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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