hey check this out.. http://www.w3schools.com/html/html_forms.asp
Hope this will serve the purpose:)
On Tuesday, 17 July 2012 01:12:02 UTC+5:30, nandita raman wrote:
Hello,
I am working on PHP/HTML/ XML-RPC.
I have a website, where one link shows the list of files as checkboxes and
a
Thanks all! It worked :)
On Thu, Jul 19, 2012 at 12:30 AM, sahil taneja sahiltanej...@gmail.comwrote:
hey check this out.. http://www.w3schools.com/html/html_forms.asp
Hope this will serve the purpose:)
On Tuesday, 17 July 2012 01:12:02 UTC+5:30, nandita raman wrote:
Hello,
I am
You have not taken care of the test case where the number is repeated more
than once in a single list and is not there in other lists.
Problem desc - the final list including only those ID numbers that
appeared in at least 2 out of the 3 lists
so it should not display the numbers which appear
@ raminder.. its clear that number cannot be specified in same list
twice..even asked in a comment from admin.
On 4/13/12, Ramindar Singh ramin...@gmail.com wrote:
You have not taken care of the test case where the number is repeated more
than once in a single list and is not there in other
@ kuldeep thanks...that wrked..bt still cud nt understand that what
difference making that global made..
On 4/13/12, Rishabh Jain rishabh@gmail.com wrote:
@ raminder.. its clear that number cannot be specified in same list
twice..even asked in a comment from admin.
On 4/13/12, Ramindar
EkoPath compiler has been open sourced
recently: http://www.pathscale.com/ekopath-compiler-suite
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Thanks alot
for the response.
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code blocks
On Jun 15, 9:05 pm, shashankreddy509 shashankreddy...@gmail.com
wrote:
Thanks alot
for the response.
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Solution:
int majorityElement(int a[], int n) {
if (a == null || a.length == 0 || n=0) return -1;
int mElement = a[0];
int count=1;
for (int i=1; i n; i++) {
if (a[i] == mElement) {
count++;
} else {
count--;
}
if (count =
a sort and another traversal would also do the same job in o( nlogn + n )
??
On Fri, Apr 15, 2011 at 12:49 AM, vishwakarma
vishwakarma.ii...@gmail.comwrote:
complexity : O(n) + O(nlogn)
Sweety wrote:
Question :Let A[1..n] be an array of integers. Design an efficient
divide and conquer
Search Topcoder Tutorial on Google..
On Apr 17, 9:06 pm, naga vinod kumar vinodkumark...@gmail.com wrote:
Hi Guys ,
Can any one give link for tutorial or videos about
segment trees. I am unable to understand the basic idea behind it .
Regards,
vinod
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complexity : O(n) + O(nlogn)
Sweety wrote:
Question :Let A[1..n] be an array of integers. Design an efficient
divide and conquer algorithm to determine if A contains a majority
element, i.e an element appears more than n/2 times in A. What is the
time complexity of your algorithm?
Answer:
I can post solution of this complexity if you want !!
On Apr 15, 12:19 am, vishwakarma vishwakarma.ii...@gmail.com wrote:
complexity : O(n) + O(nlogn)
Sweety wrote:
Question :Let A[1..n] be an array of integers. Design an efficient
divide and conquer algorithm to determine if A contains a
Yes , I also need the same...Thanks for the help .
On Fri, Apr 15, 2011 at 12:52 AM, vishwakarma
vishwakarma.ii...@gmail.comwrote:
I can post solution of this complexity if you want !!
On Apr 15, 12:19 am, vishwakarma vishwakarma.ii...@gmail.com wrote:
complexity : O(n) + O(nlogn)
Let A be the input array;
Now algorithm is follows;
struct leave{
int cand;
int count;
};
struct leave tree[120];
void build(int s,int e,int node ,int *A)
{
if(s == e){
tree[node].cand = A[s];
tree[node].count = 1;
example : let the array be { a,b,a,b,c,d,e,d,d,e,f};
now...
step 1 :pick any 2 different element and remove from the array till
array contains only same elements or any single element ...// dis
is implemented wid the above mentioned funtion build()
if there is any element whose occurence
Firstly it is srm 475, the following link has the problem
http://www.topcoder.com/stat?c=problem_statementpm=10878rd=14156
@crazysaikat : Sorry for misconstruing you. As this group is public it is
better not to post problems of a srm while it is running.
Apart from discussing it here, if you need
Hey i mean to say, please help after the contest only, i didn't mean
to cheat anyway, its only that i have to leave my net early due to
some reason thats why i posted it.
On Jul 6, 5:03 pm, Priyanka Chatterjee dona.1...@gmail.com wrote:
Contest in running and you are posting the 550 points
Hey, its on a java applet, you can download it from www.topcoder.com/tc,
and enter past competition, SRM 375, in it its the medium level
problem, i have copied all the text in the post, please help me, i
have got no idea how to tackle such type of problem.
On Jul 6, 5:20 pm, Jitendra Kushwaha
NP complete, sorry
On 3/14/09, Miroslav Balaz gpsla...@googlemail.com wrote:
NP what?
2009/3/13 Amina Maarouf amina.maar...@gmail.com
this problem was proved to be NP in a conference paper published recently.
On Fri, Mar 13, 2009 at 10:44 AM, Miroslav Balaz gpsla...@googlemail.com
I found a number of real life applications for the problem, like garbage
collection, a version of postman problem, gas system construction.. I am
looking for applications for the problem in networks, if any one can help.
