Thank you for the code snippet.
What I don't understand is
if( (double)rand() / RAND_MAX 1 / k )
Here, Why do we use less than inequality?
Why won't Udit's solution of just using the minimum random number work?
On Tuesday, 1 January 2013 20:57:47 UTC+5:30, Dave wrote:
@Doom: Yes. It is
Hi Dave
Please help me with some additional clarification regarding the
implementation of this algo. How do we make use of random number generator?
And is it necessary to traverse the list until the end?
On Friday, 28 December 2012 19:35:07 UTC+5:30, Dave wrote:
@Sanjeev: Set a pointer p
@Doom: Yes. It is necessary to go through the entire list. The code could
look like this:
int* p = head;
int k = 1;
while( head )
{
head = head - next;
k++;
if( k * (double)rand() RAND_MAX )
p = head;
}
Dave
On Sunday, December 30,
@Sanjeev: Set a pointer p to the first node. Traverse the list. When at the
kth node, set p to the kth node with probability 1/k. When you reach the
end of the list, return p, which will point to a random node with uniform
probability.
Dave
On Thursday, December 27, 2012 11:18:20 PM UTC-6,
@Dave: U said that the node p returned will be assigned to some kth
node with probability 1/k. then the probability for p to be assigned
to 1,2,3... nodes would be like 1/1, 1/2, 1/3, and so on.. the how is
the node p is assigned with uniform probability.
On Fri, Dec 28, 2012 at 7:35 PM, Dave
@Udit: Here is the proof: At step k, node k is selected with probability
1/k. On the (k+1)st step, node k is replaced with probability 1/(k+1). Thus
it is retained with probability 1 - 1/(k+1) = k/(k+1). On subsequent steps,
node k is retained with probabilities (k+1)/(k+2), (k+2)/(k+3), ...,
Given a linked list of infinite length. Write a function to return a random
node.
Constraints:
1 You can traverse a singly linked list only once.
2 You can not use any extra space.
Thanks in advance.
--
Well my algo will be Something like this
1 Get a Random number. Perhaps You can have the function like Randon(List
*head, int Randomnumber)
2 Use the function argument Randomnumber to loop the list.
i.e. for(int count=0;count=Randomnumber;count++ ){
head =
But suppose a random number generate a value 5 and your linked list has
only four elements. In that case what would be the answer ???
On Thu, Dec 27, 2012 at 4:03 PM, Prem Krishna Chettri hprem...@gmail.comwrote:
Well my algo will be Something like this
1 Get a Random number. Perhaps You can
Those are the Cornor Test Cases.. Well we here talk about Algos and
Complexity (Atmost Pseudo code ) not about unit test cases and source
code.. :)
Ofcourse U can include those checks either when U call the function or
inside the functions..
On Thu, Dec 27, 2012 at 4:06 PM, naveen shukla
You said : Given a linked list of infinite length
On Thu, Dec 27, 2012 at 4:06 PM, naveen shukla
naveenshuklasweetdrea...@gmail.com wrote:
But suppose a random number generate a value 5 and your linked list has only
four elements. In that case what would be the answer ???
On Thu, Dec 27,
If the length of the linked is infinite then the above algo would do
the needful.
In case you have a finite length linked list, then you can normalize
the random value using following:
Suppose length of linked list is: l
random number is: r; and r l
then new random number would be: r1 = r%l;
now
i mean the length of the linked list is not known to us.
@udit how can we do this in single traversal ? i think we need to traverse
the linked list twice in your case.
Please reply if i am wrong ?
On Thu, Dec 27, 2012 at 10:48 PM, Udit Agarwal uditii...@gmail.com wrote:
If the length of the
no.. i din't know that in ur case infinite number of nodes means
unknown number of the finite nodes. I thought it to be like we have as
much number of node as we want.
and as u said if the length is know to be 4 and random number is 5
then for that i gave my approach like u can now generate a new
thanks udit :) i think this approach will surely work.
On Fri, Dec 28, 2012 at 1:19 PM, Udit Agarwal uditii...@gmail.com wrote:
no.. i din't know that in ur case infinite number of nodes means
unknown number of the finite nodes. I thought it to be like we have as
much number of node as we
Hi we need find a node in linked list in O(logn). You can change the list
as u like.
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not possible i suppose..
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On Mon, Jun 4, 2012 at 3:00 PM,
This is possible only if Linked List is sorted then we can apply Merge Sort
for Linked List which would be in place.
Otherwise the time complexity would be O(n logn).
