Minimum edges to be added in a directed graph so that any node can be reachable from a given node

Given a directed graph and a node X. The task is to find the minimum number of edges that must be added to the graph such that any node can be reachable from the given node.
Examples:

Input: X = 0 

Output: 3
Input: X = 4 

Output: 1 

Approach: First, let’s mark all the vertices reachable from X as good, using a simple DFS. Then, for each bad vertex (vertices which are not reachable from X) v, count the number of bad vertices reachable from v (it also can be done by simple DFS). Let this number be cnt_v. Now, iterate over all bad vertices in non-increasing order of cnt_v. For the current bad vertex v, if it is still not marked as good, run a DFS from it, marking all the reachable vertices as good, and increase the answer by 1 (in fact, we are implicitly adding the edge (s, v)). It can be proved that this solution gives an optimal answer.
Below is the implementation of the above approach:

C++

// C++ implementation of the approach

#include <bits/stdc++.h>
using namespace std;

const int N = 5010;

int n, x;

vector<int> g[N];

// To check if the vertex has been
// visited or not
bool vis[N];

// To store if vertex is reachable
// from source or not
bool good[N];

int cnt;

void ADD_EDGE(int u, int v)
{
    g[u].push_back(v);
}

// Function to find all good vertices
void dfs1(int v)
{
    good[v] = true;
    for (auto to : g[v])
        if (!good[to])
            dfs1(to);
}

// Function to find cnt of all unreachable vertices
void dfs2(int v)
{
    vis[v] = true;
    ++cnt;
    for (auto to : g[v])
        if (!vis[to] && !good[to])
            dfs2(to);
}

// Function to return the minimum edges required
int Minimum_Edges()
{

    // Find all vertices reachable from the source
    dfs1(x);

    // To store all vertices with their cnt value
    vector<pair<int, int> > val;

    for (int i = 0; i < n; ++i) {

        // If vertex is bad i.e. not reachable
        if (!good[i]) {
            cnt = 0;
            memset(vis, false, sizeof(vis));

            // Find cnt of this vertex
            dfs2(i);
            val.push_back(make_pair(cnt, i));
        }
    }

    // Sort all unreachable vertices in
    // non-decreasing order of their cnt values
    sort(val.begin(), val.end());
    reverse(val.begin(), val.end());

    // Find the minimum number of edges
    // needed to be added
    int ans = 0;
    for (auto it : val) {
        if (!good[it.second]) {
            ++ans;
            dfs1(it.second);
        }
    }

    return ans;
}

// Driver code
int main()
{
    // Number of nodes and source node
    n = 5, x = 4;

    // Add edges to the graph
    ADD_EDGE(0, 1);
    ADD_EDGE(1, 2);
    ADD_EDGE(2, 3);
    ADD_EDGE(3, 0);

    cout << Minimum_Edges();

    return 0;
}

Java

// Java implementation of the approach
import java.util.*;

class GFG
{

// pair
static class pair
{
    int first,second;
    pair(int a,int b)
    {
        first = a;
        second = b;
    }
}

static int N = 5010;

static int n, x;

static Vector<Vector<Integer>> g = new Vector<Vector<Integer>>();

// To check if the vertex has been
// visited or not
static boolean vis[] = new boolean[N];

// To store if vertex is reachable
// from source or not
static boolean good[] = new boolean[N];

static int cnt;

static void ADD_EDGE(int u, int v)
{
    g.get(u).add(v);
}

// Function to find all good vertices
static void dfs1(int v)
{
    good[v] = true;
    for (int to = 0; to < g.get(v).size(); to++)
        if (!good[g.get(v).get(to)])
            dfs1(g.get(v).get(to));
}

// Function to find cnt of all unreachable vertices
static void dfs2(int v)
{
    vis[v] = true;
    ++cnt;
    for (int to = 0; to < g.get(v).size(); to++)
        if (!vis[g.get(v).get(to)] && !good[g.get(v).get(to)])
            dfs2(g.get(v).get(to));
}

// Function to return the minimum edges required
static int Minimum_Edges()
{

    // Find all vertices reachable from the source
    dfs1(x);

    // To store all vertices with their cnt value
    Vector<pair> val = new Vector<pair>();

    for (int i = 0; i < n; ++i)
    {

        // If vertex is bad i.e. not reachable
        if (!good[i])
        {
            cnt = 0;
            for(int j = 0; j < vis.length; j++)
                vis[j] = false;

