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Construct an N-ary Tree having no pair of adjacent nodes with same weight from given weights
  • Difficulty Level : Easy
  • Last Updated : 14 Dec, 2020

Given an array weights[] consisting of N positive integer, where weights[i] denotes the weight of ith node, the task is to construct an N-ary tree such that no two directly connected nodes have same weight. If it is possible to make such a tree, then print “Yes” along with their edges. Otherwise, print “No”.

Examples:

Input: weights[] = {1 2 1 2 5}
Output:
Yes
1 2
1 4
1 5
2 3
Explanation:
Index: 1 2 3 4 5
Weight :  1 2 1 2 5
The constructed Tree is shown in the following diagram: 
 

Input: weights[] = {1 1 1}
Output: No
Explanation: Since all weights are already same, no such tree can be constructed.



Approach: The idea to solve this problem is to first check if all nodes are assigned with same weight or not. If found to be true, then required tree cannot be constructed. Otherwise, such a tree can be constructed. Therefore traverse the array weights[] and check if all values are the same or not. If found to be true, then print “No”. Otherwise, print “Yes” and construct a tree using the following steps:

  • Take any node and make it the root node.
  • Now, connect all other nodes with weights not equal to that of the root to the root node. Now the remaining nodes are the nodes that have a value equal to the root node.
  • Choose any child node of the root node and connect all remaining nodes to them. Therefore, there exists no direct edge between nodes of same weight.
  • To check which nodes have not been included yet, keep track of visited nodes using an auxiliary array visited[]. If a node is visited, join a node with it, but do not join the visited node with another node as joining an unvisited node with a visited node is possible, but not vice versa.

Below is the implementation of the above approach:

C++

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// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
 
// Keep track of visited nodes
int visited[N];
 
// Function to construct a tree such
// that there are no two adjacent
// nodes with the same weight
void construct_tree(int weights[], int n)
{
    int minimum = *min_element(weights, weights + n);
    int maximum = *max_element(weights, weights + n);
 
    // If minimum and maximum
    // elements are equal, i.e.
    // array contains one distinct element
    if (minimum == maximum) {
 
        // Tree cannot be constructed
        cout << "No";
        return;
    }
 
    // Otherwise
    else {
 
        // Tree can be constructed
        cout << "Yes" << endl;
    }
 
    // Find the edges below
 
    // Choose weights[0] as root
    int root = weights[0];
 
    // First Node is visited
    visited[1] = 1;
 
    // Traverse the array
    for (int i = 0; i < n; i++) {
 
        // If current element has the
        // same weight as root and if
        // the node is visited, then
        // do not make an edge
 
        // Otherwise, make an edge
        if (weights[i] != root
            && visited[i + 1] == 0) {
            cout << 1 << " "
                 << i + 1 << " "
                 << endl;
 
            // Mark this node as visited
            visited[i + 1] = 1;
        }
    }
 
    // Find a weight not same as the
    // root & make edges with that node
    int notroot = 0;
    for (int i = 0; i < n; i++) {
 
        if (weights[i] != root) {
            notroot = i + 1;
            break;
        }
    }
 
    // Join non-roots with remaining nodes
    for (int i = 0; i < n; i++) {
 
        // Check if current node's weight
        // is same as root node's weight
        // and if it is not visited or not
        if (weights[i] == root
            && visited[i + 1] == 0) {
            cout << notroot << " "
                 << i + 1 << endl;
            visited[i + 1] = 1;
        }
    }
}
 
// Driver Code
int main()
{
    int weights[] = { 1, 2, 1, 2, 5 };
    int N = sizeof(weights) / sizeof(weights[0]);
 
    // Function Call
    construct_tree(weights, N);
}

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Java

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// Java program for the above approach
import java.lang.*;
import java.io.*;
import java.util.*;
 
class GFG{
   
static int N = 100000 + 5;
  
// Keep track of visited nodes
static int visited[] = new int[N];
  
// Function to construct a tree such
// that there are no two adjacent
// nodes with the same weight
static void construct_tree(int weights[], int n)
{
    int minimum = Arrays.stream(weights).min().getAsInt();
    int maximum = Arrays.stream(weights).max().getAsInt();
  
    // If minimum and maximum
    // elements are equal, i.e.
    // array contains one distinct element
    if (minimum == maximum)
    {
         
        // Tree cannot be constructed
        System.out.println("No");
        return;
    }
  
    // Otherwise
    else
    {
         
        // Tree can be constructed
        System.out.println("Yes");
    }
  
    // Find the edges below
  
    // Choose weights[0] as root
    int root = weights[0];
  
    // First Node is visited
    visited[1] = 1;
  
    // Traverse the array
    for(int i = 0; i < n; i++)
    {
         
        // If current element has the
        // same weight as root and if
        // the node is visited, then
        // do not make an edge
  
        // Otherwise, make an edge
        if (weights[i] != root &&
            visited[i + 1] == 0)
        {
            System.out.println(1 + " " +
                              (i + 1) + " ");
  
