473,231 Members | 2,494 Online
Bytes | Software Development & Data Engineering Community
Post Job

Home Posts Topics Members FAQ

Join Bytes to post your question to a community of 473,231 software developers and data experts.

Linked List

function that would: return the 5th element from the end in a singly
linked list of integers, in one pass, and then provide a set of test
cases
against that function.

And please dont think this is my homework.
Can anybody simple tell me how this can be done. I am preparing for an
interview and wanted to know how to go about implementing this.
Or atleast tell me where I can find few examples on traversing Lists.
Even if I can get any good resources which explains lists properly my
job will be done.
Jul 22 '05 #1
32 5648
"source" <su*********@hotmail.com> wrote in message
function that would: return the 5th element from the end in a singly
linked list of integers, in one pass, and then provide a set of test
cases
against that function.


Show us your best effort so far. Also try to write a function that finds
the 5th element in a container matching a certain condition (for example,
the element is >10).

Jul 22 '05 #2
"source" <su*********@hotmail.com> wrote in message
function that would: return the 5th element from the end in a singly
linked list of integers, in one pass, and then provide a set of test
cases
against that function.


Show us your best effort so far. Also try to write a function that finds
the 5th element in a container matching a certain condition (for example,
the element is >10).

Jul 22 '05 #3
source writes:
function that would: return the 5th element from the end in a singly
linked list of integers, in one pass, and then provide a set of test
cases
against that function.

And please dont think this is my homework.
Can anybody simple tell me how this can be done. I am preparing for an
interview and wanted to know how to go about implementing this.
Or atleast tell me where I can find few examples on traversing Lists.
Even if I can get any good resources which explains lists properly my
job will be done.


If I understand the question, it can't be done; at least on a bare bones
linked list. You will have to count the elements so you can subtract five
from that. And then traverse the list again, which would be easiest from
the head end. I call that two passes, not one that you asked for. My
default meaning for end is the far end; I call the other end the beginning.

For test cases, test for within range, each valid boundary and one beyond
each valid boundary. Valid because numbering might start with zero or one.
Jul 22 '05 #4
source writes:
function that would: return the 5th element from the end in a singly
linked list of integers, in one pass, and then provide a set of test
cases
against that function.

And please dont think this is my homework.
Can anybody simple tell me how this can be done. I am preparing for an
interview and wanted to know how to go about implementing this.
Or atleast tell me where I can find few examples on traversing Lists.
Even if I can get any good resources which explains lists properly my
job will be done.


If I understand the question, it can't be done; at least on a bare bones
linked list. You will have to count the elements so you can subtract five
from that. And then traverse the list again, which would be easiest from
the head end. I call that two passes, not one that you asked for. My
default meaning for end is the far end; I call the other end the beginning.

For test cases, test for within range, each valid boundary and one beyond
each valid boundary. Valid because numbering might start with zero or one.
Jul 22 '05 #5
"osmium" <r1********@comcast.net> wrote in message news:c4pnvn$2kbbqu$1@ID-
function that would: return the 5th element from the end in a singly
linked list of integers, in one pass, and then provide a set of test
cases against that function.
If I understand the question, it can't be done; at least on a bare bones
linked list. You will have to count the elements so you can subtract five
from that. And then traverse the list again, which would be easiest from
the head end. I call that two passes, not one that you asked for. My
default meaning for end is the far end; I call the other end the beginning.

You can do it in one pass. The first way to implement it could actually be
slower than the 2 pass method, and the second way would probably be faster.

For test cases, test for within range, each valid boundary and one beyond
each valid boundary. Valid because numbering might start with zero or

one.

I'd say the empty list, list of 1 element, list of 5 elements, list of >5
elements.
Jul 22 '05 #6
"osmium" <r1********@comcast.net> wrote in message news:c4pnvn$2kbbqu$1@ID-
function that would: return the 5th element from the end in a singly
linked list of integers, in one pass, and then provide a set of test
cases against that function.
If I understand the question, it can't be done; at least on a bare bones
linked list. You will have to count the elements so you can subtract five
from that. And then traverse the list again, which would be easiest from
the head end. I call that two passes, not one that you asked for. My
default meaning for end is the far end; I call the other end the beginning.

You can do it in one pass. The first way to implement it could actually be
slower than the 2 pass method, and the second way would probably be faster.

