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Minimum cost to connect weighted nodes represented as array

Given an array of N elements(nodes), where every element is weight of that node.Connecting two nodes will take a cost of product of their weights.You have to connect every node with every other node(directly or indirectly).Output the minimum cost required.

Examples:

Input : a[] = {6, 2, 1, 5}
Output :  13
Explanation : 
Here, we connect the nodes as follows:
connect a[0] and a[2], cost = 6*1 = 6,
connect a[2] and a[1], cost = 1*2 = 2,
connect a[2] and a[3], cost = 1*5 = 5.
every node is reachable from every other node:
Total cost = 6+2+5 = 13.

Input  : a[] = {5, 10}
Output : 50
Explanation : connections:
connect a[0] and a[1], cost = 5*10 = 50,
Minimum cost = 50. 



We need to make some observations that we have to make a connected graph with N-1 edges. As the output will be sum of products of two numbers, we have to minimize the product of every term in that sum equation. How can we do it? Clearly, choose the minimum element in the array and connect it with every other. In this way we can reach every other node from a particular node.
Let, minimum element = a[i], (let it’s index be 0)
Minimum Cost

    =a[i]*a[1]+a[i]*a[2]+....+a[i]*a[n-1],
    = a[i]*(a[1]+a[2]+....+a[n-1]);

So, answer is product of minimum element and sum of all the elements except minimum element.

C++

// cpp code for Minimum Cost Required to connect weighted nodes
#include <bits/stdc++.h>
using namespace std;
int minimum_cost(int a[], int n)
{
    int mn = INT_MAX;
    int sum = 0;
    for (int i = 0; i < n; i++) {
  
        // To find the minimum element 
        mn = min(a[i], mn);
  
        // sum of all the elements 
        sum += a[i]; 
    }
  
    return mn * (sum - mn);
}
  
// Driver code
int main()
{
    int a[] = { 4, 3, 2, 5 };
    int n = sizeof(a) / sizeof(a[0]);
    cout << minimum_cost(a, n) << endl;
    return 0;
}

Java

// Java code for Minimum Cost Required to
// connect weighted nodes
  
import java.io.*;
  
class GFG {
      
    static int minimum_cost(int a[], int n)
    {
          
        int mn = Integer.MAX_VALUE;
        int sum = 0;
          
        for (int i = 0; i < n; i++) {
  
            // To find the minimum element
            mn = Math.min(a[i], mn);
  
            // sum of all the elements
            sum += a[i];
        }
  
        return mn * (sum - mn);
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int a[] = { 4, 3, 2, 5 };
        int n = a.length;
          
        System.out.println(minimum_cost(a, n));
    }
}
  
// This code is contributed by vt_m.

Python 3

# Python 3 code for Minimum Cost
# Required to connect weighted nodes
import sys

def minimum_cost(a, n):

mn = sys.maxsize
sum = 0
for i in range(n):

# To find the minimum element
mn = min(a[i], mn)

# sum of all the elements
sum += a[i]

return mn * (sum – mn)

# Driver code
if __name__ == “__main__”:

a = [ 4, 3, 2, 5 ]
n = len(a)
print(minimum_cost(a, n))

# This code is contributed
# by ChitraNayal

C#

// C# code for Minimum Cost Required 
// to connect weighted nodes
using System;
  
class GFG {
      
    // Function to calculate minimum cost
    static int minimum_cost(int []a, int n)
    {
          
        int mn = int.MaxValue;
        int sum = 0;
          
        for (int i = 0; i < n; i++) 
        {
  
            // To find the minimum element
            mn = Math.Min(a[i], mn);
  
            // sum of all the elements
            sum += a[i];
        }
  
        return mn * (sum - mn);
    }
  
    // Driver code
    public static void Main()
    {
        int []a = {4, 3, 2, 5};
        int n = a.Length;
          
        Console.WriteLine(minimum_cost(a, n));
    }
}
  
// This code is contributed by vt_m.

PHP

<?php
// PHP code for Minimum Cost Required 
// to connect weighted nodes 
function minimum_cost($a, $n
    $mn = PHP_INT_MAX; 
    $sum = 0; 
    for ($i = 0; $i < $n; $i++) 
    
  
        // To find the minimum element 
        $mn = min($a[$i], $mn); 
  
        // sum of all the elements 
        $sum += $a[$i]; 
    
  
    return $mn * ($sum - $mn); 
  
// Driver code 
$a = array( 4, 3, 2, 5 ); 
$n = sizeof($a); 
echo minimum_cost($a, $n), " ";
  
// This code is contributed
// by Kiit_Tush
?>


Output:

  24

Time complexity: O(n).

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

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