How is this code dealing with the scenario where two arrays have common
number. For example where two arrays are same. A[] = {1,2,3,4} and B[] =
{1,2,3,4}. Here the median position is not (4+4)/2 rather it is 4/2.
On Sun, Jun 9, 2013 at 3:12 PM, rahul sharma wrote:
> @doncan u plz explain t
When malloc allocates memory it assigns a size field in the header to
indicate the size of the memory it allocates at a stretch. Can we not use
this information effectively?
On Jan 31, 2013 11:44 PM, "Piyush" wrote:
> it will always work if array is statically allocated
> Not if dynamically allo
I think you have to think about the manual way of doing it - how to
handle the integer part and decimal part. You can find this in any
standard book. Then try out the algorithm.
On Mon, Jan 30, 2012 at 12:26 AM, Rahul Kumar wrote:
> ur subject is decimal to binary but in content u have written
If you have Android sdk setup then you can find samples in that only.
If you haven't done that please find necessary information in
http://developer.android.com/index.html
- Abhirup
On Tue, Jan 31, 2012 at 5:20 PM, saurabh singh wrote:
>
> Saurabh Singh
> B.Tech (Computer Science)
> MNNIT
> blo
If the subsequence is not to be continuous then
sort the given array. O(nlogn)
and then go on adding from the least element until the sum gets >= k
and then denote the array elements till the previous one as the set of
desired elements. O(n)
Total complexity O(nlogn).
- Abhirup
On Tue, Aug 17,
We can build a wrapper object having two fields one th actual integer
in the array and the count o the integer in the given array. Then
build an array of those objects. Range of this array can be found
easily by finding max and min of the array in O(n) time. We can build
the auxiliary array in O(n)
I think this can be implemented with queue data structure. Whenever an
element is used, remove it from the queue use it and then again insert
it in the queue at the back. So the front element in the queue is the
least recently used one.
-Abhirup
On Fri, Jul 2, 2010 at 10:23 PM, jaladhi dave wro
;
>
> On Fri, Jul 2, 2010 at 3:29 PM, Abhirup Ghosh 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 s
Can you please elaborate on the solution you have with auxiliary array?
On Fri, Jul 2, 2010 at 3:53 AM, jalaj jaiswal wrote:
>
> we are given with Numerator and Denominator. After division we might get a
> recurring decimal points float as the answer.
> For example 23.34563456 ...
> return 345
I think those two sensors should not be exactly opposite to each other
to make the decision meaningful.
On Fri, Jul 2, 2010 at 11:58 AM, Jitendra Kushwaha
wrote:
> I think two sensors beside one another is enough to find the direction of
> rotation.
> At some time both will be sensing the same
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 ca
n>max then max = n;
>
> so the range would lie between min and max..
>
> correct me if I am wrong..
>
> On Mon, Jun 28, 2010 at 2:20 PM, nisha goyal
> wrote:
>>
>> @abhirup: range is required in counting sort how can you decide that??
>> please elabor
We can sort the array in O(n) time using counting sort. And then take
the difference of consecutive elements in O(n) time to get the minimum
one.
-Abhirup
On Mon, Jun 28, 2010 at 10:36 AM, Jagadish M wrote:
> In the general case, we can reduce Element-Distinctness(ED) problem to
> this problem
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