No, time complexity for 1st is O(n).
for 2nd, a quick solution is to make one pass through the array to store the
frequency of 1,2,3,4,5.
Then using this information, a second pass to fill the array based on freq.

On Wed, Jun 1, 2011 at 10:53 PM, Abhishek Goswami <zeal.gosw...@gmail.com>wrote:

> for second question explanation.
>
> if you have array for 100 element a[100]. and you can enter element into
> array from 1 to 5 range upto a[100].So you have array
> a[100] which contain number from 1 to 5. what will be efficient algorithm
> for arranging number in ascending order..
> ex 1 2 5 5 5 3 4 1 1 upto a[100]
>
> output : 1 1 1  2 3 4 5 5 5
>
> @First question.
> What will be time complexity. I believe it will be o(n^2) can it be
> implemented in more efficient way..
>
>
>
>
> On Wed, Jun 1, 2011 at 10:27 PM, Harshal <hc4...@gmail.com> wrote:
>
>> the simplest way is to use a hashmap, or an array arr[10000] and keep
>> track of the number seen so far, eg. by making arr[num]=1, and checking
>> while inserting new elements.
>> second question is a bit unclear
>>
>>
>> On Wed, Jun 1, 2011 at 10:04 PM, Abhishek Goswami <zeal.gosw...@gmail.com
>> > wrote:
>>
>>> I am sure this question has come already but can anyone point me answer
>>> again.
>>>
>>> 1.if user enter number 0 to 10000. than what will we algorithum to
>>>    determine duplicate number (Note : user can not enter more than
>>>     10000 number)
>>>
>>> 2. if we have a number in the range of  [1...5] than we insert number
>>>     randomly into array. how will we arrange element in sorted order.
>>>
>>>      Ex.  1 1 2 4 5 5 5 3 4
>>>
>>>             1 1 2 3  4 4 5 5
>>>
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>>
>>
>>
>> --
>> Harshal Choudhary,
>> III Year B.Tech CSE,
>> NIT Surathkal, Karnataka, India.
>>
>>
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-- 
Harshal Choudhary,
III Year B.Tech CSE,
NIT Surathkal, Karnataka, India.

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