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Check if every node can me made accessible from a node of a Tree by at most N/2 given operations

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  • Last Updated : 26 May, 2021

Given a Directed Tree consisting of N nodes, the task is to check if there exists a node in the given tree such that all other nodes are reachable by removing any directed edge from the Tree and adding another directed edge between any pair of nodes in the Tree at most floor(N/2) times. If there exists any such node, then print “Yes”. Otherwise, print “No”.

Examples:

Input: N = 3

Output: Yes
Explanation: 
Remove the edge 2 -> 3 and insert an edge 1 -> 3. 

Therefore, both the remaining nodes (2, 3) are now accessible from the node 1.
Count of operations required is 1, which is <= floor(3/2) (= 1).
Input: N = 5

Output: No

 

Approach: The idea to solve this problem is based on the following observations:

  • Every node should have at least one parent node, i.e. every node should have at least 1 indegree to make the tree accessible from the required node.
  • It can be concluded that if every node has at least 1 indegree, then all other nodes can be accessed.
  • Therefore, the task is reduced to finding the number of nodes having 0 in-degrees and checking if it is at most N / 2 or not.

Follow the steps below to solve the problem: 

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;
 
void findNode(map<int, int> mp, int n)
{
     
    // Store the indegree
    // of every node
    int a[n];
 
    for(int i = 0; i < n; i++)
    {
        a[i] = mp[i + 1];
    }
 
    // Store the nodes having
    // indegree equal to 0
    int count0 = 0;
 
    // Traverse the array
    for(int i = 0; i < n; i++)
    {
         
        // If the indegree
        // of i-th node is 0
        if (a[i] == 0)
        {
             
            // Increment count0 by 1
            count0++;
        }
    }
 
    count0 -= 1;
 
    // If the number of operations
    // needed is at most floor(n/2)
    if (count0 <= floor(((double)n) /
                        ((double)2)))
    {
        cout << "Yes";
    }
 
    // Otherwise
    else
        cout << "No";
}
 
// Driver Code
int main()
{
     
    // Given number of nodes
    int N = 3;
 
    // Given Directed Tree
    map<int, int> mp;
    mp[1] = 0;
    mp[2] = 2;
    mp[3] = 0;
 
    findNode(mp, N);
}
 
// This code is contributed by SURENDRA_GANGWAR

Java




// Java program for the above approach
import java.io.*;
import java.util.HashMap;
 
class GFG {
 
    // Function to check if there is a
    // node in tree from where all other
    // nodes are accessible or not
    public static void
    findNode(HashMap<Integer, Integer> map,
             int n)
    {
 
        // Store the indegree
        // of every node
        int[] a = new int[n];
 
        for (int i = 0; i < n; i++) {
            a[i] = map.getOrDefault(i + 1, 0);
        }
 
        // Store the nodes having
        // indegree equal to 0
        int count0 = 0;
 
        // Traverse the array
        for (int i = 0; i < n; i++) {
 
            // If the indegree
            // of i-th node is 0
            if (a[i] == 0) {
 
                // Increment count0 by 1
                count0++;
            }
        }
 
        count0 -= 1;
 
        // If the number of operations
        // needed is at most floor(n/2)
        if (count0
            <= Math.floor(((double)n)
                          / ((double)2))) {
            System.out.println("Yes");
        }
 
        // Otherwise
        else
            System.out.println("No ");
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        // Given number of nodes
        int N = 3;
 
        // Given Directed Tree
        HashMap<Integer, Integer> map
            = new HashMap<>();
 
        map.put(1, 0);
        map.put(2, 2);
        map.put(3, 0);
 
        findNode(map, N);
    }
}

Python3




# python 3 program for the above approach
 
 
def findNode(mp, n):
 
    # Store the indegree
    # of every node
    a = [0]*n
 
    for i in range(n):
 
        a[i] = mp[i + 1]
 
    # Store the nodes having
    # indegree equal to 0
    count0 = 0
 
    # Traverse the array
    for i in range(n):
 
        # If the indegree
        # of i-th node is 0
        if (a[i] == 0):
 
            # Increment count0 by 1
            count0 += 1
 
    count0 -= 1
 
    # If the number of operations
    # needed is at most floor(n/2)
    if (count0 <= (n) /
            (2)):
 
        print("Yes")
 
    # Otherwise
    else:
        print("No")
 
 
# Driver Code
if __name__ == "__main__":
 
    # Given number of nodes
    N = 3
 
    # Given Directed Tree
    mp = {}
    mp[1] = 0
    mp[2] = 2
    mp[3] = 0
 
    findNode(mp, N)

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
public class GFG{
  
    // Function to check if there is a
    // node in tree from where all other
    // nodes are accessible or not
    public static void
    findNode(Dictionary<int, int> map,
             int n)
    {
 
        // Store the indegree
        // of every node
        int[] a = new int[n];
 
        for (int i = 0; i < n; i++) {
            if(map.ContainsKey(i+1))
            a[i] = map[i + 1];
            else
            a[i] = 0;
        }
 
        // Store the nodes having
        // indegree equal to 0
        int count0 = 0;
 
        // Traverse the array
        for (int i = 0; i < n; i++) {
 
            // If the indegree
            // of i-th node is 0
            if (a[i] == 0) {
 
                // Increment count0 by 1
                count0++;
            }
        }
 
        count0 -= 1;
 
        // If the number of operations
        // needed is at most floor(n/2)
        if (count0
            <= Math.Floor(((double)n)
                          / ((double)2))) {
            Console.WriteLine("Yes");
        }
 
        // Otherwise
        else
            Console.WriteLine("No ");
    }
  
    static public void Main ()
    {
       
   // Given number of nodes
        int N = 3;
 
        // Given Directed Tree
        Dictionary<int, int> map
            = new Dictionary<int, int>();
 
       map[1]= 0;
        map[2] = 2;
        map[3] = 0;
 
        findNode(map, N);
    }
}
 
// This code is contributed by offbeat

Javascript




<script>
 
// Javascript program for the above approach
 
function findNode(mp, n)
{
     
    // Store the indegree
    // of every node
    var a = new Array(n);
     
    var i;
    for(i = 0; i < n; i++)
    {
        a[i] = mp[i + 1];
    }
 
    // Store the nodes having
    // indegree equal to 0
    var count0 = 0;
 
    // Traverse the array
    for(i = 0; i < n; i++)
    {
         
        // If the indegree
        // of i-th node is 0
        if (a[i] == 0)
        {
             
            // Increment count0 by 1
            count0++;
        }
    }
 
    count0 -= 1;
 
    // If the number of operations
    // needed is at most floor(n/2)
    if (count0 <= parseInt(n/2))
    {
        document.write("Yes");
    }
 
    // Otherwise
    else
        document.write("No");
}
 
// Driver Code
  
    // Given number of nodes
    var N = 3;
 
    // Given Directed Tree
    var mp = new Map();
    mp.set(1,0);
    mp.set(2,2);
    mp.set(3,0);
    mp[1] = 0;
    mp[2] = 2;
    mp[3] = 0;
 
    findNode(mp, N);
 
</script>
Output: 
Yes

 

Time Complexity: O(N)
Auxiliary Space: O(N)
 


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