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Diameters for each node of Tree after connecting it with given disconnected component
  • Last Updated : 02 Sep, 2020

Given a Tree having N nodes connected by N − 1 edge and a single disconnected node, the task is to find the diameters for each node of the given Tree after connecting it with the given disconnected component.

Example:

Input:

 
Output: 3 3 4 4 4 4 
Explanation: 
Initially diameter of tree = 3
If an edge is added between node 1 and node 7, then the diameter of the tree is equal to 3(7 -> 1 -> 2 -> 5). 
If an edge is added between node 2 and node 7, then the diameter of the tree is equal to 3(3 -> 1 -> 2 -> 5). 
If an edge is added between node 3 and node 7, then the diameter of the tree is equal to 4(7 -> 3 -> 1 -> 2 -> 5). 
If an edge is added between node 4 and node 7, then the diameter of the new tree is equal to 4(7-> 4 -> 2 -> 1 -> 3). 
If an edge is added between node 5 and node 7, then the diameter of the new tree will be 4(7-> 5 -> 2 -> 1 -> 3). 
If an edge is added between node 6 and node 7, then the diameter of the new tree will be 4((7-> 6 -> 2 -> 1 -> 3)). 
 



Input:

 
Output: 3 2 3 
Explanation: 
Initial diameter of the tree = 2 
If an edge is added between node 1 and node 4, then the diameter of the tree is equal to 3(4 -> 1 -> 2 -> 3). 
If an edge is added between node 2 and node 4, then the diameter of the tree is equal to 2(4 -> 2 -> 3). 
If an edge is added between node 3 and node 4, then the diameter of the tree is equal to 3(4 -> 3 -> 2 -> 1). 
 

Approach: To solve the problem, the following observations need to be made: 
 

  • The diameter increases by 1 when the disconnected node is connected to an edge which forms the end of the diameter.
  • For all other nodes, the diameter remains unchanged on connecting the disconnected node.

Follow the steps given below to solve the problem based on the above observations:  

  1. Perform the DFS traversal of the given tree.
  2. While traversing, keep track of the farthest distance traveled and the farthest node.
  3. Now, perform DFS from the farthest node obtained from the above step and keep track of the farthest node from this node.
  4. Now, perform DFS and keep adding nodes in a Map that is farthest from the two nodes obtained from the above steps separately.

Below is the implementation of the above approach:

C++




// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Keeps track of the farthest
// end of the diameter
int X = 1;
 
// Keeps track of the length of
// the diameter
int diameter = 0;
 
// Stores the nodes which are at
// ends of the diameter
map<int, bool> mp;
 
// Perform DFS on the given tree
void dfs(int current_node, int prev_node,
         int len, bool add_to_map,
         vector<vector<int> >& adj)
{
    // Update diameter and X
    if (len > diameter) {
        diameter = len;
        X = current_node;
    }
 
    // If current node is an end of diameter
    if (add_to_map && len == diameter) {
        mp[current_node] = 1;
    }
 
    // Traverse its neighbors
    for (auto& it : adj[current_node]) {
        if (it != prev_node)
            dfs(it, current_node, len + 1,
                add_to_map, adj);
    }
}
 
// Function to call DFS for the
// required purposes
void dfsUtility(vector<vector<int> >& adj)
{
    // DFS from a random node and find
    // the node farthest from it
    dfs(1, -1, 0, 0, adj);
 
    int farthest_node = X;
 
    // DFS from X to calculate diameter
    dfs(farthest_node, -1, 0, 0, adj);
 
    // DFS from farthest_node to find
    // the farthest node(s) from it
    dfs(farthest_node, -1, 0, 1, adj);
 
    // DFS from X (other end of diameter) and
    // check the farthest node(s) from it
    dfs(X, -1, 0, 1, adj);
}
 
void printDiameters(vector<vector<int> >& adj)
{
    dfsUtility(adj);
 
    for (int i = 1; i <= 6; i++) {
 
        // If node i is the end of
        // a diameter
        if (mp[i] == 1)
 
            // Increase diameter by 1
            cout << diameter + 1 << ",  ";
 
        // Otherwise
        else
 
            // Remains unchanged
            cout << diameter << ",  ";
    }
}
 
// Driver Code
int main()
{
 
    /* constructed tree is
            1
           / \
          2   3    7
         /|\
        / | \
       4  5  6    */
 
    vector<vector<int> > adj(7);
 