For test cases, test for within range, each valid boundary and one beyond
each valid boundary. Valid because numbering might start with zero or

one.

I'd say the empty list, list of 1 element, list of 5 elements, list of >5
elements.
Jul 22 '05 #7
On 4 Apr 2004 12:05:51 -0700 in comp.lang.c++, su*********@hotmail.com
(source) wrote,
function that would: return the 5th element from the end in a singly
linked list of integers, in one pass, and then provide a set of test
cases
against that function.


This sentence no verb.

What I consider the usual approach to that kind of thing is what I call
a "following pointer". That is, you have two iterators, one points to
where you are looking in the list and the other five items behind it.
You advance both of them at the same time. When you get to what you
were looking for (in this case, the end) the "following" iterator points
to what you wanted.

No doubt you can write the C++ code based on that.

Jul 22 '05 #8
On 4 Apr 2004 12:05:51 -0700 in comp.lang.c++, su*********@hotmail.com
(source) wrote,
function that would: return the 5th element from the end in a singly
linked list of integers, in one pass, and then provide a set of test
cases
against that function.


This sentence no verb.

What I consider the usual approach to that kind of thing is what I call
a "following pointer". That is, you have two iterators, one points to
where you are looking in the list and the other five items behind it.
You advance both of them at the same time. When you get to what you
were looking for (in this case, the end) the "following" iterator points
to what you wanted.

No doubt you can write the C++ code based on that.

Jul 22 '05 #9
source wrote:
function that would: return the 5th element from the end in a singly
linked list of integers, in one pass, and then provide a set of test
cases
against that function.

And please dont think this is my homework.
Can anybody simple tell me how this can be done. I am preparing for an
interview and wanted to know how to go about implementing this.
Or atleast tell me where I can find few examples on traversing Lists.
Even if I can get any good resources which explains lists properly my
job will be done.


P-code, and without error checking (assumes list is at least 5 long):

node* node = yourfirstnode;
node* node5 = node->next->next->next->next;

for(;;)
{
if(node5->next == 0)
{
return node->val;
}
node = node->next;
node5 = node5->next;
}

- Pete
Jul 22 '05 #10
source wrote:
function that would: return the 5th element from the end in a singly
linked list of integers, in one pass, and then provide a set of test
cases
against that function.

And please dont think this is my homework.
Can anybody simple tell me how this can be done. I am preparing for an
interview and wanted to know how to go about implementing this.
Or atleast tell me where I can find few examples on traversing Lists.
Even if I can get any good resources which explains lists properly my
job will be done.


P-code, and without error checking (assumes list is at least 5 long):

node* node = yourfirstnode;
node* node5 = node->next->next->next->next;

for(;;)
{
if(node5->next == 0)
{
return node->val;
}
node = node->next;
node5 = node5->next;
}

- Pete
Jul 22 '05 #11
osmium wrote:
source writes:

function that would: return the 5th element from the end in a singly
linked list of integers, in one pass,

If I understand the question, it can't be done; at least on a bare bones
linked list. You will have to count the elements so you can subtract five
from that. And then traverse the list again, which would be easiest from
the head end. I call that two passes, not one that you asked for. My
default meaning for end is the far end; I call the other end the beginning.


Keep two pointers. Begin traversing the list with the first. 5 elements
in, begin traversing with the second. Once the first reaches the end,
the second will be 5 elements from the end.

-Kevin
--
My email address is valid, but changes periodically.
To contact me please use the address from a recent posting.
Jul 22 '05 #12
osmium wrote:
source writes:

function that would: return the 5th element from the end in a singly
linked list of integers, in one pass,

If I understand the question, it can't be done; at least on a bare bones
linked list. You will have to count the elements so you can subtract five
from that. And then traverse the list again, which would be easiest from
the head end. I call that two passes, not one that you asked for. My
default meaning for end is the far end; I call the other end the beginning.


Keep two pointers. Begin traversing the list with the first. 5 elements
in, begin traversing with the second. Once the first reaches the end,
the second will be 5 elements from the end.

-Kevin
--
My email address is valid, but changes periodically.
To contact me please use the address from a recent posting.
Jul 22 '05 #13
"Pete" <x@x.x> wrote in message news:yKZbc.14870
node* node = yourfirstnode;
node* node5 = node->next->next->next->next;


What happens on a list of 0 to 4 elements?

