@Hassan Monfared

was reading that link only. it's strange how he assume's we can leave
half of element's ??

– If A[n/2-1]>A[n/2]  then search for a peak among A[1]… A[n/2-1] ?? why ??

eg. 12 12 12 11 1 2 5 3

m i missing something ??

On Sat, Jun 23, 2012 at 10:02 PM, Hassan Monfared <[email protected]> wrote:
> just found a good resource for 1d and 2D version :
> http://courses.csail.mit.edu/6.006/spring11/lectures/lec02.pdf
>
>
> On Sun, Jun 24, 2012 at 3:31 AM, sengar.mahi <[email protected]> wrote:
>>
>> @adarsh :its nt dat easy as u see it..
>>
>>
>> On Sun, Jun 24, 2012 at 1:45 AM, Sourabh Singh <[email protected]>
>> wrote:
>>>
>>> @adarsh kumar
>>>
>>> are u sure it's worst case will be O (log n) ??
>>> i think iff array is fully sorted O(n) will be required to say "NO
>>> such element present"
>>>
>>> On Sat, Jun 23, 2012 at 1:11 PM, adarsh kumar <[email protected]> wrote:
>>> > This is a variation of binary search, the difference being that we have
>>> > to
>>> > search for an element that is greater than its immediate left one and
>>> > lesser
>>> > than its immediate right one. Just implement binary search with these
>>> > additional constraints, thereby giving O(log n).
>>> > In case of any difficulty/error, let me know.
>>> >
>>> > On Sun, Jun 24, 2012 at 1:27 AM, Hassan Monfared <[email protected]>
>>> > wrote:
>>> >>
>>> >> Given an array of integers find a peak element of array in log(n)
>>> >> time.
>>> >> for example if A={3,4,6,5,10} then peak element is 6  ( 6>5 & 6>4 ).
>>> >>
>>> >> Regards.
>>> >>
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>>
>>
>>
>> --
>> Regards
>> Mahendra Pratap Singh Sengar
>> B-tech 4/4
>> NIT Warangal.
>>
>> Facebook ID
>>
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