            // Find cnt of this vertex
            dfs2(i);
            val.add(new pair(cnt, i));
        }
    }

    // Sort all unreachable vertices in
    // non-decreasing order of their cnt values
    Collections.sort(val,new Comparator<pair>()
    {
            public int compare(pair p1, pair p2)
            {
                return p1.first - p2.first;
            }
    });

    Collections.reverse(val);

    // Find the minimum number of edges
    // needed to be added
    int ans = 0;
    for (int it = 0; it < val.size(); it++)
    {
        if (!good[val.get(it).second])
        {
            ++ans;
            dfs1(val.get(it).second);
        }
    }

    return ans;
}

// Driver code
public static void main(String args[])
{
    // Number of nodes and source node
    n = 5; x = 4;

    for(int i = 0; i < N + 1; i++)
    g.add(new Vector<Integer>());

    // Add edges to the graph
    ADD_EDGE(0, 1);
    ADD_EDGE(1, 2);
    ADD_EDGE(2, 3);
    ADD_EDGE(3, 0);

    System.out.println( Minimum_Edges());
}
}

// This code is contributed by Arnab Kundu

Python3

# Python3 implementation of the approach
N = 5010
g = [[] for i in range(N)]

# To check if the vertex
# has been visited or not
vis = [False for i in range(N)]

# To store if vertex is reachable
# from source or not
good = [False for i in range(N)]

def ADD_EDGE(u, v):

    g[u].append(v)

# Function to find all good vertices
def dfs1(v):

    good[v] = True
    for to in g[v]:
        if not good[to]:
            dfs1(to)

# Function to find cnt of
# all unreachable vertices
def dfs2(v):

    global cnt
    vis[v] = True
    cnt += 1
    for to in g[v]:
        if not vis[to] and not good[to]:
            dfs2(to)

# Function to return
# the minimum edges required
def Minimum_Edges():

    global vis, cnt

    # Find all vertices reachable
    # from the source
    dfs1(x)

    # To store all vertices
    # with their cnt value
    val = []

    for i in range(0, n):

        # If vertex is bad i.e. not reachable
        if not good[i]:
            cnt = 0
            vis = [False for i in range(N)]

            # Find cnt of this vertex
            dfs2(i)
            val.append((cnt, i))

    # Sort all unreachable vertices
    # in non-decreasing order of
    # their cnt values
    val.sort(reverse = True)

    # Find the minimum number of edges
    # needed to be added
    ans = 0
    for it in val:
        if not good[it[1]]:
            ans += 1
            dfs1(it[1])

    return ans

# Driver code
if __name__ == "__main__":

    # Number of nodes and source node
    n, x = 5, 4

    # Add edges to the graph
    ADD_EDGE(0, 1)
    ADD_EDGE(1, 2)
    ADD_EDGE(2, 3)
    ADD_EDGE(3, 0)

    print(Minimum_Edges())

# This code is contributed by Rituraj Jain

C#

// C# implementation of the approach
using System;
using System.Collections;
using System.Collections.Generic;

class GFG
{

// pair
class pair
{
    public int first,second;
    public pair(int a,int b)
    {
        first = a;
        second = b;
    }
}

static int N = 5010;

static int n, x;

static ArrayList g = new ArrayList();

// To check if the vertex has been
// visited or not
static bool []vis = new bool[N];

// To store if vertex is reachable
// from source or not
static bool []good = new bool[N];

static int cnt;

static void Add_EDGE(int u, int v)
{
    ((ArrayList)g[u]).Add(v);
}

// Function to find all good vertices
static void dfs1(int v)
{
    good[v] = true;
    for (int to = 0; to < ((ArrayList)g[v]).Count; to++)
        if (!good[(int)((ArrayList)g[v])[to]])
            dfs1((int)((ArrayList)g[v])[to]);
}

// Function to find cnt of all unreachable vertices
static void dfs2(int v)
{
    vis[v] = true;
    ++cnt;
    for (int to = 0; to < ((ArrayList)g[v]).Count; to++)
        if (!vis[(int)((ArrayList)g[v])[to]] && !good[(int)((ArrayList)g[v])[to]])
            dfs2((int)((ArrayList)g[v])[to]);
}

class sortHelper : IComparer
{
   int IComparer.Compare(object a, object b)
   {
      pair first = (pair)a;
      pair second = (pair)b;

      return first.first - second.first;
   }
}

// Function to return the minimum edges required
static int Minimum_Edges()
{

    // Find all vertices reachable from the source
    dfs1(x);

