lol
On Mon, Aug 15, 2011 at 12:40 AM, aditya kumar <aditya.kumar130...@gmail.com
> wrote:
> @aditi : sry i dint realise that n > log n .:P
>
> On Mon, Aug 15, 2011 at 12:38 AM, aditi garg <aditi.garg.6...@gmail.com>wrote:
>
>> @aditya : dis is obviously correct bt here complexity will be O(n) bt we
>> are asked to gv O(log n) solution
>>
>> On Mon, Aug 15, 2011 at 12:37 AM, aditya kumar <
>> aditya.kumar130...@gmail.com> wrote:
>>
>>> for(j=0;j<n;j++)
>>> {
>>> if(a[j]==j)
>>> return j;
>>> else
>>> continue ;
>>> }
>>>
>>> this shud also be correct right ??
>>>
>>> On Mon, Aug 15, 2011 at 12:31 AM, Akash Mukherjee <akash...@gmail.com>wrote:
>>>
>>>> just my 2 cents ---- in d binary search, replacing key with mid, ie
>>>> if(a[mid] > mid)
>>>> check lower half
>>>> else upper half
>>>>
>>>> should work??
>>>>
>>>>
>>>> On Mon, Aug 15, 2011 at 12:26 AM, aditi garg <aditi.garg.6...@gmail.com
>>>> > wrote:
>>>>
>>>>> Given an ordered array A[1…n] with numbers in strictly increasing
>>>>> order. Find a ‘j’ such that A [j]=j or -1 if no such number exist in
>>>>> o (log n).
>>>>>
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>>
>>
>>
>> --
>> Aditi Garg
>> Undergraduate Student
>> Electronics & Communication Divison
>> NETAJI SUBHAS INSTITUTE OF TECHNOLOGY
>> Sector 3, Dwarka
>> New Delhi
>>
>>
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>
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--
Aditi Garg
Undergraduate Student
Electronics & Communication Divison
NETAJI SUBHAS INSTITUTE OF TECHNOLOGY
Sector 3, Dwarka
New Delhi
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