Maximize the sum of products of the degrees between any two vertices of the tree

Given an integer N, the task is to construct a tree such that the sum of for all ordered pairs (u, v) is the maximum where u != v. Print the maximum possible sum.

Examples:

Input: N = 4
Output: 26
      1
     /
    2
   /
  3
 /
4
For node 1, 1*2 + 1*2 + 1*1 = 5
For node 2, 2*1 + 2*2 + 2*1 = 8
For node 3, 2*1 + 2*2 + 2*1 = 8
For node 4, 1*1 + 1*2 + 1*2 = 5
Total sum = 5 + 8 + 8 + 5 = 26

Input: N = 6
Output: 82 

Approach: We know that sum of the degree of all nodes in a tree is (2 * N) – 2 where N is the number of nodes in the tree. As we have to maximize the sum, we have to minimize the number of leaf nodes, as the leaf nodes have the minimum degree among all the nodes of the tree and the tree will be of the form:  

      1
     /
    2
   /
  ...
 /
N

where only the root and the only leaf node will have a degree of 1 and all the other nodes will have degree 2.

Below is the implementation of the above approach:  

C++

// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
#define ll long long int

// Function to return the maximum possible sum
ll maxSum(int N)
{
    ll ans = 0;

    for (int u = 1; u <= N; u++) {
        for (int v = 1; v <= N; v++) {
            if (u == v)
                continue;

            // Initialize degree for node u to 2
            int degreeU = 2;

            // If u is the leaf node or the root node
            if (u == 1 || u == N)
                degreeU = 1;

            // Initialize degree for node v to 2
            int degreeV = 2;

            // If v is the leaf node or the root node
            if (v == 1 || v == N)
                degreeV = 1;

            // Update the sum
            ans += (degreeU * degreeV);
        }
    }

    return ans;
}

// Driver code
int main()
{
    int N = 6;
    cout << maxSum(N);
}

Java

// Java implementation of above approach
class GFG
{

// Function to return the maximum possible sum
static int maxSum(int N)
{
    int ans = 0;

    for (int u = 1; u <= N; u++)
    {
        for (int v = 1; v <= N; v++)
        {
            if (u == v)
                continue;

            // Initialize degree for node u to 2
            int degreeU = 2;

            // If u is the leaf node or the root node
            if (u == 1 || u == N)
                degreeU = 1;

            // Initialize degree for node v to 2
            int degreeV = 2;

            // If v is the leaf node or the root node
            if (v == 1 || v == N)
                degreeV = 1;

            // Update the sum
            ans += (degreeU * degreeV);
        }
    }

    return ans;
}

// Driver code
public static void main(String[] args)
{
    int N = 6;
    System.out.println(maxSum(N));
}
}

// This code is contributed by Code_Mech

Python3

# Python3 implementation of above approach

# Function to return the maximum possible sum
def maxSum(N) :
    ans = 0;

    for u in range(1, N + 1) :
        for v in range(1, N + 1) :
            if (u == v) :
                continue;

            # Initialize degree for node u to 2
            degreeU = 2;

            # If u is the leaf node or the root node
            if (u == 1 or u == N) :
                degreeU = 1;

            # Initialize degree for node v to 2
            degreeV = 2;

            # If v is the leaf node or the root node
            if (v == 1 or v == N) :
                degreeV = 1;

            # Update the sum
            ans += (degreeU * degreeV);

    return ans;

# Driver code
if __name__ == "__main__" :

    N = 6;
    print(maxSum(N));

# This code is contributed by Ryuga

C#

// C# implementation of above approach
using System;
class GFG
{

// Function to return the maximum possible sum
static int maxSum(int N)
{
    int ans = 0;

    for (int u = 1; u <= N; u++)
    {
        for (int v = 1; v <= N; v++)
        {
            if (u == v)
                continue;

            // Initialize degree for node u to 2
            int degreeU = 2;

            // If u is the leaf node or the root node
            if (u == 1 || u == N)
                degreeU = 1;

            // Initialize degree for node v to 2
            int degreeV = 2;

            // If v is the leaf node or the root node
            if (v == 1 || v == N)
                degreeV = 1;

            // Update the sum
            ans += (degreeU * degreeV);
        }
    }

    return ans;
}

// Driver code
static void Main()
{
    int N = 6;
    Console.WriteLine(maxSum(N));
}
}

// This code is contributed by Chandan_jnu

PHP

<?php
// PHP implementation of above approach

// Function to return the maximum
// possible sum
function maxSum($N)
{
    $ans = 0;

    for ($u = 1; $u <= $N; $u++)
    {
        for ($v = 1; $v <= $N; $v++)
        {
            if ($u == $v)
                continue;

            // Initialize degree for node u to 2
            $degreeU = 2;

            // If u is the leaf node or the
            // root node
            if ($u == 1 || $u == $N)
                $degreeU = 1;

            // Initialize degree for node v to 2
            $degreeV = 2;

            // If v is the leaf node or the
            // root node
            if ($v == 1 || $v == $N)
                $degreeV = 1;

            // Update the sum
            $ans += ($degreeU * $degreeV);
        }
    }

    return $ans;
}

// Driver code
$N = 6;
echo maxSum($N);

// This code is contributed
// by Akanksha Rai
?>

Javascript

<script>

// Javascript implementation of above approach

// Function to return the maximum possible sum
function maxSum(N)
{
    var ans = 0;

    for (var u = 1; u <= N; u++) {
        for (var v = 1; v <= N; v++) {
            if (u == v)
                continue;

            // Initialize degree for node u to 2
            var degreeU = 2;

            // If u is the leaf node or the root node
            if (u == 1 || u == N)
                degreeU = 1;

            // Initialize degree for node v to 2
            var degreeV = 2;

            // If v is the leaf node or the root node
            if (v == 1 || v == N)
                degreeV = 1;

            // Update the sum
            ans += (degreeU * degreeV);
        }
    }

    return ans;
}

// Driver code
var N = 6;
document.write( maxSum(N));


</script>

Output

82

Complexity Analysis:

• Time Complexity: O(N²)
• Auxiliary Space: O(1)