    // To store all vertices with their cnt value
    ArrayList val = new ArrayList();

    for (int i = 0; i < n; ++i)
    {

        // If vertex is bad i.e. not reachable
        if (!good[i])
        {
            cnt = 0;
            for(int j = 0; j < vis.Length; j++)
                vis[j] = false;

            // Find cnt of this vertex
            dfs2(i);
            val.Add(new pair(cnt, i));
        }
    }

    // Sort all unreachable vertices in
    // non-decreasing order of their cnt values
    val.Sort(new sortHelper());

    // Find the minimum number of edges
    // needed to be Added
    int ans = 0;
    for (int it = 0; it < val.Count; it++)
    {
        if (!good[((pair)val[it]).second])
        {
            ++ans;
            dfs1(((pair)val[it]).second);
        }
    }

    return ans;
}

// Driver code
public static void Main(string []args)
{
    // Number of nodes and source node
    n = 5; x = 4;

    for(int i = 0; i < N + 1; i++)
        g.Add(new ArrayList());

    // Add edges to the graph
    Add_EDGE(0, 1);
    Add_EDGE(1, 2);
    Add_EDGE(2, 3);
    Add_EDGE(3, 0);

    Console.WriteLine(Minimum_Edges());
}
}

// This code is contributed by rutvik_56

Javascript

<script>
// Javascript implementation of the approach

class pair
{
    constructor(a,b)
    {
        this.first=a;
        this.second=b;
    }
}

let N = 5010;

let n, x;

let g = [];

// To check if the vertex has been
// visited or not
let vis= new Array(N);

// To store if vertex is reachable
// from source or not
let good=new Array(N);

for(let i=0;i<N;i++)
{
    vis[i]=false;
    good[i]=false;
}

let cnt;

function ADD_EDGE(u,v)
{
    g[u].push(v);
}

// Function to find all good vertices
function dfs1(v)
{
    good[v] = true;
    for (let to = 0; to < g[v].length; to++)
        if (!good[g[v][to]])
            dfs1(g[v][to]);
}

// Function to find cnt of all unreachable vertices
function dfs2(v)
{
    vis[v] = true;
    ++cnt;
    for (let to = 0; to < g[v].length; to++)
        if (!vis[g[v][to]] && !good[g[v][to]])
            dfs2(g[v][to]);
}

// Function to return the minimum edges required
function Minimum_Edges()
{
    // Find all vertices reachable from the source
    dfs1(x);

    // To store all vertices with their cnt value
    let val = [];

    for (let i = 0; i < n; ++i)
    {

        // If vertex is bad i.e. not reachable
        if (!good[i])
        {
            cnt = 0;
            for(let j = 0; j < vis.length; j++)
                vis[j] = false;

            // Find cnt of this vertex
            dfs2(i);
            val.push(new pair(cnt, i));
        }
    }

    // Sort all unreachable vertices in
    // non-decreasing order of their cnt values
    val.sort(

            function(p1,p2)
            {
                return p1.first - p2.first;
            }
    );

    val.reverse();

    // Find the minimum number of edges
    // needed to be added
    let ans = 0;
    for (let it = 0; it < val.length; it++)
    {
        if (!good[val[it].second])
        {
            ++ans;
            dfs1(val[it].second);
        }
    }

    return ans;
}

// Driver code
// Number of nodes and source node
n = 5; x = 4;

for(let i = 0; i < N + 1; i++)
    g.push([]);

// Add edges to the graph
ADD_EDGE(0, 1);
ADD_EDGE(1, 2);
ADD_EDGE(2, 3);
ADD_EDGE(3, 0);

document.write( Minimum_Edges());

// This code is contributed by avanitrachhadiya2155
</script>

Output: 

1

Time Complexity: O(V+E)

The time complexity of the above solution is O(V+E), where V is the number of vertices and E is the number of edges in the graph. We traverse over all the edges and vertices of the graph.

Space Complexity: O(V+E)

The space complexity of the above solution is also O(V+E), as we need to store all the edges and vertices of the graph.