour: adj.get(curr)){
                        //original edges are to be assigned a weight of 0
                        newAdj.get(curr).add(new Node(neighbour, 0));
                        //make a fake edge and assign a weight of 1
                        newAdj.get(neighbour).add(new Node(curr, 1));
                       
                        if(visited[neighbour]){
                          //if the neighbour was visited, then dont
                         // add it again in the queue
                          continue;
                        }
                        visited[neighbour] = true;
                        queue.offer(neighbour);
                    }
                }
            }
        }
        return newAdj;
    }
}

Output: 

2

Time Complexity: O(V+E)

Space Complexity: O(V+2*E)

This article is contributed by Utkarsh Trivedi. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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