Jul 22 '05 #14
"Pete" <x@x.x> wrote in message news:yKZbc.14870
node* node = yourfirstnode;
node* node5 = node->next->next->next->next;


What happens on a list of 0 to 4 elements?

Jul 22 '05 #15
Siemel Naran wrote:
"Pete" <x@x.x> wrote in message news:yKZbc.14870

node* node = yourfirstnode;
node* node5 = node->next->next->next->next;

What happens on a list of 0 to 4 elements?


As he noted:

"P-code, and without error checking (assumes list is at least 5 long):"

mark

Jul 22 '05 #16
Siemel Naran wrote:
"Pete" <x@x.x> wrote in message news:yKZbc.14870

node* node = yourfirstnode;
node* node5 = node->next->next->next->next;

What happens on a list of 0 to 4 elements?


As he noted:

"P-code, and without error checking (assumes list is at least 5 long):"

mark

Jul 22 '05 #17
"Pete" <x@x.x> wrote in message news:yKZbc.14870
node* node = yourfirstnode;
node* node5 = node->next->next->next->next;

for(;;)
{
if(node5->next == 0)
{
return node->val;
}
node = node->next;
node5 = node5->next;
}


Is it really faster than the 2 pass method? In the 2 pass method, first
visit all N elements for a total of N calls to const_iterator::operator++.
Then you perform N-5 more calls to operator++. But in the 1 pass method,
you perform N calls to const_iterator::operator++ on node, and N-5 on node5.
So don't they both have the same running time?
Jul 22 '05 #18
"Pete" <x@x.x> wrote in message news:yKZbc.14870
node* node = yourfirstnode;
node* node5 = node->next->next->next->next;

for(;;)
{
if(node5->next == 0)
{
return node->val;
}
node = node->next;
node5 = node5->next;
}


Is it really faster than the 2 pass method? In the 2 pass method, first
visit all N elements for a total of N calls to const_iterator::operator++.
Then you perform N-5 more calls to operator++. But in the 1 pass method,
you perform N calls to const_iterator::operator++ on node, and N-5 on node5.
So don't they both have the same running time?
Jul 22 '05 #19
Siemel Naran wrote:
"Pete" <x@x.x> wrote in message news:yKZbc.14870
node* node = yourfirstnode;
node* node5 = node->next->next->next->next;

for(;;)
{
if(node5->next == 0)
{
return node->val;
}
node = node->next;
node5 = node5->next;
}


Is it really faster than the 2 pass method? In the 2 pass method,
first visit all N elements for a total of N calls to
const_iterator::operator++. Then you perform N-5 more calls to
operator++. But in the 1 pass method, you perform N calls to
const_iterator::operator++ on node, and N-5 on node5. So don't they
both have the same running time?


No, it really wouldn't be faster much if at all, but it's much more
"elegant" IMHO.

- Pete
Jul 22 '05 #20
Siemel Naran wrote:
"Pete" <x@x.x> wrote in message news:yKZbc.14870
node* node = yourfirstnode;
node* node5 = node->next->next->next->next;

for(;;)
{
if(node5->next == 0)
{
return node->val;
}
node = node->next;
node5 = node5->next;
}


Is it really faster than the 2 pass method? In the 2 pass method,
first visit all N elements for a total of N calls to
const_iterator::operator++. Then you perform N-5 more calls to
operator++. But in the 1 pass method, you perform N calls to
const_iterator::operator++ on node, and N-5 on node5. So don't they
both have the same running time?


No, it really wouldn't be faster much if at all, but it's much more
"elegant" IMHO.

- Pete
Jul 22 '05 #21
On Mon, 05 Apr 2004 00:42:42 GMT in comp.lang.c++, "Siemel Naran"
<Si*********@REMOVE.att.net> wrote,
Is it really faster than the 2 pass method? In the 2 pass method, first
visit all N elements for a total of N calls to const_iterator::operator++.
Then you perform N-5 more calls to operator++. But in the 1 pass method,
you perform N calls to const_iterator::operator++ on node, and N-5 on node5.
So don't they both have the same running time?


Nobody said it was faster. The problem spec said "in one pass".
The question is, with both pointers running through the list like that
does it count as one pass, or two passes running not quite in parallel?

