@Kirubakaran : still space complexity is O(n) due to stack.Can it be solved
in space complexity O(1).

On Wed, Jun 20, 2012 at 8:00 PM, Kirubakaran D <kirubakara...@gmail.com>wrote:

> You could use recursion.
>
> def reverse_Q q
> if !q.isEmpty?
>   el = q.dequeue
>   nQ = reverse_Q(q)
>   nQ.enqueue el
>   return nQ
> end
> return q
> end
>
>
>
> On Wednesday, June 20, 2012 6:57:23 PM UTC+5:30, Navin Kumar wrote:
>>
>> Use only standard operation of Queue like: EnQueue, DeQueue, IsEmptyQueue
>> etc
>>
>> On Wed, Jun 20, 2012 at 6:50 PM, amrit harry <dabbcomput...@gmail.com>wrote:
>>
>>> can we create other methods or we have to use only enqueue and
>>> dequeue...? if yes then simply
>>> for(i=0;i<=n/2;i++)
>>> swap(i,n-i);
>>>
>>>
>>>
>>> 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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>>>>>
>>>>>
>>>>>
>>>>> --
>>>>> Thanks & Regards,
>>>>> Saurabh
>>>>>
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>>>
>>>
>>>
>>> --
>>> Thanks & Regards
>>> Amritpal singh
>>>
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