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

Last Updated : 25 Apr, 2023

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:

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

C#

using System;
using System.Collections.Generic;
// C# program to find minimum edge reversal to get
// atleast one path from source to destination
namespace ConsoleApp
{
     
     
// This class represents a directed graph using
// adjacency list representation
    class Pair
    {
        public int First { get; set; }
        public int Second { get; set; }
 
        public Pair(int first, int second)
        {
            First = first;
            Second = second;
        }
    }
 
    class Graph
    {
        private const int INF = 0x3f3f3f3f;
         
// No. of vertices
        public int V { get; }
        // In a weighted graph, we need to store vertex
// and weight pair for every edge
        private List<Pair>[] adj;
// Allocates memory for adjacency list
        public Graph(int V)
        {
            this.V = V;
            adj = new List<Pair>[V];
 
            for (int i = 0; i < V; i++)
            {
                adj[i] = new List<Pair>();
            }
        }
// Function adds a directed edge from
// u to v with weight w
        public void addEdge(int u, int v, int w)
        {
            adj[u].Add(new Pair(v, w));
        }
// Prints shortest paths from 
// src to all other vertices
        public int[] shortestPath(int src)
        {
                // Create a set to store vertices
    // that are being preprocessed
            var setds = new HashSet<Pair>();
            // Create a vector for distances and 
    // initialize all distances as infinite(INF)
            int[] dist = new int[V];
            Array.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.Count > 0)
            {
                 // The first vertex in Set is the minimum
        // distance vertex, extract it from set.
                var itr = setds.GetEnumerator();
                itr.MoveNext();
                 // 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)
                var tmp = itr.Current;
                setds.Remove(tmp);
 
                int u = tmp.Second;
 
        // 'i' is used to get all adjacent 
        // vertices of a vertex
                foreach (var p in 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
    {
        private const int INF = 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)
        {
            var g = new Graph(V);
            for (int i = 0; i < E; i++)
            { // Original edge : weight 0
                g.addEdge(edge[i][0], edge[i][1], 0);
                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)
        {
            var g = modelGraphWithEdgeWeight(edge, E, V);
            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
        static void Main(string[] args)
        {
            int V = 7;
            int[][] edge = { new [] { 0, 1 }, new [] { 2, 1 },
                             new [] { 2, 3 }, new [] { 5, 1 },
                             new [] { 4, 5 }, new [] { 6, 4 },
                             new [] { 6, 3 } };
            int E = edge.Length;
 
            int minEdgeToReverse = getMinEdgeReversal(
                edge, E, V, 0, 6);
 
            Console.WriteLine(minEdgeToReverse);
        }
    }
}

Javascript

// Javascript Program to find minimum edge reversal to get
// atleast one path from source to destination
 
      let INF = Number.MAX_VALUE;
 
      // This class represents a directed graph using
      // adjacency list representation
      class Graph {
        // Constructor:
        constructor(V) {
          this.V = V;
          this.graph = Array.from(Array(V), () => new Array());
        }
 
        // Adding edges into the graph:
        addEdge(u, v, w) {
          this.graph[u].push([v, w]);
        }
 
        // Returns shortest path from source to all other vertices.
        shortestPath(source) {
          // Create a set to store vertices that are being preprocessed
 
          let setds = new Set();
 
          // Create a vector for distances and initialize all
          // distances as (INF)
 
          let distance = new Array(V);
          distance.fill(INF);
 
          // Insert source itself in Set and initialize its distance as 0.
          setds.add([0, source]);
          distance = 0;
 
          /* Looping till all shortest distance are finalized
                 then setds will become empty */
          while (setds.size != 0) {
            // The first vertex in Set is the minimum distance
            // vertex, extract it from set.
            let it = setds.values();
            //get first entry:
            let tm = it.next();
            let tmp = tm.value;
 
            setds.delete(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)
            let u = tmp[1];
 
            for (let j in this.graph[u]) {
              // Get vertex label and weight of current adjacent
              // of u.
              let i = this.graph[u][j];
              let v = i[0];
              let weight = i[1];
 
              //  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.delete([distance[v], v]);
 
                // Updating distance of v
                distance[v] = distance[u] + weight;
 
                setds.add([distance[v], v]);
              }
            }
          }
          return distance;
        }
 
        modelGraphWithEdgeWeight(edge, E, V) {
          let g = new Graph(V);
          for (let 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;
        }
 
        getMinEdgeReversal(edge, E, V, source, destination) {
          // get modified graph with edge weight.
          let g = this.modelGraphWithEdgeWeight(edge, E, V);
 