Jul 22 '05 #22
On Mon, 05 Apr 2004 00:42:42 GMT in comp.lang.c++, "Siemel Naran"
<Si*********@REMOVE.att.net> wrote,
Is it really faster than the 2 pass method? In the 2 pass method, first
visit all N elements for a total of N calls to const_iterator::operator++.
Then you perform N-5 more calls to operator++. But in the 1 pass method,
you perform N calls to const_iterator::operator++ on node, and N-5 on node5.
So don't they both have the same running time?


Nobody said it was faster. The problem spec said "in one pass".
The question is, with both pointers running through the list like that
does it count as one pass, or two passes running not quite in parallel?

Jul 22 '05 #23
Siemel Naran wrote:
"Pete" <x@x.x> wrote in message news:yKZbc.14870
node* node = yourfirstnode;
node* node5 = node->next->next->next->next;

for(;;)
{
if(node5->next == 0)
{
return node->val;
}
node = node->next;
node5 = node5->next;
}


Is it really faster than the 2 pass method? In the 2 pass method,
first visit all N elements for a total of N calls to
const_iterator::operator++. Then you perform N-5 more calls to
operator++. But in the 1 pass method, you perform N calls to
const_iterator::operator++ on node, and N-5 on node5. So don't they
both have the same running time?


BTW, this exercise is a good example of a good reason to use a doubly linked
list, because it's easy to go either forwards or back depending on the
situation. The extra memory used for the extra pointer per node really isn't
an issue, usually.

- Pete
Jul 22 '05 #24
Siemel Naran wrote:
"Pete" <x@x.x> wrote in message news:yKZbc.14870
node* node = yourfirstnode;
node* node5 = node->next->next->next->next;

for(;;)
{
if(node5->next == 0)
{
return node->val;
}
node = node->next;
node5 = node5->next;
}


Is it really faster than the 2 pass method? In the 2 pass method,
first visit all N elements for a total of N calls to
const_iterator::operator++. Then you perform N-5 more calls to
operator++. But in the 1 pass method, you perform N calls to
const_iterator::operator++ on node, and N-5 on node5. So don't they
both have the same running time?


BTW, this exercise is a good example of a good reason to use a doubly linked
list, because it's easy to go either forwards or back depending on the
situation. The extra memory used for the extra pointer per node really isn't
an issue, usually.

- Pete
Jul 22 '05 #25
* su*********@hotmail.com (source) schriebt:
function that would: return the 5th element from the end in a singly
linked list of integers, in one pass, and then provide a set of test
cases
against that function.


What is your definition of "one pass", and does this need to be
non-destructive?

--
A: Because it messes up the order in which people normally read text.
Q: Why is top-posting such a bad thing?
A: Top-posting.
Q: What is the most annoying thing on usenet and in e-mail?
Jul 22 '05 #26
* su*********@hotmail.com (source) schriebt:
function that would: return the 5th element from the end in a singly
linked list of integers, in one pass, and then provide a set of test
cases
against that function.


What is your definition of "one pass", and does this need to be
non-destructive?

--
A: Because it messes up the order in which people normally read text.
Q: Why is top-posting such a bad thing?
A: Top-posting.
Q: What is the most annoying thing on usenet and in e-mail?
Jul 22 '05 #27
"Pete" <x@x.x> wrote in message news:yF2cc.15284
Siemel Naran wrote: node* node = yourfirstnode;
node* node5 = node->next->next->next->next;

for(;;)
{
if(node5->next == 0)
{
return node->val;
}
node = node->next;
node5 = node5->next;
}
Is it really faster than the 2 pass method? In the 2 pass method,
first visit all N elements for a total of N calls to
const_iterator::operator++. Then you perform N-5 more calls to
operator++. But in the 1 pass method, you perform N calls to
const_iterator::operator++ on node, and N-5 on node5. So don't they
both have the same running time?


Seems the 2 pass method you maintain an index for looping from [0, N-5) but
in the 1 pass method you don't. So based on that the 1 pass method might be
faster, though the improvement is probably negligible.

BTW, this exercise is a good example of a good reason to use a doubly linked list, because it's easy to go either forwards or back depending on the
situation. The extra memory used for the extra pointer per node really isn't an issue, usually.


If going back is not a common operation, the single list could be better.
Anyway, the point of these 'clever' interview questions is not always to
mimic reality, but to see how you think through problems, especially the
logic questions.

