Yes, I agree that problem seems to be incomplete.
The best we can do is to make better the O(n) search.
For searching a number k in array A[n];
Starting from index 0, we can skip the indexes which are within the
difference between A[i] and k, ie:

i=0;
while(i<n) {
  if(A[i]==k) return i;
  else  i= i+ | k-A[i] |;
}
The order is still O(n).

To see why divide and conquer wont work (in case trying a log(n) solution),
see the example:
Array:  4,5,4,5,4,5,4,5,4,5,4,5,4,5,4
k=6.
Even if we divide array into half, we have to look into both of the halves
since 6 can occur in both of them ( ie we cant discard a particular half on
the basis of range of numbers a segment of array can have), hence O(n).

On Sun, Mar 27, 2011 at 11:00 PM, kunal srivastav <
kunal.shrivas...@gmail.com> wrote:

> Also please do provide some test cases to comprehend this problem better
>
>
> On Sun, Mar 27, 2011 at 10:34 PM, Raunak Agrawal 
> <raunak.ra...@gmail.com>wrote:
>
>> Hi Ankit,
>>
>> Please correct me if I am wrong:
>>
>> 1. There is no need of recursion and also I cant see any base
>> condition....so this is basically a iterative problem.
>>
>> 2. Suppose the element is at last index of array...so the worst case order
>> would be of O(n)
>>
>> 3.  *|n-a[0]| >size: I am not able to get the logic...I mean is there any
>> relation between n and size?*
>>
>>
>> On Sun, Mar 27, 2011 at 10:24 PM, ankit sambyal 
>> <ankitsamb...@gmail.com>wrote:
>>
>>> For the following question  :
>>>  There is an array and the distance between any two consequent elements
>>> is one(+1 or -1) and given a number. You have to check whether the number is
>>> in array or not with minimum complexity.
>>>
>>> Assuming the array may not be sorted, the following algo can be used:
>>> Let a[] be the array of size "size" and n be the element to be searched.
>>> 1. If a[0]==n
>>>        return yes
>>>   else if( |n-a[0]| >size)
>>>       return no
>>>   else
>>>       call this function recursively with pointer to array pointing to
>>> the next element.
>>>
>>>
>>> Any problems with this solution, Plz let me know
>>>
>>>
>>>
>>>
>>> On Wed, Mar 23, 2011 at 9:07 PM, balaji a <peshwa.bal...@gmail.com>wrote:
>>>
>>>> First I had a paper pen coding round. The questions were:
>>>>   1) There are two sorted linked lists. Write a code to return the
>>>> merged linked list which is also sorted. No additional nodes must be used.
>>>>
>>>>   2) Design a DS that would do Push(),Pop(), and GetMax() elements at
>>>> complexity O(1)
>>>>   3) Do a BFS in given binary tree do find whether the given element in
>>>> the tree or not.
>>>>
>>>> I got shortlisted and I attended three interview rounds.
>>>> Round 1:
>>>>    It was a kind of debugging round. The questions were:
>>>>        1) Consider you are given a mobile alarm application how will u
>>>> test it
>>>>        2) Consider ur gmail chat box is not working for a particular
>>>> person alone, wht will u do to find the problem
>>>>        3) I was given a program (without the code - black box testing)
>>>> and asked to write the test cases for it
>>>>        4) A program was given and asked to debug - was simple only
>>>>
>>>>   Round 2:
>>>>     It was an algorithm designing round.
>>>>         1) Consider there is an array with duplicates and u r given two
>>>> numbers as input and u have to return the minimum distance between the two
>>>> in the array with minimum complexity.
>>>>         2) For a normal Binary tree write the code for inorder traversal
>>>> without recursion
>>>>         3) Given an array and an number find all the pairs in the array
>>>> that would add up to the given number. This also with minimum complexity.
>>>>
>>>>  Round 3:
>>>>     It was also coding round
>>>>     1) There is an array and the distance between any two consequent
>>>> elements is one(+1 or -1) and given a number. You have to check whether the
>>>> number is in array or not with minimum complexity.
>>>>
>>>>
>>>>
>>>> On Thu, Mar 24, 2011 at 12:11 AM, Akash Mukherjee 
>>>> <akash...@gmail.com>wrote:
>>>>
>>>>> kul man...wud appreciate if u cud post your question
>>>>>
>>>>>  On Wed, Mar 23, 2011 at 11:28 PM, balaji a 
>>>>> <peshwa.bal...@gmail.com>wrote:
>>>>>
>>>>>> hi i got till the third round of technical interview out of the four
>>>>>> rounds and got eliminated in third round.....anyways thnx for ur support
>>>>>> dude :-)
>>>>>>
>>>>>>
>>>>>> On Tue, Mar 22, 2011 at 12:51 PM, balaji a 
>>>>>> <peshwa.bal...@gmail.com>wrote:
>>>>>>
>>>>>>> Thnx :-) I am from SSN College of Engineering,Chennai....
>>>>>>>
>>>>>>>
>>>>>>> On Tue, Mar 22, 2011 at 12:28 PM, Akash Mukherjee <
>>>>>>> akash...@gmail.com> wrote:
>>>>>>>
>>>>>>>> u r welcome :), nd all the best for ur test.....btw, which clg??
>>>>>>>>
>>>>>>>>
>>>>>>>> On Tue, Mar 22, 2011 at 11:45 AM, guru <peshwa.bal...@gmail.com>wrote:
>>>>>>>>
>>>>>>>>> Thank you very much for the info friend....And sure will give u a
>>>>>>>>> treat :-)
>>>>>>>>>
>>>>>>>>> On Mar 22, 11:02 am, Akash Mukherjee <akash...@gmail.com> wrote:
>>>>>>>>> > hey, dis is what i was told by a friend working @ amazon -
>>>>>>>>> >
>>>>>>>>> > Sometimes they do go to the level of the subject basics like OS
>>>>>>>>> or DS but
>>>>>>>>> > you should be able to tackle these if you had studied well. No
>>>>>>>>> separate prep
>>>>>>>>> > is needed.
>>>>>>>>> >
>>>>>>>>> > Few Favs DS & Algos ( i should get treat for revealing this.;)...
>>>>>>>>> )
>>>>>>>>> > 1) All Trees (Binary for sure)
>>>>>>>>> > 2) Graphs
>>>>>>>>> > 3) Sorting Algos
>>>>>>>>> > 4) Heaps
>>>>>>>>> > "Let us C" ... though clichéd gives a good insight. If you can
>>>>>>>>> find time.
>>>>>>>>> >
>>>>>>>>> > can u tell a bit more about your profile?? fresher??
>>>>>>>>> >
>>>>>>>>> > On Tue, Mar 22, 2011 at 11:20 AM, guru <peshwa.bal...@gmail.com>
>>>>>>>>> wrote:
>>>>>>>>> > > Hi geeks,
>>>>>>>>> > >    tomorrow i am having Amazon.com's Coding round followed by
>>>>>>>>> > > Interview...pls suggest some tips to help me out...
>>>>>>>>> >
>>>>>>>>> > > --
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>>>>>>>>>
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>>>>>>>
>>>>>>> --
>>>>>>> A.Balaji
>>>>>>>
>>>>>>>
>>>>>>
>>>>>>
>>>>>> --
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>>>>>
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>>>>
>>>>
>>>> --
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>>>>
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>>>
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>
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