On Mon, Jun 4, 2012 at 3:00 PM, VIHARRI viharri@gmail.com wrote:
Hi we need find a node in linked list in O(logn). You can
can be done using skip lists
On Mon, Jun 4, 2012 at 3:03 PM, Jeevitesh jeeviteshshekha...@gmail.comwrote:
This is possible only if Linked List is sorted then we can apply Merge
Sort for Linked List which would be in place.
Otherwise the time complexity would be O(n logn).
On Mon, Jun 4,
as question says you can change the list as u like...i guess skip list is
the answer.
On Mon, Jun 4, 2012 at 3:51 PM, SANDEEP CHUGH sandeep.aa...@gmail.comwrote:
can be done using skip lists
On Mon, Jun 4, 2012 at 3:03 PM, Jeevitesh jeeviteshshekha...@gmail.comwrote:
This is possible only
Hi ,
I think the only possiblity is to make it doubly linked list and then
consider next prev as left and right child like tree and then perform
search as we in tree , it would serve the purpose .
let me know if iam wrong .
On Mon, Jun 4, 2012 at 3:51 PM, SANDEEP CHUGH
well converting single linked list to balanced BST...this would also work
On Mon, Jun 4, 2012 at 4:29 PM, Nishant Pandey nishant.bits.me...@gmail.com
wrote:
Hi ,
I think the only possiblity is to make it doubly linked list and then
consider next prev as left and right child like tree and
Why don't you use templates ?
template class T
class LNode
{
public:
LNode(T pData,LNode *pNext):data(pData),next(pNext){}
T data;
LNode *next;
};
template class T
class LList
{
protected:
LNodeT *head;
LNodeT *tail;
public:
LList()
{
head=tail=NULL;
}
void push_back(T pData)
{
this way u can do it in c creation and printing of generic lisk list .
void(List **p,void *data, unsigned int n)
{
List *temp;
int i;
/* Error check is ignored */
temp = malloc(sizeof(List));
temp-data = malloc(n);
for (i = 0; i n; i++)
*(char *)(temp-data +
in case of generic link list when u have to create the node u have to copy
the data part character by character ie always one -one byte and in this
way whatever be the data type of ur data u can easily get the data and
u will keep doing this until the size of the data that to entered .
On Thu,
how to implement generioc linked list..using void pointer...i havent
used void pointer much so, m not able to use it properly in linked
list..please help asap !!!
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design a linked list that has only one pointer per node yet can be
traversed in forward or reverse direction.
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
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Algorithm
XOR Linked Lists
On Sat, Jan 14, 2012 at 7:06 PM, Ashish Goel ashg...@gmail.com wrote:
design a linked list that has only one pointer per node yet can be
traversed in forward or reverse direction.
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
yup XOR link list
On Sat, Jan 14, 2012 at 7:13 PM, Ankur Garg ankurga...@gmail.com wrote:
XOR Linked Lists
On Sat, Jan 14, 2012 at 7:06 PM, Ashish Goel ashg...@gmail.com wrote:
design a linked list that has only one pointer per node yet can be
traversed in forward or reverse direction.
XOR linked lists...
dis willbe
On Sat, Jan 14, 2012 at 7:13 PM, Ankur Garg ankurga...@gmail.com wrote:
XOR Linked Lists
helpful if not known..
http://www.geeksforgeeks.org/archives/12367
Regards,
PAYAL GUPTA,
NIT-BHOPAL.
On Sat, Jan 14, 2012 at 7:06 PM, Ashish Goel ashg...@gmail.com
sorry 4 d error in last post..
dis link vud b helpful..if not known previously...
http://www.geeksforgeeks.org/archives/12367
Regards,
PAYAL GUPTA,
NIT-BHOPAL.
On Sat, Jan 14, 2012 at 7:29 PM, payal gupta gpt.pa...@gmail.com wrote:
XOR linked lists...
dis willbe
On Sat, Jan 14, 2012 at
http://cslibrary.stanford.edu/105/LinkedListProblems.pdf
-Moheed
On Sun, Dec 25, 2011 at 11:39 PM, atul anand atul.87fri...@gmail.comwrote:
@ashish : please provide link for that page
On Sun, Dec 25, 2011 at 10:36 PM, Ashish Goel ashg...@gmail.com wrote:
refer stanford page
1,2,3,4,5,6
refer stanford page
1,2,3,4,5,6
will become
5,3,1
6,4,2
void MoveNode(struct node** destRef, struct node** sourceRef) {
struct node* newNode = *sourceRef; // the front source node
assert(newNode != NULL);
*sourceRef = newNode-next; // Advance the source pointer
newNode-next = *destRef; // Link
@ashish : please provide link for that page
On Sun, Dec 25, 2011 at 10:36 PM, Ashish Goel ashg...@gmail.com wrote:
refer stanford page
1,2,3,4,5,6
will become
5,3,1
6,4,2
void MoveNode(struct node** destRef, struct node** sourceRef) {
struct node* newNode = *sourceRef; // the front
@ashish : acc to Karthikeyan question ...output should not be in reverse
order.
there is not much difference b/w both implementation.
it will become little interesting if you try to solve using recursion .
