@above sort the array using any inplace algorithm and then chk adjacent for
duplicates... this seems to me only inplace solution for chking duplicates

On Sat, Jul 3, 2010 at 5:57 PM, Vikas Jayaprakash <
[email protected]> wrote:

> I dint understand answer to rotation question.
>
> Also is there any way of identifying duplicates in an array without using
> any extra memory (like creating hash table takes extra memory).
>
> Any reference w.r.t book (or google book) to seek the concept behind two
> questions?
>
> Please let me know.
>
> Regards,
> Vikas J
>
>
> On Sat, Jul 3, 2010 at 12:47 PM, Abhirup Ghosh <[email protected]>wrote:
>
>> How can you find smallest element in the array in O(logn) time? Can
>> you please elaborate?
>>
>> -Abhirup
>>
>>
>>
>> On Fri, Jul 2, 2010 at 3:44 PM, jalaj jaiswal <[email protected]>
>> wrote:
>> > second question can be done in O(logn)
>> > do a modified binary search to find the starting index of the rotated
>> array
>> > i.e the smallest element.
>> >
>> > after that you have two sorted arrays..
>> > for eg 789 1236 (your modified binary search should return index of 1)
>> > now check if the elemnent you are searching in greater then a[min] then
>> do
>> > binary search in 1st array
>> > else do in second array
>> >
>> >
>> >
>> >
>> > On Fri, Jul 2, 2010 at 3:29 PM, Abhirup Ghosh <[email protected]>
>> wrote:
>> >>
>> >> 1. (1) Maintain a hash table and insert the elements in the table when
>> >> passing through the array. And if there is a collision then that
>> >> element is a duplicate element in the array.Time - O(n) and the space
>> >> is O(n).
>> >> (2) Duplicate is detected by the above process. Then it can  be easily
>> >> removed. I can't get what it means that remove duplicates without
>> >> hampering the array! The hash table anyway would contain all the
>> >> distinct elements.
>> >>
>> >> 2. Pass the array checking if the current element is lower than the
>> >> previous one. If found such element The current element is the start
>> >> of the original sorted array. If such element is not found then the
>> >> first element of the present is the desired element. Note down the
>> >> index. Takes O(n).
>> >>
>> >> Now do binary search. The mid point of the array is manipulated by
>> >> considering the index that we have saved. One trivial method could be
>> >> keep track of the array in each recursion and then get the half length
>> >> and then calculate the index according to the placement of the
>> >> starting index at that recursion. This takes O(logn).
>> >>
>> >> So overall Time O(n). Space O(1) [no extra space].
>> >>
>> >>
>> >> Correct me if I am wrong.
>> >>
>> >> -Abhirup
>> >>
>> >>
>> >>
>> >>
>> >> On Fri, Jul 2, 2010 at 1:50 AM, Vikas Jayaprakash
>> >> <[email protected]> wrote:
>> >> > Hi AlgoGeeks,
>> >> > Can anyone provide me answers for the below.
>> >> >
>> >> > 1. given an array of random integers write a program to (1) detect
>> >> > duplicate (2) remove duplicate (array should not get hampered at any
>> >> > cost!).
>> >> > 2 - In a sorted array some of the elements say N are rotated. for
>> >> > example initially 1 2 3 6 7 8 9 after rotation of 3 elements 7 8 9 1
>> 2
>> >> > 3 6.So how do you search an element in this array?
>> >> >
>> >> >
>> >> > Regards,
>> >> > Vikas J
>> >> >
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>> >
>> >
>> >
>> > --
>> > With Regards,
>> > Jalaj Jaiswal
>> > +919026283397
>> > B.TECH IT
>> > IIIT ALLAHABAD
>> >
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>
>
> --
> Thanks & Regards,
> Vikas Jayaprakash
>
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-- 
With Regards,
Jalaj Jaiswal
+919026283397
B.TECH IT
IIIT ALLAHABAD

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