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[0], represent length of 1st wire

6 -> Value of Arr[1], represent length of 2nd wire

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

6 -> Value of Arr[3], represent length of 4th wire

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

1 -> Value of Arr[5], represent length of 6th wire

**Output :**

68

**Example 2:**

**Input **

4 -> Value of N, represents size of Arr

12 -> Value of Arr[0], represents length of 1st wire

2 -> Value of Arr[1], represent length of 2nd wire

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

5 -> Value of Arr[3], 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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