    // creating undirected edges
    adj[1].push_back(2);
    adj[2].push_back(1);
    adj[1].push_back(3);
    adj[3].push_back(1);
    adj[2].push_back(4);
    adj[4].push_back(2);
    adj[2].push_back(5);
    adj[5].push_back(2);
    adj[2].push_back(6);
    adj[6].push_back(2);
 
    printDiameters(adj);
 
    return 0;
}

Java




// Java Program to implement
// the above approach
import java.util.*;
class GFG{
 
// Keeps track of the farthest
// end of the diameter
static int X = 1;
 
// Keeps track of the length of
// the diameter
static int diameter = 0;
 
// Stores the nodes which are at
// ends of the diameter
static HashMap<Integer,
                 Boolean> mp = new HashMap<>();
 
// Perform DFS on the given tree
static void dfs(int current_node, int prev_node,
                 int len, boolean add_to_map,
                Vector<Integer> []  adj)
{
    // Update diameter and X
    if (len > diameter)
    {
        diameter = len;
        X = current_node;
    }
 
    // If current node is an end of diameter
    if (add_to_map && len == diameter)
    {
        mp.put(current_node, true);
    }
 
    // Traverse its neighbors
    for (int it : adj[current_node])
    {
        if (it != prev_node)
            dfs(it, current_node, len + 1,
                add_to_map, adj);
    }
}
 
// Function to call DFS for the
// required purposes
static void dfsUtility(Vector<Integer> []  adj)
{
    // DFS from a random node and find
    // the node farthest from it
    dfs(1, -1, 0, false, adj);
 
    int farthest_node = X;
 
    // DFS from X to calculate diameter
    dfs(farthest_node, -1, 0, false, adj);
 
    // DFS from farthest_node to find
    // the farthest node(s) from it
    dfs(farthest_node, -1, 0, true, adj);
 
    // DFS from X (other end of diameter) and
    // check the farthest node(s) from it
    dfs(X, -1, 0, true, adj);
}
 
static void printDiameters(Vector<Integer> [] adj)
{
    dfsUtility(adj);
 
    for (int i = 1; i <= 6; i++)
    {
 
        // If node i is the end of
        // a diameter
        if (mp.containsKey(i) &&
            mp.get(i) == true)
 
            // Increase diameter by 1
            System.out.print(diameter + 1 + ",  ");
 
        // Otherwise
        else
 
            // Remains unchanged
            System.out.print(diameter + ",  ");
    }
}
 
// Driver Code
public static void main(String[] args)
{
 
    /* constructed tree is
            1
           / \
          2   3    7
         /|\
        / | \
       4  5  6    */
 
    Vector<Integer> []adj = new Vector[7];
    for (int i = 0; i < adj.length; i++)
        adj[i] = new Vector<Integer>();
   
    // creating undirected edges
    adj[1].add(2);
    adj[2].add(1);
    adj[1].add(3);
    adj[3].add(1);
    adj[2].add(4);
    adj[4].add(2);
    adj[2].add(5);
    adj[5].add(2);
    adj[2].add(6);
    adj[6].add(2);
 
    printDiameters(adj);
}
}
 
// This code is contributed by PrinciRaj1992

Python3




# Python3 program to implement
# the above approach
from collections import defaultdict
 
# Keeps track of the farthest
# end of the diameter
X = 1
 
# Keeps track of the length of
# the diameter
diameter = 0
 
# Stores the nodes which are at
# ends of the diameter
mp = defaultdict(lambda :0)
 
# Perform DFS on the given tree
def dfs(current_node, prev_node,
        len, add_to_map, adj):
             
    global diameter, X
     
    # Update diameter and X
    if len > diameter:
        diameter = len
        X = current_node
     
    # If current node is an end of diameter
    if add_to_map and len == diameter:
        mp[current_node] = 1
     
    # Traverse its neighbors
    for it in adj[current_node]:
        if it != prev_node:
            dfs(it, current_node, len + 1,
                add_to_map, adj)
             
