Hi,
I am highly interested in such type of work.I would like to
work with you
On Jan 24, 2008 5:22 PM, Albert Sanchez [EMAIL PROTECTED] wrote:
Hi,
Anyone interested in road traffic strategies? Flow optimization, time
dependent shortest paths problems?
Albert
#2 is weird, start by observing that if you've seen n/2+2 items then
you're done.
You want n/2+1 equal numbers. So there needs to be a number that is
repeated once after seeing n/2+2 numbers for this to be possible(and
the rest must all be the same), now divide and conquer. For all
possible
wht exactly we hve 2 do?
On 1/24/08, Albert Sanchez [EMAIL PROTECTED] wrote:
Hi,
Anyone interested in road traffic strategies? Flow optimization, time
dependent shortest paths problems?
Albert
--~--~-~--~~~---~--~~
You received this message because
2.
keep a counter n
take the first element of the array say e and initialize the counter n=1
for each remaining number in the array
if n==0, e=current element, n=1
if current element is equal to e, n++
else n--
if n==0 there is no number
otherwise check the number of occurrences of e in the array
Hey I am interested too...pls do let me know what do we have to do..
On 1/24/08, Albert Sanchez [EMAIL PROTECTED] wrote:
Hi,
Anyone interested in road traffic strategies? Flow optimization, time
dependent shortest paths problems?
Albert
--
Ciao,
Ajinkya
Hi,
I also wish to do it!! Let us know more about it..
Sumedh
On Jan 25, 2008 8:44 PM, Ajinkya Kale [EMAIL PROTECTED] wrote:
Hey I am interested too...pls do let me know what do we have to do..
On 1/24/08, Albert Sanchez [EMAIL PROTECTED] wrote:
Hi,
Anyone interested in road traffic
If you take k balls out of a bunch of N balls, the probability of
having selected k balls is always one(1.0). It takes no calculation,
because after you sum that expression up, it add up to one by
definition/law of probability.
On Jan 23, 5:12 pm, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote:
I
*1) find the second biggest number in a given array of size n. you have to
use n+logn number of searches or less
By this you want to have a complexity less than or equal to n + log n rite?
Consider an array { 17 , 4, 3, 18 , 9 , 15, 6 }
max1 max2 are index in array.
Iteration 1:-i=1 ( as
#2 is weird, start by observing that if you've seen n/2+2 items then
you're done.
You want n/2+1 equal numbers. So there needs to be a number that is
repeated once after seeing n/2+2 numbers for this to be possible(and
the rest must all be the same), now divide and conquer. For all
possible
