It can be done easily by counting sort....
On Wed, Dec 15, 2010 at 5:36 AM, Soumya Prasad Ukil
<[email protected]>wrote:

> Have a look : http://geeksforgeeks.org/?p=1488
>
>
> On 15 December 2010 05:19, Saurabh Koar <[email protected]> wrote:
>
>> @ Bittu:
>>
>> Lets analyze your code with iterations:
>>
>> the array contains  1 3 3 1 2 3 5 2 3
>> count contains       0 2 2 4 0 1 0 0 0
>>
>> now 1st iteration:
>> i=8,7,6 the inner loop will nt be executed;
>>
>> i=5; so j=0 only; as count[5]=1;
>> Now arr[pos](i.e array[0])=5
>>
>> i=4; the inner loop fails
>>
>> i=3; so j=0 to 3 as count[3]=4
>> Now array[1]=array[2]=array[3]=array[4]=3
>>
>> i=2; so j=0 to 1 as count[2]=2
>> so array[5]=array[6]=2
>>
>> and likewise array[7]=array[8]=1
>>
>> so the final array 5 3 3 3 3 2 2 1 1
>>
>> It doesnt match with the desired output.
>>
>> Correct me if I m missing something.
>>
>>
>>
>> On 12/15/10, bittu <[email protected]> wrote:
>> > @ankur,saurabh,soumya......
>> >
>> > ya ankur i forget to remove range from dare also no need to find
>> > range  for this..\
>> >
>> > put size-1 intead of range so that malloc will alocate the memory to
>> > all  elements in array ..no hope its fine...
>> >
>> >
>> > what i did is that
>> > i took counter array  thta cvontains the  no of times particular
>> > elements  comes
>> > then i have sort d array in reverse  order...&  prints accoding to
>> > count array..thats it... and the last line  of actual question says..
>> >
>> > """..in case of a tie  print the number which occurred first ""..i
>> > don't think d way i hav solved dis......
>> >
>> > still please let me no..if m wrong..
>> >
>> > Regards
>> > Shashank Mani Narayan " Don't b evil U can earn while u Learn"
>> > Computer Science & enginerring
>> > BIrla Institute of Technology,Mesra
>> > Cell 9166674831
>> >
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>
>
> --
> regards,
> soumya prasad ukil
>
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