Hi pharta :
" find the point where it is rotated to get 14<->1 O(log(n)) " how can you
find rotation point in log(n) ?

On Fri, Jun 8, 2012 at 6:57 PM, partha sarathi Mohanty <
[email protected]> wrote:

> It is easy.. find the point where it is rotated to get 14<->1 O(log(n))
> since 2<14 that means u have to find it in subarray [123].. do a binary
> search here o(long(n))
> final 2*O(log(n))...
>
>
> On Fri, Jun 8, 2012 at 7:44 PM, Dave <[email protected]> wrote:
>
>> @Hassan: This is not possible without additional conditions. E.g., you
>> cannot find the 1 in the array {0,0,0,...,0,1,0,0,...,0} in less than O(n)
>> time.
>>
>> But with the condition that the elements of the array are pairwise
>> distinct, you can use a binary search to find k such that a[k-1] > a[0] and
>> a[k] < a[0]. Then if x < a[k], do a binary search to find x in {a[k] ...
>> a[n-1]}; otherwise do a binary search to find x in {a[0] ... a[k]}.
>>
>> Dave
>>
>> On Friday, June 8, 2012 8:41:47 AM UTC-5, Hassan Monfared wrote:
>>
>>> A sorted array of integers is rotated unknown times. find an item in
>>> O(logn) time and O(1) space complexity.
>>> for example : 1,2,3,7,10,14 -------rotated 3 times------> 7,10,14,1,2,3
>>> find 2 in O(logn) time in O(1) space complexity
>>>
>>> Regards
>>> Hassan H. Monfared
>>>
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