@sanjiv : if x is the majority element then it is ok, but if not(x is
present and not majority element), u'r method doesn't work ..

On Sun, Apr 1, 2012 at 9:38 PM, rahul sharma <[email protected]>wrote:

> http://www.geeksforgeeks.org/archives/503
>
>
> On Sun, Apr 1, 2012 at 6:54 PM, sanjiv yadav <[email protected]>wrote:
>
>> if all values x are known to be contiguous then I think it will take o(1)
>> time because it will be majority element if it occurs more than n/2 times
>> as x must be at the middle.
>>
>> so if mid=(low+high)/2 and x==a[mid]then x is majority element otherwise
>> not.
>>
>> just check....let me know.
>>
>> On Thu, Feb 9, 2012 at 11:40 PM, Don <[email protected]> wrote:
>>
>>> It can't be done in O(log n) unless the array is sorted or there is
>>> some other special property, for example, if all values x are known to
>>> be contiguous, allowing you to use a binary search to find the first
>>> and last location of x.
>>>
>>> In the general case it is impossible in O(log n) because you have to
>>> examine at least n/2 elements to reach a conclusion. In some cases you
>>> must examine all n elements.
>>>
>>> Don
>>>
>>> On Feb 8, 1:45 pm, Prakhar Jain <[email protected]> wrote:
>>> > Hello everyone,
>>> >
>>> > suppose we have an array of size n and a number, say x, and we have to
>>> > determine whether the number x is present in the array more then n/2
>>> times
>>> > or not....?
>>> > can we have an O(log n) algo for determining it......?..pls help...!!!
>>> >
>>> > --
>>> > --
>>> > Prakhar Jain
>>> > IIIT Allahabad
>>> > B.Tech IT 3rd Year
>>> > Mob no: +91 9454992196
>>> > E-mail: [email protected]
>>> >           [email protected]
>>>
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>>
>>
>> --
>> Regards....
>>
>> Sanjiv Yadav
>>
>> MobNo.-  8050142693
>>
>> Email Id-  [email protected]
>>
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