You only need to take a numeric array [0 .. 9], counting each number
of input incrementing the array index. Then put in order this array,
then print in reverse order.

Luciano Pinheiro.

2010/7/6 Abhirup Ghosh <abhiru...@gmail.com>:
> We can build a wrapper object having two fields one th actual integer
> in the array and the count o the integer in the given array. Then
> build an array of those objects. Range of this array can be found
> easily by finding max and min of the array in O(n) time. We can build
> the auxiliary array in O(n) time. Then we can sort that array on the
> basis of count field using counting sort in O(n). As counting sort is
> stable sort if two counts are equal then the sequence is maintained.
> So the whole process is done in O(n). But the space complexity is O(n)
> as two auxiliary arrays are needed.
>
> -Abhirup
>
>
> On Sun, Jul 4, 2010 at 3:20 AM, Amir hossein Shahriari
> <amir.hossein.shahri...@gmail.com> wrote:
>> we can apply that case in the comparator or sort them after counting again
>> in nlogn (with respect to number of occurrences and their first index)
>> since the first occurrence of a number happens when it's not inserted in bst
>> we can do this easily
>>
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-- 
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Luciano Soares Pinheiro Jr.
Analista desenvolvedor Sr.

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