Minimum sum

PROBLEM :

Given an array of digits (values are from 0 to 9), find the minimum possible sum of two numbers formed from digits of the array. All digits of given array must be used to form the two numbers.

Input:

The first line of input contains a single integer T denoting the number of test cases. Then T test cases follow. Each test case consist of two lines. The first line of each test case consists of an integer N, where N is the size of array.
The second line of each test case contains N space separated integers denoting array elements.

Output:

Corresponding to each test case, in a new line, print the minimum possible sum of two numbers formed from digits of the array.

Constraints:

1 = T = 100
1 = N = 30
1 = A[i] = 9

Example:

Input
1
5
5 3 0 7 4

Output
82

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SIMPLE c++ IMPLEMENTATION :
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#include<iostream>
using namespace std;
long long Minimum_sum(int [],int ) ;
void bubble(int [],int ) ;
int main()
 {
int t,no,arr[30],i ;
long long ans ;
cin>>t ;
while(t--)
{
   cin>>no ;
   for(i=0;i<no;i++)
       cin>>arr[i] ;
     
   ans=Minimum_sum(arr,no) ;
   cout<<ans<<endl ;
}
return 0;
}

long long Minimum_sum(int arr[],int no)
{
    bubble(arr,no) ;
    long long a,b ;
    int i ;
    a=0 ;
    b=0 ;
   
    for(i=0;i<no;i++)
    {
        if(i%2==0)
            a=a*10+arr[i] ;
        else
            b=b*10+arr[i] ;
    }
    return a+b ;
}

void bubble(int arr[],int no)
{
    int i,j,temp ;
   
    for(i=0;i<no-1;i++)
    {
        for(j=0;j<no-i-1;j++)
        {
            if(arr[j]>arr[j+1])
            {
                temp=arr[j] ;
                arr[j]=arr[j+1] ;
                arr[j+1]=temp ;
            }
        }
    }
}

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