          // distance vector stores shortest path.
          let 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];
        }
      }
 
      let V = 7;
      let g = new Graph(V);
 
      let edge = [
        [0, 1],
        [2, 1],
        [2, 3],
        [5, 1],
        [4, 5],
        [6, 4],
        [6, 3],
      ];
      let E = edge.length;
 
      let minEdgeToReverse = g.getMinEdgeReversal(edge, E, V, 0, 6);
      if (minEdgeToReverse != -1) console.log(minEdgeToReverse);
      else console.log("Not Possible.");

Output

Time complexity:
Therefore, the overall time complexity of the program is O(E*log(V)).

Space complexity:
Therefore, the overall space complexity of the program is O(E + V).

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

Below is the implementation of that algorithm:

C++

// C++ code to find minimum edge reversal to get
// atleast one path from source to destination using 0-1 BFS
// Code By: Sparsh_CBS
#include <bits/stdc++.h>
using namespace std;
 
// Creating a node
class Node {
private:
    int val;
    int weight;
    int parent;
 
public:
    Node(int val, int weight)
    {
        this->val = val;
        this->weight = weight;
        parent = -1;
    }
 
    // 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, int parent)
    {
        this->val = val;
        this->weight = distance;
        this->parent = parent;
    }
    int getVal() { return val; }
    int getWeight() { return weight; }
    int getParent() { return parent; }
};
 
int getMinRevEdges(vector<vector<int> >& adj, int src,
                   int dest)
{
    int n = adj.size();
 
    // Create the given graph into bidirectional graph
    vector<vector<Node> > newAdj(n);
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < adj[i].size(); j++) {
            int neighbour = adj[i][j];
            // original edges are to be assigned a weight of
            // 0
            newAdj[i].push_back(Node(neighbour, 0));
            // make a fake edge and assign a weight of 1
            newAdj[neighbour].push_back(Node(i, 1));
        }
    }
 
    // 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;
    // Here Node is made up of : Node(int node_val, int
    // node_distance, int node_parent)
    dq.push_front(Node(src, 0, -1));
    vector<int> dist(n, INT_MAX);
    // Set the distance of all nodes to infinity(INT_MAX)
    // set distance of source node as 0
    dist[src] = 0;
 
    while (!dq.empty()) {
        Node curr = dq.front();
        dq.pop_front();
        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 (auto neighbourNode : newAdj[currVal]) {
            int neighbour = neighbourNode.getVal();
            if (neighbour == currParent)
                continue;
 
            int wt = neighbourNode.getWeight();
            if (wt == 0 && dist[neighbour] > currWeight) {
                dist[neighbour] = currWeight;
                dq.push_front(
                    Node(neighbour, currWeight, currVal));
            }
            else if (dist[neighbour] > currWeight + wt) {
                dist[neighbour] = currWeight + wt;
                dq.push_back(Node(
                    neighbour, currWeight + wt, currVal));
            }
        }
    }
    return INT_MAX;
}
 
// Driver code
int main()
{
    vector<vector<int> > adj = { { 1 }, {},    { 1, 3 }, {},
                                 { 5 }, { 1 }, { 3, 4 } };
    int ans = getMinRevEdges(adj, 0, 6);
    if (ans == INT_MAX)
        cout << -1;
    else
        cout << ans;
    return 0;
}

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;
    }
}

Python3

# Python Code to find minimum edge reversal to get
# atleast one path from source to destination using 0-1 BFS
# Code By: Sparsh_CBS
 
 
class Node:
    def __init__(self, val, weight, parent=None):
        self.val = val
        self.weight = weight
        self.parent = parent
 
 
def getMinRevEdges(adj, src, dest):
    n = len(adj)
    # Create the given graph into bidirectional graph
    newAdj = getBiDirectionalGraph(adj)
    # Now, Apply 0-1 BFS using Deque to get the shortest path
    dq = []
    dq.append(Node(src, 0, 0))
    dist = [float("inf")] * n
    # Set the distance of all nodes to infinity(Integer.MAX_VALUE)
    dist[src] = 0
    # set distance of source node as 0
    while len(dq):
        curr = dq.pop(0)
        currVal = curr.val
        currWeight = curr.weight
        currParent = curr.parent
        # If we encounter the destination node, we return
        if currVal == dest:
            return currWeight
        # Iterate over the neighbours of the current Node
        for neighbourNode in newAdj[currVal]:
            neighbour = neighbourNode.val
            if neighbour == currParent:
                continue
            wt = neighbourNode.weight
            if wt == 0 and dist[neighbour] > currWeight:
                dist[neighbour] = currWeight
                dq.append(Node(neighbour, currWeight, currVal))
            elif dist[neighbour] > currWeight + wt:
                dist[neighbour] = currWeight + wt
                dq.append(Node(neighbour, currWeight + wt, currVal))
    return float("inf")
 