On Sun, Dec 25, 2011 at 10:36 PM, Ashish Goel ashg...@gmail.com wrote:
refer stanford
Segregate even and odd psoitioned nodes in a linked list
Eg:
2-3-1-9-7-5
The output should be two separate lists
2-1-7
3-9-5
*My code is:*
void segregate(node* head)
{
int i=1;
node* sec=head-next;
node *odd=head,*even=head-next;
while(even)
{
if(i%2)
{
odd-next=even-next;
even-next=NULL;
because you are doing odd=odd-next; and even=even-next;
you will lose head pointers for the two linked list formed once you come
out of the loop.
void segregate(node* head)
{
node *temp,*odd,*even;
toggle=1;
if(head==NULL)
{
return;
}
odd=head;
As you mentioned, the numbers designating the gaps can only be repeated
so in your example 2 20 can be broken into 19 1 but 19 is already there in
the list (the
first couplet) and it's not the one which describes the gap. can you please
clarify?
On Mon, Nov 21, 2011 at 5:23 AM, Ankur Garg
a linked list contains
2 19 _ _ 3 47 _ _ _ 2 20 _ _ ..and so on
I have to fill those empty nodes with numbers whose sum is equal to the
numbers
occurring just before the gaps and the number of gaps is determined by the
node
which is at 2 distance before the gaps with the limitation
There are two linked list of length m and n. There is some common data
at the end of both. Find the starting position in both the linked
list. I could suggest two methods
1) Reverse the list and check .
2) Use recursion to go to the last element and move back from there.
Is there any other way ?
consider the length of list1 is 7(ie m) and length of list2 is 5 (ie n)..now
find (m-n) ie 7-5=2..since length of list2 is smaller than list1,,have a
pointer pointing to the head of the list2 and move that pointer (m-n)
positions ie (here 2 positions)..now have a pointer pointing to the head of
@prasanth but how can u know that it will start after the difference
of the number of nodes (m-n) . they may be similar even after one
element also ??
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@annarao : as the common data is at the end of the linked lists so the
length of common data can not be greater than the length of shorter linked
list . so i think prasanth is right !!
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Computer Engg.
NIT Kurukshetra
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@Ankuj Gupta: your algo wont work
see,what happens when ur algo run.
list1 1-2-3-4-5-6
list2 9-8-4-5-6
reverse list 1
list1 6-5-4-3-2-1
6-5-4
^
|
list2 8-9
reverse list 2
list1 6-5-4-9-8
list2 4-9-8
Thank you,
Sid.
On Fri, Sep 30, 2011 at 9:37
find length of linked list 1 say m
find length of linked list2 say n
take mod(m-n) say k
traverse k nodes in bigger lis
after that both listts one position at a time until the pointers are equal
On Fri, Sep 30, 2011 at 9:34 AM, partik madan partikma...@gmail.com wrote:
@annarao : as the common
delete all the numbers found in list2 from list1 recursively or iteratively
Also optimize ur algo when list1 is sorted in ascending order and list2 is
sorted in descending order
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count number of elements from both lists and reverse list with minimum
number of elements, go ahead with checking and deleting linearly
surender
On Fri, Jul 15, 2011 at 10:38 PM, Nishant Mittal mittal.nishan...@gmail.com
wrote:
delete all the numbers found in list2 from list1 recursively or
@surender thanx for ur algo but tell me about the 1st part when numbers are
not sorted.
i think if we apply merge sort on both the list then it would be easy to
delete after sorting.
correct me if i m wrong
On Sat, Jul 16, 2011 at 12:40 AM, surender sanke surend...@gmail.comwrote:
count
if lists are unordered, make a map of list2 and during traversal of list1
search for map, if found delete that node
surender
On Sat, Jul 16, 2011 at 2:02 AM, Nishant Mittal
mittal.nishan...@gmail.comwrote:
@surender thanx for ur algo but tell me about the 1st part when numbers are
not
*if i have to write a code for finding middle element in a linked list..
then for even length linked which will be the middle element both n/2-1 and
n/2+1 or just n/2+1???
if its both then how can i make my function two elements??