# Function to call DFS for the
# required purposes
def dfsUtility(adj):
     
    # DFS from a random node and find
    # the node farthest from it
    dfs(1, -1, 0, 0, adj)
     
    farthest_node = X
     
    # DFS from X to calculate diameter
    dfs(farthest_node, -1, 0, 0, adj)
     
    # DFS from farthest_node to find
    # the farthest node(s) from it
    dfs(farthest_node, -1, 0, 1, adj)
     
    # DFS from X (other end of diameter) and
    # check the farthest node(s) from it
    dfs(X, -1, 0, 1, adj)
     
def printDiameters(adj):
     
    global diameter
    dfsUtility(adj)
     
    for i in range(1, 6 + 1):
         
        # If node i is the end of
        # a diameter
        if mp[i] == 1:
             
            # Increase diameter by 1
            print(diameter + 1, end = ", ")
             
        # Otherwise
        else:
             
            # Remains unchanged
            print(diameter, end = ", ")
             
# Driver code
     
# constructed tree is
#         1
#        / \
#      2   3
#    / | \
#   4  5  6
#      |
#      7
 
adj = []
for i in range(7):
    adj.append([])
     
# Creating undirected edges
adj[1].append(2)
adj[2].append(1)
adj[1].append(3)
adj[3].append(1)
adj[2].append(4)
adj[4].append(2)
adj[2].append(5)
adj[5].append(2)
adj[2].append(6)
adj[6].append(2)
 
printDiameters(adj)
 
# This code is contributed by Stuti Pathak

C#




// C# Program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG {
 
    // Keeps track of the farthest
    // end of the diameter
    static int X = 1;
 
    // Keeps track of the
    // length of the diameter
    static int diameter = 0;
 
    // Stores the nodes which are at
    // ends of the diameter
    static Dictionary<int, Boolean> mp
        = new Dictionary<int, Boolean>();
 
    // Perform DFS on the given tree
    static void dfs(int current_node, int prev_node,
                    int len, bool add_to_map,
                    List<int>[] adj)
    {
        // Update diameter and X
        if (len > diameter)
        {
            diameter = len;
            X = current_node;
        }
 
        // If current node is an end of diameter
        if (add_to_map && len == diameter)
        {
            mp.Add(current_node, true);
        }
 
        // Traverse its neighbors
        foreach(int it in adj[current_node])
        {
            if (it != prev_node)
                dfs(it, current_node, len + 1,
                    add_to_map, adj);
        }
    }
 
    // Function to call DFS for
    // the required purposes
    static void dfsUtility(List<int>[] adj)
    {
 
        // DFS from a random node and find
        // the node farthest from it
        dfs(1, -1, 0, false, adj);
 
        int farthest_node = X;
 
        // DFS from X to calculate diameter
        dfs(farthest_node, -1, 0, false, adj);
 
        // DFS from farthest_node to find
        // the farthest node(s) from it
        dfs(farthest_node, -1, 0, true, adj);
 
        // DFS from X (other end of diameter) and
        // check the farthest node(s) from it
        dfs(X, -1, 0, true, adj);
    }
 
    static void printDiameters(List<int>[] adj)
    {
        dfsUtility(adj);
 
        for (int i = 1; i <= 6; i++)
        {
 
            // If node i is the end
            // of a diameter
            if (mp.ContainsKey(i) && mp[i] == true)
 
                // Increase diameter by 1
                Console.Write(diameter + 1 + ",  ");
 
            // Otherwise
            else
 
                // Remains unchanged
                Console.Write(diameter + ",  ");
        }
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
 
        /* constructed tree is
            1
           / \
          2   3    7
         /|\
        / | \
       4  5  6    */
 
        List<int>[] adj = new List<int>[ 7 ];
        for (int i = 0; i < adj.Length; i++)
            adj[i] = new List<int>();
 
        // creating undirected edges
        adj[1].Add(2);
        adj[2].Add(1);
        adj[1].Add(3);
        adj[3].Add(1);
        adj[2].Add(4);
        adj[4].Add(2);
        adj[2].Add(5);
        adj[5].Add(2);
        adj[2].Add(6);
        adj[6].Add(2);
        printDiameters(adj);
    }
}
 
// This code is contributed by Amit Katiyar
Output: 
3,  3,  4,  4,  4,  4,




 

Time Complexity: O(V + E), where V is the number of vertices and E is the number of edges in the graph. 
Auxiliary Space: O(V)
 

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