Sorting Elements of an Array by Frequency

PROBLEM :

Given an array of integers, sort the array according to frequency of elements. For example, if the input array is {2, 3, 2, 4, 5, 12, 2, 3, 3, 3, 12}, then modify the array to {3, 3, 3, 3, 2, 2, 2, 12, 12, 4, 5}.

If frequencies of two elements are same, print them in increasing order.



Input:
The first line of input contains an integer T denoting the number of test cases. The description of T test cases follows. The first line of each test case contains a single integer N denoting the size of array. The second line contains N space-separated integers A1, A2, ..., AN denoting the elements of the array.


Output:
Print each sorted array in a seperate line. For each array its numbers should be seperated by space.


Constraints:
1 = T = 70
30 = N = 130
1 = A [ i ] = 60


Example:
Input:
1
5
5 5 4 6 4

Output:
4 4 5 5 6

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SIMPLE c++ IMPLEMENTATION :(Using Auxilary arry)
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#include<iostream>
using namespace std;
void Array_by_Frequency(int [],int ) ;
int main()
 {
int t,no,arr[200],i ;
cin>>t ;
while(t--)
{
   cin>>no ;
   for(i=0;i<no;i++)
       cin>>arr[i] ;
 
    Array_by_Frequency(arr,no) ;
   
    for(i=0;i<no;i++)
       cout<<arr[i]<<" " ;
    cout<<endl ;
}
return 0;
}

void Array_by_Frequency(int arr[],int no)
{
    int i,max,ele,k ;
    int hash[200]={0} ;
    int b[200] ;
   
    for(i=0;i<no;i++)
        hash[arr[i]]++ ;
 
    max=-1 ;
    ele=-1 ;
    k=0 ;
   
    lable :
    for(i=0;i<no;i++)
    {
        if(hash[arr[i]]>=max)
        {
            if(hash[arr[i]]==max&&arr[i]<ele)
            {
                max=hash[arr[i]] ;
                ele=arr[i] ;
            }
            else if(hash[arr[i]]>max)
            {
                max=hash[arr[i]] ;
                ele=arr[i] ;
            }
        }
    }
   
    for(i=k;i<max+k;i++)
    b[i]=ele ;
   
k=i ;
     
    hash[ele]=-1 ;
    ele=-1 ;
    max=0 ;
   
    if(k<no)
        goto lable ;
       
    for(i=0;i<no;i++)
    arr[i]=b[i] ;
}

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