# Given N gold wires, each wire has a length associated with it. At a time, only two

Given N gold wires, each wire has a length associated with it. At a time, only two adjacent small wires are assembled at the end of a large wire and the cost of forming is the sum of their length. Find the minimum cost when all wires are assembled to form a single wire.

For Example:

Suppose, Arr[]={7,6,8,6,1,1,}

{7,6,8,6,1,1}-{7,6,8,6,2} , cost =2

{7,6,8,6,2}- {7,6,8,8}, cost = 8

{7,6,8,8} – {13,8,8}, cost=13

{13,8,8} -{13,16}, cost=16

{13, 16} – {29}, cost =29

2+8+13+16+29=68

Hence , the minimum cost to assemble all gold wires is 68.

Constraints

• 1<=N<=30
• 1<= Arr[i]<=100

Example 1:

Input

6  -> Value of N, represent size of Arr

7  -> Value of Arr, represent length of 1st wire

6 -> Value of Arr, represent length of 2nd wire

8 -> Value of Arr , represent length of 3rd wire

6 -> Value of Arr, represent length of 4th wire

1 -> Value of Arr, represent length of 5th wire

1 -> Value of Arr, represent length of 6th wire

Output :

68

Example 2:

Input

4   -> Value of N, represents size of Arr

12  -> Value of Arr, represents length of 1st wire

2   -> Value of Arr, represent length of 2nd wire

2   -> Value of Arr, represent length of 3rd wire

5  -> Value of Arr, represent length of 4th wire

Output :

34

## Solution In Java

`import java.util.*;class Solution{   public static int solve (int arr[], int n)   {       PriorityQueue <Integer> queue = new PriorityQueue<Integer>();             for (int i = 0; i < n; i++)           queue.add (arr[i]);             int sum = 0, temp1, temp2;             while (queue.size () >= 2)       {            temp1 = queue.poll ();            temp2 = queue.poll ();            sum += temp1 + temp2;            queue.add (temp1 + temp2);       }       return sum;   }   public static void main (String[]args)   {       Scanner sc = new Scanner (System.in);       int n = sc.nextInt ();       int arr[] = new int[n];         for (int i = 0; i < n; i++)           arr[i] = sc.nextInt ();             System.out.println (solve (arr, n));   }}`

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