Anyway, I think if you just store the value of the pointer in one pass, you
could get better
node* node = yourfirstnode;
node* node5 = node->next->next->next->next;

T * last[5] = { &node->val, &node->next->val, ... };

for(int i=0; ; i = (i==5 ? 0 : 5))
{
if(node5->next == 0)
{
return last[i];
}
last[i] = &node5->val;
node5 = node5->next;
}
Jul 22 '05 #28
"Pete" <x@x.x> wrote in message news:yF2cc.15284
Siemel Naran wrote: node* node = yourfirstnode;
node* node5 = node->next->next->next->next;

for(;;)
{
if(node5->next == 0)
{
return node->val;
}
node = node->next;
node5 = node5->next;
}
Is it really faster than the 2 pass method? In the 2 pass method,
first visit all N elements for a total of N calls to
const_iterator::operator++. Then you perform N-5 more calls to
operator++. But in the 1 pass method, you perform N calls to
const_iterator::operator++ on node, and N-5 on node5. So don't they
both have the same running time?


Seems the 2 pass method you maintain an index for looping from [0, N-5) but
in the 1 pass method you don't. So based on that the 1 pass method might be
faster, though the improvement is probably negligible.

BTW, this exercise is a good example of a good reason to use a doubly linked list, because it's easy to go either forwards or back depending on the
situation. The extra memory used for the extra pointer per node really isn't an issue, usually.


If going back is not a common operation, the single list could be better.
Anyway, the point of these 'clever' interview questions is not always to
mimic reality, but to see how you think through problems, especially the
logic questions.

Anyway, I think if you just store the value of the pointer in one pass, you
could get better
node* node = yourfirstnode;
node* node5 = node->next->next->next->next;

T * last[5] = { &node->val, &node->next->val, ... };

for(int i=0; ; i = (i==5 ? 0 : 5))
{
if(node5->next == 0)
{
return last[i];
}
last[i] = &node5->val;
node5 = node5->next;
}
Jul 22 '05 #29
> Is it really faster than the 2 pass method?

If the time for retrieval of one element is high, then a one pass approach
is faster than a two pass.
However the method of using two pointers are actually a two pass algorithm
in that case (each element is loaded/passed twice).
A one pass algorithm, that would save time, would cache the elements once
loaded and maintain a local 5 element list.

Niels Dybdahl
Jul 22 '05 #30
> Is it really faster than the 2 pass method?

If the time for retrieval of one element is high, then a one pass approach
is faster than a two pass.
However the method of using two pointers are actually a two pass algorithm
in that case (each element is loaded/passed twice).
A one pass algorithm, that would save time, would cache the elements once
loaded and maintain a local 5 element list.

Niels Dybdahl
Jul 22 '05 #31
osmium wrote:

source writes:
function that would: return the 5th element from the end in a singly
linked list of integers, in one pass, and then provide a set of test
cases
against that function.

And please dont think this is my homework.
Can anybody simple tell me how this can be done. I am preparing for an
interview and wanted to know how to go about implementing this.
Or atleast tell me where I can find few examples on traversing Lists.
Even if I can get any good resources which explains lists properly my
job will be done.


If I understand the question, it can't be done; at least on a bare bones
linked list. You will have to count the elements so you can subtract five
from that. And then traverse the list again, which would be easiest from
the head end. I call that two passes, not one that you asked for. My
default meaning for end is the far end; I call the other end the beginning.


Or simply:
Use a queue to store exactly 5 pointers to nodes as you traverse the list.
When you have reached the end of the list, the queue will contain pointers
to the last 5 list elements.
--
Karl Heinz Buchegger
kb******@gascad.at
Jul 22 '05 #32
osmium wrote:

source writes:
function that would: return the 5th element from the end in a singly
linked list of integers, in one pass, and then provide a set of test
cases
against that function.

And please dont think this is my homework.
Can anybody simple tell me how this can be done. I am preparing for an
interview and wanted to know how to go about implementing this.
Or atleast tell me where I can find few examples on traversing Lists.
Even if I can get any good resources which explains lists properly my
job will be done.


If I understand the question, it can't be done; at least on a bare bones
linked list. You will have to count the elements so you can subtract five
from that. And then traverse the list again, which would be easiest from
the head end. I call that two passes, not one that you asked for. My
default meaning for end is the far end; I call the other end the beginning.