            // Mark this node as visited
            visited[i + 1] = 1;
        }
    }
  
    // Find a weight not same as the
    // root & make edges with that node
    int notroot = 0;
    for(int i = 0; i < n; i++)
    {
        if (weights[i] != root)
        {
            notroot = i + 1;
            break;
        }
    }
  
    // Join non-roots with remaining nodes
    for(int i = 0; i < n; i++)
    {
         
        // Check if current node's weight
        // is same as root node's weight
        // and if it is not visited or not
        if (weights[i] == root &&
            visited[i + 1] == 0)
        {
            System.out.println(notroot + " " +
                                    (i + 1));
            visited[i + 1] = 1;
        }
    }
}
   
// Driver Code
public static void main(String[] args)
{
    int weights[] = { 1, 2, 1, 2, 5 };
    int N = weights.length;
     
    // Function Call
    construct_tree(weights, N);
}
}
 
// This code is contributed by sanjoy_62

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Python3

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#Python3 program for the above approach
 
N = 10**5 + 5
 
#Keep track of visited nodes
visited=[0]*N
 
#Function to construct a tree such
#that there are no two adjacent
#nodes with the same weight
def construct_tree(weights, n):
    minimum = min(weights)
    maximum = max(weights)
 
    #If minimum and maximum
    #elements are equal, i.e.
    #array contains one distinct element
    if (minimum == maximum):
 
        #Tree cannot be constructed
        print("No")
        return
 
    #Otherwise
    else:
        print("Yes")
 
    #Find the edges below
 
    #Choose weights[0] as root
    root = weights[0]
 
    #First Node is visited
    visited[1] = 1
 
    #Traverse the array
    for i in range(n):
 
        #If current element has the
        #same weight as root and if
        #the node is visited, then
        #do not make an edge
 
        #Otherwise, make an edge
        if (weights[i] != root
            and visited[i + 1] == 0):
            print(1,i+1)
 
            #Mark this node as visited
            visited[i + 1] = 1
 
    #Find a weight not same as the
    #root & make edges with that node
    notroot = 0
    for i in range(n):
 
        if (weights[i] != root):
            notroot = i + 1
            break
 
    #Join non-roots with remaining nodes
    for i in range(n):
 
        #Check if current node's weight
        #is same as root node's weight
        #and if it is not visited or not
        if (weights[i] == root
            and visited[i + 1] == 0):
            print(notroot,i + 1)
            visited[i + 1] = 1
 
#Driver Code
if __name__ == '__main__':
    weights=[1, 2, 1, 2, 5]
 
    N = len(weights)
 
    #Function Call
    construct_tree(weights, N)

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C#

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// C# program to implement
// the above approach 
using System;
using System.Linq;
 
class GFG{
    
static int N = 100000 + 5;
   
// Keep track of visited nodes
static int[] visited = new int[N];
   
// Function to construct a tree such
// that there are no two adjacent
// nodes with the same weight
static void construct_tree(int[] weights, int n)
{
    int minimum = weights.Min();
    int maximum = weights.Max();
   
    // If minimum and maximum
    // elements are equal, i.e.
    // array contains one distinct element
    if (minimum == maximum)
    {
          
        // Tree cannot be constructed
        Console.WriteLine("No");
        return;
    }
   
    // Otherwise
    else
    {
          
        // Tree can be constructed
        Console.WriteLine("Yes");
    }
   
    // Find the edges below
   
    // Choose weights[0] as root
    int root = weights[0];
   
    // First Node is visited
    visited[1] = 1;
   
    // Traverse the array
    for(int i = 0; i < n; i++)
    {
          
        // If current element has the
        // same weight as root and if
        // the node is visited, then
        // do not make an edge
   
        // Otherwise, make an edge
        if (weights[i] != root &&
            visited[i + 1] == 0)
        {
             Console.WriteLine(1 + " " + (i + 1) + " ");
   
            // Mark this node as visited
            visited[i + 1] = 1;
        }
    }
   
    // Find a weight not same as the
    // root & make edges with that node
    int notroot = 0;
    for(int i = 0; i < n; i++)
    {
        if (weights[i] != root)
        {
            notroot = i + 1;
            break;
        }
    }
   
    // Join non-roots with remaining nodes
    for(int i = 0; i < n; i++)
    {
          
        // Check if current node's weight
        // is same as root node's weight
        // and if it is not visited or not
        if (weights[i] == root &&
            visited[i + 1] == 0)
        {
            Console.WriteLine(notroot + " " +
                                    (i + 1));
            visited[i + 1] = 1;
        }
    }
}
    
// Driver Code
public static void Main()
{
    int[] weights = { 1, 2, 1, 2, 5 };
    int N = weights.Length;
      
    // Function Call
    construct_tree(weights, N);
}
}
 
// This code is contributed by code_hunt.

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Output: 

Yes
1 2 
1 4 
1 5 
2 3

 

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

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