*
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i have considered it (n/2+1)th
and didn't bothered much :P
On Mon, Jul 11, 2011 at 4:57 PM, Anika Jain anika.jai...@gmail.com wrote:
*if i have to write a code for finding middle element in a linked list..
then for even length linked which will be the middle element both n/2-1 and
n/2+1 or
dnt worry about that write a code in which initialy 2 ptrs .move 1 ptr 2
nodes at a time and other ptr 1 time.when the first ptr reaches end then
seconf ptr will be at middle
On Mon, Jul 11, 2011 at 6:15 PM, vaibhav shukla vaibhav200...@gmail.comwrote:
i have considered it (n/2+1)th
and
moreover n/2-1 will never be the middle element. in case of even no. of
nodes ,middle shall be n/2 and n/2+1
On Mon, Jul 11, 2011 at 6:19 PM, chandy jose paul jpchaa...@gmail.comwrote:
dnt worry about that write a code in which initialy 2 ptrs .move 1 ptr 2
nodes at a time and other ptr 1
yeah.
it totally depends on the no. of data fields :)
On Mon, Jul 4, 2011 at 2:03 PM, sunny agrawal sunny816.i...@gmail.comwrote:
@Sagar
if it has a large no of data fields than don't u think just changing
pointers will be much better than swapping all the data contained in the
node
On
@Sagar
if it has a large no of data fields than don't u think just changing
pointers will be much better than swapping all the data contained in the
node
On Mon, Jul 4, 2011 at 11:13 AM, sagar pareek sagarpar...@gmail.com wrote:
@Anantha Krishnan
Well be specific
just read the question
@Anantha Krishnan
Well be specific
just read the question again. it has only to reorder numbers...
what problem will occur if it has more than one data fields
we can traverse with help of digits and then swap all the data fields
On Thu, Jun 30, 2011 at 5:03 PM, Anantha Krishnan
http://geeksforgeeks.org/?p=12000
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Here is one approach which works :)
Traverse the list from a pointer name ptr
when encounter odd then start traversing from a tmp pointer (start point is
ptr-next) to find even no. when find it swap the numbers.
then again start forwarding ptr. If during traversing ptr==tmp make
flag=0(at start
Check this
http://ideone.com/1I40z
On Wed, Jun 29, 2011 at 8:04 PM, Nishant Mittal
mittal.nishan...@gmail.comwrote:
segregate even and odd nodes in a singly linked list.Order of even and
odd numbers must be same...
e.g:-
i/p list is 4-1-3-6-12-8-7-NULL
o/p list 4-6-12-8-1-3-7-NULL
--
@sagar pareek
If there are more fields in the node like:
struct node{
int data;
float mark;
char ch;
struct node *link;
};
Here swapping *data* alone will corrupt the list right!!
On Thu, Jun 30, 2011 at 4:38 PM, sagar pareek sagarpar...@gmail.com wrote:
Here is one
segregate even and odd nodes in a singly linked list.Order of even and
odd numbers must be same...
e.g:-
i/p list is 4-1-3-6-12-8-7-NULL
o/p list 4-6-12-8-1-3-7-NULL
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maintain two pointers one at the tail of even number list one at tail of odd
Number list
traverse the list and add the number at required list
On Wed, Jun 29, 2011 at 8:04 PM, Nishant Mittal
mittal.nishan...@gmail.comwrote:
segregate even and odd nodes in a singly linked list.Order of even and
node *segregate(node *head)
{
node *even,*odd,*even1,*odd1;
even=odd=NULL;
while(head)
{
if((head-data)%2)
{
if(!odd)
{
odd = head;
can nodes of linked list in c/c++ contain different types of
datameans suppose 1st node of the list contains an integer value,
2nd contains a float,3rd has a string and so on..
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1) You can use void* as the data part and then allocate memory separately
and use.
struct node
{
void* dataPtr;
struct node* next;
};
2) or Use embeddable lists:
http://isis.poly.edu/kulesh/stuff/src/klist/
Thanks,
Balaji.
On Wed, Mar 16, 2011 at 8:27 PM, himanshu kansal
Given a linked list structure where every node represents a linked
list and contains two pointers of its type:
(i) pointer to next node in the main list.
(ii) pointer to a linked list where this node is head.
Write a C function to flatten the list into a single linked list.
Eg.
If the given
Node *flatten(Node *node) {
if (!node)
return NULL;
Node *head = node;
Node *next = node-next;
node-next = NULL;
if (node-down)
node-next = flatten(node-down);
while (node-next != NULL)
node = node-next;
node-next = flatten(next);
return head;
}
On Wed, Feb 16, 2011 at 8:38 PM, bittu
Hi,
Could anyone help me representing linked list in the form a binary
tree ?
Thanks !!