 
def getBiDirectionalGraph(adj):
    n = len(adj)
    newAdj = [[] for _ in range(n)]
    visited = [False] * n
    queue = []
    for i in range(n):
        if not visited[i]:
            visited[i] = True
            queue.append(i)
            while len(queue):
                curr = queue.pop(0)
                for neighbour in adj[curr]:
                    # original edges are to be assigned a weight of 0
                    newAdj[curr].append(Node(neighbour, 0))
                    # make a fake edge and assign a weight of 1
                    newAdj[neighbour].append(Node(curr, 1))
                    if visited[neighbour]:
                        continue  # if the neighbour was visited, then dont
                    # add it again in the queue
                    visited[neighbour] = True
                    queue.append(neighbour)
    return newAdj
 
 
if __name__ == "__main__":
    adj = [[] for _ in range(7)]
    adj[0].append(1)
    adj[2].append(1)
    adj[5].append(1)
    adj[2].append(3)
    adj[6].append(3)
    adj[6].append(4)
    adj[4].append(5)
 
    ans = getMinRevEdges(adj, 0, 6)
 
    if ans == float("inf"):
        print(-1)
    else:
        print(ans)

C#

using System;
using System.Collections.Generic;
 
class Node {
    private int val;
    private int weight;
    private int ? parent;
    public 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
    public Node(int val, int distance, int? parent)
    {
        this.val = val;
        this.weight = distance;
        this.parent = parent;
    }
 
    public int GetVal() { return val; }
 
    public int GetWeight() { return weight; }
 
    public int ? GetParent() { return parent; }
}
 
public class Gfg {
    public static void Main(string[] args)
    {
        List<List<int> > adj = new List<List<int> >();
        for (int i = 0; i < 7; i++)
            adj.Add(new List<int>());
 
        adj[0].Add(1);
 
        adj[2].Add(1);
 
        adj[5].Add(1);
 
        adj[2].Add(3);
 
        adj[6].Add(3);
 
        adj[6].Add(4);
 
        adj[4].Add(5);
 
        int ans = GetMinRevEdges(adj, 0, 6);
 
        if (ans == int.MaxValue)
            Console.WriteLine(-1);
        else
            Console.WriteLine(ans);
    }
 
    private static int GetMinRevEdges(List<List<int> > adj,
                                      int src, int dest)
    {
        int n = adj.Count;
 
        // 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
        Queue<Node> dq = new Queue<Node>();
 
        // Here Node is made up of : Node(int node_val, int
        // node_distance, int? node_parent)
        dq.Enqueue(new Node(src, 0, null));
        int[] dist = new int[n];
        // Set the distance of all nodes to
        // infinity(int.MaxValue)
        Array.Fill(dist, int.MaxValue);
        // set distance of source node as 0
        dist[src] = 0;
 
        while (dq.Count > 0) {
            Node curr = dq.Dequeue();
            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
            foreach(Node neighbourNode in newAdj[currVal])
            {
                int neighbour = neighbourNode.GetVal();
                if (neighbour == currParent)
                    continue;
 
                int wt = neighbourNode.GetWeight();
                if (wt == 0
                    && dist[neighbour] > currWeight) {
                    dist[neighbour] = currWeight;
                    dq.Enqueue(new Node(
                        neighbour, currWeight, currVal));
                }
                else if (dist[neighbour]
                         > currWeight + wt) {
                    dist[neighbour] = currWeight + wt;
                    dq.Enqueue(new Node(neighbour,
                                        currWeight + wt,
                                        currVal));
                }
            }
        }
        // If destination is not reachable from source,
        // return -1
        return int.MaxValue;
    }
 
