I meant do it for n-1 times (my focus was on time complexity).Try with more
examples and you will know :)

On Wed, Jun 20, 2012 at 6:50 PM, Navin Kumar <algorithm.i...@gmail.com>wrote:

> For ex: let only two element are in queue: 1 2 (1 at front and rear is at
> 2).
> looping two times:
>
> first time: delete from front and adding to rear: queue will be: 2 1(front
> at 2 , rear at 1)
>
> second iteration: deleting 2 and adding to queue :result will be: 1 2
> (front 1, rear 2)
>
>
> On Wed, Jun 20, 2012 at 6:46 PM, Navin Kumar <algorithm.i...@gmail.com>wrote:
>
>> @Saurabh: queue will be remain unchanged according to your algorithm.
>> Because if you will delete an element from front and add at rear no change
>> will be there. After n iteration front will be pointing to same element and
>> rear will also point to same element.
>>
>> Correct me if i am wrong. :)
>>
>>
>> On Wed, Jun 20, 2012 at 6:39 PM, saurabh singh <saurabh.n...@gmail.com>wrote:
>>
>>> count the size of queue : O(n)
>>> loop for n and do remove and add in queue : O(n)
>>>
>>> Total : O(n)
>>>
>>>
>>> On Wed, Jun 20, 2012 at 6:34 PM, Navin Kumar 
>>> <algorithm.i...@gmail.com>wrote:
>>>
>>>> How to reverse a Queue .
>>>>
>>>> Constraints: Time complexity O(n). space complexity: O(1)
>>>>
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>>>
>>>
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>>> Thanks & Regards,
>>> Saurabh
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Thanks & Regards,
Saurabh

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