ignore my comment...

On Thu, Feb 9, 2012 at 4:15 PM, Ashish Goel <ashg...@gmail.com> wrote:

> log n is impossible. the other solution i thought was of building a tree
> where each node contains value and its count. and then building a heap out
> of these counts, but this will be overkill.
>
> the fact that rest of the n/2 elements are not unique is the killer in the
> logic otherwise only n/2+1 elements are sufficient
>
> Best Regards
> Ashish Goel
> "Think positive and find fuel in failure"
> +919985813081
> +919966006652
>
>
>
> On Thu, Feb 9, 2012 at 2:53 PM, atul anand <atul.87fri...@gmail.com>wrote:
>
>> i guess can be done using binary indexed tree.
>>
>>
>> On Thu, Feb 9, 2012 at 2:07 PM, Prakhar Jain <jprakha...@gmail.com>wrote:
>>
>>> But in this post, we don't have prior information about what can be
>>> possible majority element.
>>>
>>> According to my question, we know that either x is the majority element
>>> or there is no majority element.
>>> Can we use that information to reduce complexity to O(log n)..???
>>>
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