    // Utility function to convert given graph into
    // bidirectional graph
    private static List<List<Node> >
    GetBiDirectionalGraph(List<List<int> > adj)
    {
        int n = adj.Count;
        List<List<Node> > newAdj = new List<List<Node> >();
 
        for (int i = 0; i < n; i++)
            newAdj.Add(new List<Node>());
 
        for (int i = 0; i < n; i++) {
            foreach(int neighbour in adj[i])
            {
                newAdj[i].Add(new Node(neighbour, 0));
                newAdj[neighbour].Add(new Node(i, 1));
            }
        }
 
        return newAdj;
    }
}

Javascript

// JavaScript Code to find minimum edge reversal to get
// atleast one path from source to destination using 0-1 BFS
// Code By: Sparsh_CBS
class Node {
    constructor(val, weight, parent = null) {
        this.val = val;
        this.weight = weight;
        this.parent = parent;
    }
}
 
function getMinRevEdges(adj, src, dest) {
    let n = adj.length;
    // Create the given graph into bidirectional graph
    let newAdj = getBiDirectionalGraph(adj);
    // Now, Apply 0-1 BFS using Deque to get the shortest path
    let dq = [];
    dq.push(new Node(src, 0, 0));
    let dist = new Array(n).fill(Number.MAX_VALUE);
    // Set the distance of all nodes to infinity(Number.MAX_VALUE)
    dist[src] = 0;
    // set distance of source node as 0
    while (dq.length) {
        let curr = dq.shift();
        let currVal = curr.val;
        let currWeight = curr.weight;
        let currParent = curr.parent;
        // If we encounter the destination node, we return
        if (currVal == dest) {
            return currWeight;
        }
        // Iterate over the neighbours of the current Node
        for (let neighbourNode of newAdj[currVal]) {
            let neighbour = neighbourNode.val;
            if (neighbour == currParent) {
                continue;
            }
            let wt = neighbourNode.weight;
            if (wt == 0 && dist[neighbour] > currWeight) {
                dist[neighbour] = currWeight;
                dq.push(new Node(neighbour, currWeight, currVal));
            } else if (dist[neighbour] > currWeight + wt) {
                dist[neighbour] = currWeight + wt;
                dq.push(new Node(neighbour, currWeight + wt, currVal));
            }
        }
    }
    return Number.MAX_VALUE;
}
 
function getBiDirectionalGraph(adj) {
    let n = adj.length;
    let newAdj = new Array(n);
    for (let i = 0; i < n; i++) {
        newAdj[i] = [];
    }
    let visited = new Array(n).fill(false);
    let queue = [];
    for (let i = 0; i < n; i++) {
        if (!visited[i]) {
            visited[i] = true;
            queue.push(i);
            while (queue.length) {
                let curr = queue.shift();
                for (let neighbour of adj[curr]) {
                    // original edges are to be assigned a weight of 0
                    newAdj[curr].push(new Node(neighbour, 0));
                    // make a fake edge and assign a weight of 1
                    newAdj[neighbour].push(new Node(curr, 1));
                    if (visited[neighbour]) {
                        continue; // if the neighbour was visited, then dont add it again in the queue
                    }
                    visited[neighbour] = true;
                    queue.push(neighbour);
                }
            }
        }
    }
    return newAdj;
}
 
let adj = [
    [],
    [],
    [],
    [],
    [],
    [],
    []
];
adj[0].push(1);
adj[2].push(1);
adj[5].push(1);
adj[2].push(3);
adj[6].push(3);
adj[6].push(4);
adj[4].push(5);
 
let ans = getMinRevEdges(adj, 0, 6);
 
if (ans == Number.MAX_VALUE) {
    console.log(-1);
} else {
    console.log(ans);
}

Output

Time Complexity: O(V+E)
Space Complexity: O(V+2*E)

Suggest improvement

Minimize the maximum difference between the heights

Find the Largest Cube formed by Deleting minimum Digits from a number

Share your thoughts in the comments

Standard Greedy algorithms

Greedy algorithm on Array

Greedy algorithm on Operating System

Greedy algorithm on Graph

Approximate Greedy algorithm for NP complete problems

Greedy algorithm for special cases of DP

Some easy problems on Greedy algorithm

Some medium level problems on Greedy algorithm

Some hard problems on Greedy algorithm

Standard Greedy algorithms

Greedy algorithm on Array

Greedy algorithm on Operating System

Greedy algorithm on Graph

Approximate Greedy algorithm for NP complete problems

Greedy algorithm for special cases of DP

Some easy problems on Greedy algorithm

Some medium level problems on Greedy algorithm

Some hard problems on Greedy algorithm

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

C++

Java

Python3

C#

Javascript

C++

Java

Python3

C#

Javascript

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