Given an array **weights[]** consisting of **N** positive integer, where **weights[i]** denotes the weight of **i**^{th} 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 3Explanation:

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:NoExplanation: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++

`// C++ program to implement` `// 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);` `}` |

## Java

`// Java program to implement` `// 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` |

## Python3

`# Python3 program to implement` `# 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)` |

## C#

`// 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.` |

## Javascript

`<script>` `// Javascript program to implement` `// the above approach` `let N = 100000 + 5;` `// Keep track of visited nodes` `let visited = ` `new` `Array(N);` `visited.fill(0);` `// Function to construct a tree such` `// that there are no two adjacent` `// nodes with the same weight` `function` `construct_tree(weights, n)` `{` ` ` `let minimum = Number.MAX_VALUE;` ` ` `let maximum = Number.MIN_VALUE;` ` ` ` ` `for` `(let i = 0; i < weights.length; i++)` ` ` `{` ` ` `minimum = Math.min(minimum, weights[i]);` ` ` `maximum = Math.max(maximum, weights[i]);` ` ` `}` ` ` `// If minimum and maximum` ` ` `// elements are equal, i.e.` ` ` `// array contains one distinct element` ` ` `if` `(minimum == maximum)` ` ` `{` ` ` ` ` `// Tree cannot be constructed` ` ` `document.write(` `"No"` `);` ` ` `return` `;` ` ` `}` ` ` ` ` `// Otherwise` ` ` `else` ` ` `{` ` ` ` ` `// Tree can be constructed` ` ` `document.write(` `"Yes"` `+ ` `"</br>"` `);` ` ` `}` ` ` `// Find the edges below` ` ` ` ` `// Choose weights[0] as root` ` ` `let root = weights[0];` ` ` `// First Node is visited` ` ` `visited[1] = 1;` ` ` `// Traverse the array` ` ` `for` `(let 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)` ` ` `{` ` ` `document.write(1 + ` `" "` `+` ` ` `(i + 1) + ` `"</br>"` `);` ` ` `// Mark this node as visited` ` ` `visited[i + 1] = 1;` ` ` `}` ` ` `}` ` ` `// Find a weight not same as the` ` ` `// root & make edges with that node` ` ` `let notroot = 0;` ` ` `for` `(let i = 0; i < n; i++)` ` ` `{` ` ` `if` `(weights[i] != root)` ` ` `{` ` ` `notroot = i + 1;` ` ` `break` `;` ` ` `}` ` ` `}` ` ` `// Join non-roots with remaining nodes` ` ` `for` `(let 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)` ` ` `{` ` ` `document.write(notroot + ` `" "` `+` ` ` `(i + 1) + ` `"</br>"` `);` ` ` `visited[i + 1] = 1;` ` ` `}` ` ` `}` `}` `// Driver code` `let weights = [ 1, 2, 1, 2, 5 ];` `let n = weights.length;` ` ` `// Function Call` `construct_tree(weights, n);` `// This code is contributed by divyeshrabadiya07` `</script>` |

**Output:**

Yes 1 2 1 4 1 5 2 3

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer **Complete Interview Preparation Course****.**

In case you wish to attend live classes with industry experts, please refer **DSA Live Classes**