Or simply:
Use a queue to store exactly 5 pointers to nodes as you traverse the list.
When you have reached the end of the list, the queue will contain pointers
to the last 5 list elements.
--
Karl Heinz Buchegger
kb******@gascad.at
Jul 22 '05 #33

This thread has been closed and replies have been disabled. Please start a new discussion.

Similar topics

11
by: C++fan | last post by:
Suppose that I define the following class: class example_class{ public: example_class(); void funtion_1(); void function_2(); protected:
5
by: Dream Catcher | last post by:
1. I don't know once the node is located, how to return that node. Should I return pointer to that node or should I return the struct of that node. 2. Also how to do the fn call in main for that...
10
by: Kent | last post by:
Hi! I want to store data (of enemys in a game) as a linked list, each node will look something like the following: struct node { double x,y; // x and y position coordinates struct enemy...
6
by: Steve Lambert | last post by:
Hi, I've knocked up a number of small routines to create and manipulate a linked list of any structure. If anyone could take a look at this code and give me their opinion and details of any...
12
by: Eugen J. Sobchenko | last post by:
Hi! I'm writing function which swaps two arbitrary elements of double-linked list. References to the next element of list must be unique or NULL (even during swap procedure), the same condition...
12
by: joshd | last post by:
Hello, Im sorry if this question has been asked before, but I did search before posting and couldnt find an answer to my problem. I have two classes each with corresponding linked lists, list1...
51
by: Joerg Schoen | last post by:
Hi folks! Everyone knows how to sort arrays (e. g. quicksort, heapsort etc.) For linked lists, mergesort is the typical choice. While I was looking for a optimized implementation of mergesort...
1
by: theeverdead | last post by:
Ok I have a file in it is a record of a persons first and last name. Format is like: Trevor Johnson Kevin Smith Allan Harris I need to read that file into program and then turn it into a linked...
0
by: Atos | last post by:
SINGLE-LINKED LIST Let's start with the simplest kind of linked list : the single-linked list which only has one link per node. That node except from the data it contains, which might be...
7
by: QiongZ | last post by:
Hi, I just recently started studying C++ and basically copied an example in the textbook into VS2008, but it doesn't compile. I tried to modify the code by eliminating all the templates then it...
0
by: VivesProcSPL | last post by:
Obviously, one of the original purposes of SQL is to make data query processing easy. The language uses many English-like terms and syntax in an effort to make it easy to learn, particularly for...
0
by: jianzs | last post by:
Introduction Cloud-native applications are conventionally identified as those designed and nurtured on cloud infrastructure. Such applications, rooted in cloud technologies, skillfully benefit from...
0
by: abbasky | last post by:
### Vandf component communication method one: data sharing ​ Vandf components can achieve data exchange through data sharing, state sharing, events, and other methods. Vandf's data exchange method...
2
isladogs
by: isladogs | last post by:
The next Access Europe meeting will be on Wednesday 7 Feb 2024 starting at 18:00 UK time (6PM UTC) and finishing at about 19:30 (7.30PM). In this month's session, the creator of the excellent VBE...
0
by: fareedcanada | last post by:
Hello I am trying to split number on their count. suppose i have 121314151617 (12cnt) then number should be split like 12,13,14,15,16,17 and if 11314151617 (11cnt) then should be split like...
0
by: stefan129 | last post by:
Hey forum members, I'm exploring options for SSL certificates for multiple domains. Has anyone had experience with multi-domain SSL certificates? Any recommendations on reliable providers or specific...
1
by: davi5007 | last post by:
Hi, Basically, I am trying to automate a field named TraceabilityNo into a web page from an access form. I've got the serial held in the variable strSearchString. How can I get this into the...
0
by: DolphinDB | last post by:
The formulas of 101 quantitative trading alphas used by WorldQuant were presented in the paper 101 Formulaic Alphas. However, some formulas are complex, leading to challenges in calculation. Take...
0
by: Aftab Ahmad | last post by:
So, I have written a code for a cmd called "Send WhatsApp Message" to open and send WhatsApp messaage. The code is given below. Dim IE As Object Set IE =...

By using Bytes.com and it's services, you agree to our Privacy Policy and Terms of Use.

To disable or enable advertisements and analytics tracking please visit the manage ads & tracking page.