i have 2 pointers fast and slow.now if tehy meet there is a loop...
now keep one ptr at meeting point and take other one to the begining of
listmove both at speed of one..they will meet at start of loophow
this happens???why they meet at start..plz tell logic behind this???thnx in
No They will not meet at the start in a case containing 5 nods and having
loop at the third node. once check this
On Fri, Mar 9, 2012 at 3:48 PM, rahul sharma rahul23111...@gmail.comwrote:
i have 2 pointers fast and slow.now if tehy meet there is a loop...
now keep one ptr at meeting
@ rahul sharma:
the linked list is a combination of a list a-b-...-p-q and a cycle
q-r-...-z-q. (z != p).
noting that the start of cycle q is the only node with 2 predecessor: p
and z.
if 2 pointers meet at some node x, different from q, in last step they
must have met at x', the predecessor
@terencei cant get..can u eleboratethnx for the sol..but plz
elaborate...
On Fri, Mar 9, 2012 at 5:59 PM, Terence technic@gmail.com wrote:
@ rahul sharma:
the linked list is a combination of a list a-b-...-p-q and a cycle
q-r-...-z-q. (z != p).
noting that the start of cycle q
suppose linked list is
a-b-c-d-e
and suppose loop starts from 'c'
according to u let one pointer be at 'c' say *q and another be at 'a' say
*p. Now if we move both at the speed of one then
After first pass
p will be at b
q will be at d
After second pass
p will be at c
q will be at e
#includeiostream
using namespace std;
class Abc
{
public :
int i;
Abc(){ i = 0;}
};
int main()
{ Abc a = new Abc();
couta.i;
}
--
Cheers,
Vicky
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write..
Abc a=Abc()
it will execute...
On Fri, Mar 9, 2012 at 8:15 PM, ~*~VICKY~*~ venkat.jun...@gmail.com wrote:
#includeiostream
using namespace std;
class Abc
{
public :
int i;
Abc(){ i = 0;}
};
int main()
{ Abc a = new Abc();
couta.i;
}
--
Cheers,
@sanjiv...how can u take q at c.they will meet at some position ...
suppose u have
1-2-3-4-5-6-7-4
now initialy
slow and fast both at 1
start incrementing slow by 1 and fast by 2
slow @2 ...fast @3
slow @3fast @5
s...@4...fast@7
slow @5...fast@5
both are equalmeans dere is
on left side of new there should be pointer that should collect the address
of the memory allocated..
write.
ABC *a=new ABC();
correct if wrng...
On Fri, Mar 9, 2012 at 8:19 PM, sanjiv yadav sanjiv2009...@gmail.comwrote:
write..
Abc a=Abc()
it will execute...
On Fri, Mar 9, 2012 at
I think u r making mistake i.e. after 7,5 will com not 4 because loop
starts from 5.check it again
On 3/9/12, rahul sharma rahul23111...@gmail.com wrote:
@sanjiv...how can u take q at c.they will meet at some position ...
suppose u have
1-2-3-4-5-6-7-4
now initialy
slow and fast
may u r confused with java
Abc a ; // this itself is creating an object of type ABC in C++ , no need
to use new keyword to allocate m/m.
where as in C++ new keyword allocate m/m and return an address ,
so to make it work do.
Abc *a=new ABC();
On Fri, Mar 9, 2012 at 8:28 PM, rahul sharma
First Rounf :
Aptitude paper :-
three section(Quant,aptitude,verabal)
20 qs each part
negative marking after 25 % of wrong attempts
quant was quite difficult :-
1.questions from area of triangle(two find a side if longest side and
other side along with area is given)
2.(loga + log b + log
itz ryt...loop starts from 4 n not from 5...
On Fri, Mar 9, 2012 at 8:58 PM, sanjiv yadav sanjiv2009...@gmail.comwrote:
I think u r making mistake i.e. after 7,5 will com not 4 because loop
starts from 5.check it again
On 3/9/12, rahul sharma rahul23111...@gmail.com wrote:
@Rahul: You have to figure out where the fast and slow pointers meet, which
will depend on the distance m from the head to the cycle and the period p
of the cycle. It is awkward to explain in words, so draw a picture and
label it according to this discussion.
Suppose the two pointers meet k
@dave..
plz tell use of variable...m and p??
p i thnik no. of nodes in cycle..???
plz tell what is p and m
On Fri, Mar 9, 2012 at 10:16 PM, Dave dave_and_da...@juno.com wrote:
@Rahul: You have to figure out where the fast and slow pointers meet,
which will depend on the distance m from
find the efficient implementation to find longest common contiguous
intersection from 2 lists provided to the function.
Example: list1: abcrfghwetf
list2: abrfghwwetxyab
Longest common intersection here is: rfghw
need the suffix array implementation in c...
thnx..in advance!!
--
You
actually how u r moving from 7 to 4?
can tell me in the detail...
On Sat, Mar 10, 2012 at 4:37 AM, Dave dave_and_da...@juno.com wrote:
@Rahul: The definitions of m and p are in the first sentence of the
posting.
Dave
On Friday, March 9, 2012 12:46:43 PM UTC-6, rahul sharma
This? http://en.wikipedia.org/wiki/Longest_common_substring_problem
On Friday, March 9, 2012 2:30:57 PM UTC-5, payal gupta wrote:
find the efficient implementation to find longest common contiguous
intersection from 2 lists provided to the function.
Example: list1: abcrfghwetf
list2:
On Monday, March 5, 2012 5:33:46 PM UTC-5, Sehaj wrote:
... I was asked this question during an interview, so I think there
must/might be some possible solution.
Ha. I wouldn't draw that conclusion. They are fishing for what you know.
Theoretical limits are good to know. So are
which college
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