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Actually, a linear data structure like a linked list is also a
specific kind of tree structure.
2010/8/23 Raj N rajn...@gmail.com:
Hi,
Could anyone help me representing linked list in the form a binary
tree ?
Thanks !!
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What will be the representation. How do you define left and right pointers
of the tree for a linked list.
On Mon, Aug 23, 2010 at 10:35 PM, Yan Wang wangyanadam1...@gmail.comwrote:
Actually, a linear data structure like a linked list is also a
specific kind of tree structure.
2010/8/23 Raj N
that's because you are creating char str[20]; on the stack so when the
function createList completes all str [20] will be deleted from stack as a
part of stack unwinding. and next you are calling display where all the
references to name is removed and you are getting gabage value.to avoid this
you
#includeiostream
#includemalloc.h
#include string.h
using namespace std;
struct node
{
void *name;
struct node *next;
};
typedef struct node Node;
void createList(Node **head )
{
char *str= (char*)malloc(20*sizeof(char));
char *p;
coutEnter a String: ;
gets (str) ;
wats d problem in my display()
#includeiostream
#includemalloc.h
#include string.h
using namespace std;
struct node
{
char *name;
struct node *next;
};
typedef struct node Node;
void createList(Node **head )
{
char str[20];
char *p;
coutEnter a String: ;
gets (str)
make declaration of char str[20] to char
*str=(char*)malloc(20*sizeof(char));
and it will work .
On Wed, Aug 18, 2010 at 11:23 PM, Raj Jagvanshi raj.jagvan...@gmail.comwrote:
wats d problem in my display()
#includeiostream
#includemalloc.h
#include string.h
using namespace std;
void display(Node *head)
{
cout\n;
for( ; head ; head=head-next)
cout\thead-name;
cout\n;
}
when head reaches last node
condition head is true , then head will become head-next which is null ,
and it will try to print the name field from of a null value which is error
On
no it head to head-next at the end of for loop block not at the beginning.
try out this
for (int t=0;t5;t++)
cout\nt;
On Thu, Aug 19, 2010 at 5:22 PM, vineeth mohan vm.vineethmo...@gmail.comwrote:
void display(Node *head)
{
cout\n;
for( ; head ; head=head-next)
my display function print garbage value from begining
On Thu, Aug 19, 2010 at 4:23 PM, ram das ramnaraya...@gmail.com wrote:
make declaration of char str[20] to char
*str=(char*)malloc(20*sizeof(char));
and it will work .
On Wed, Aug 18, 2010 at 11:23 PM, Raj Jagvanshi
my display function print garbage value from begining
On Thu, Aug 19, 2010 at 5:22 PM, vineeth mohan vm.vineethmo...@gmail.comwrote:
void display(Node *head)
{
cout\n;
for( ; head ; head=head-next)
cout\thead-name;
cout\n;
}
when head reaches last node
condition
can any one tell me in the below post
struct node
{
int info;
struct node *next;
};
typedef struct node NODE; 1. why typedef used whats d
benifet of doing this
NODE *start;
void createmptylist(NODE **start) 2. why Node **start used instead we
can also use Node *start get same re.
{
Keep a pointer list1 at the beginning of one of the lists, and call the
function below on the other list.
int reverseCheck(NODE *p)
{
list2=p;
if(list2-next==NULL)
return;
reverseCheck(list2-next);
if(list2-next-data==list1-data)
list1=list1-next;
else
flag=0;
return
if we dont want to change original list..
is there ny efficient method for that?
On 23 June 2010 06:49, ANUJ KUMAR kumar.anuj...@gmail.com wrote:
reverse one of the linked lists in O(n) time and then see if the
resulting one is same as the other one
On Wed, Jun 23, 2010 at 1:56 AM, divya
this is same as finding palindrome in a given linked list..
it may help
http://geeksforgeeks.org/?p=1072
On Wed, Jun 23, 2010 at 6:00 PM, divya jain sweetdivya@gmail.comwrote:
if we dont want to change original list..
is there ny efficient method for that?
On 23 June 2010 06:49, ANUJ
Two link lists are given ...check if one is reverse of other. Do it
without using extra space.
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1. Given a single link list with one info part containing single character
and a link. Check whether the link list is a palindrome or not.
The algo should run in Linear time only. You can't use any array or string
to store the string of link-list.
2. You are given a Double Link List with one
what does linked list corruption mean ..
how to detect linked list corruption
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is there any algo that can search a linked list in less than O(n) time
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what is the most efficient way to search a linked list in sorted order ?
is there an algorithm that performs faster than O(n)
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what is the most efficient way to search a linked list in sorted order ?
is there an algorithm that performs faster than O(n)
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