On 3/14/09, Miroslav Balaz gpsla...@googlemail.com wrote:
NP what?
I don't know any concrete real life problem, of course everyting real life
problem that is in NP can be reducet to this problem.
That includes timetable sheduling. | think there is no such problem in
networks, because you need to know topology ad it has to be fixed. And if
you want to make
I think there is a 4^k kernel for TSP ..
On Sat, Mar 14, 2009 at 4:49 PM, Miroslav Balaz gpsla...@googlemail.comwrote:
I don't know any concrete real life problem, of course everyting real life
problem that is in NP can be reducet to this problem.
That includes timetable sheduling. | think
Hi,
I guess he is going by the assumption that the queries will be large
enough that precomputing the all-pairs shortest paths is better.
Anyways..mukesh..this is what you need..
In the place of your output
while(cin.get(c) c!='\n')
{
cin.unget();
On Feb 20, 11:18 pm, ramtin [EMAIL PROTECTED] wrote:
Hi,
could anybody help me to solve this problem
http://acm.zju.edu.cn/show_problem.php?pid=1002
This is a neat variant of the n-Queens problem: place n Queens on an
nxn chess board so that no two queens attack each other. The most
This is easy BackTrack problem ,I think
On 2/20/07, ramtin [EMAIL PROTECTED] wrote:
Hi,
could anybody help me to solve this problem
http://acm.zju.edu.cn/show_problem.php?pid=1002
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Won't brute force does the trick? Looks like it's at most 4x4 (correct
me if I'm wrong). You could play around with bitmasks probably, ... treat
each cell as one bit, ... probably also worthwhile to see what the bitmask
corresponds to the controlled cells if a rook is placed in certain
I'm sorry but I couldn't understand your code
and I think we can't use BackTrack because it is useful for problem
that we want to find a good leaf
but in this problem we must check all possible combinations
On Feb 20, 11:34 pm, Lego Haryanto [EMAIL PROTECTED] wrote:
Won't brute force does
Ok, ... it's easier to view it this way ... consider a maximum 4x4 board.
So, this is the same as 16 cells. In each cell, obviously it could either
be 0 or 1, ... 0 being empty, and 1 being occupied.
Just looking at the fact above, ... you can see we can have at most 2^16
combinations or
I am assuming that by random you mean uniformly at random (that
is, the probability of generating i is 1/5)
map the following pairs of numbers (1, 1), (1, 2), (1, 3), (1, 4), (1,
5), (2, 1), (2, 2) to 1, 2, .. , 7 respectively
let p_i denote the pair corresponding to the number i (e.g. p_1 = (1,
random generator is
uniformly distributed
-Original Message-
From: algogeeks@googlegroups.com [mailto:[EMAIL PROTECTED] On
Behalf Of Sandesh
Sent: Tuesday, January 30, 2007 11:09 PM
To: Algorithm Geeks
Subject: [algogeeks] Re: (need help) How to solve this random number
generatioin
Sent: Tuesday, January 30, 2007 11:09 PM
To: Algorithm Geeks
Subject: [algogeeks] Re: (need help) How to solve this random number
generatioin problem?
suppose given function returns the random numbers between 1 -5 then
you can have
(given + given) % 7 +1
which
On Jan 30, 8:57 pm, Jialin [EMAIL PROTECTED] wrote:
Question:
Given a program which can generate one of {1, 2, 3, 4, 5} randomly.
How can we get another generator which can generate one of
{1,2,3,4,5,6,7} randomly?
Thank you!
If you generate random 1 to 5 twice, there are 25 equally
Why can't we simply take the 1..5 random number, multiply by 7 and
divide by 5.
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Then you will get only the following numbers
(1*7)/5 = 1
(2*7)/5 = 2
(3*7)/5 = 4
(4*7)/5 = 5
(5*7)/5 = 7
What you will do to get 3 and 6?
On 2/1/07, pramod [EMAIL PROTECTED] wrote:
Why can't we simply take the 1..5 random number, multiply by 7 and
divide by 5.
/* You can do it as shown below. */
#include iostream
#include stdlib.h
using namespace std;
int random(int low,int high)
{
srand(time(NULL));
int n=(high-low+1);
return low + (rand()%n);
}
int main()
{
coutrandom(1,7);
return 0;
}
On 1/31/07, Jialin [EMAIL PROTECTED] wrote:
Question:
suppose given function returns the random numbers between 1 -5 then
you can have
(given + given) % 7 +1
which will generate between 1 and 7 .
-Sandesh Hegde
On Jan 31, 6:57 am, Jialin [EMAIL PROTECTED] wrote:
Question:
Given a program which can generate one of {1, 2, 3, 4,
It's not uniformly distributed, suppose the given random generator is
uniformly distributed
-Original Message-
From: algogeeks@googlegroups.com [mailto:[EMAIL PROTECTED] On
Behalf Of Sandesh
Sent: Tuesday, January 30, 2007 11:09 PM
To: Algorithm Geeks
Subject: [algogeeks] Re: (need help
Hi Minhaz
thnkx for help and pointing me to my mistake but now one thing i know
... is the number of points in the input will be distinct or there may
be some duplicate point . like suppose the input is like this...
9
0 1
0 2
1 2
1 3
2 3
3 3
3 0
1 0
1 1
now the output of this program will
(0
Hi Minhaz
thnkx for help and pointing me to my mistake but now one thing i know
... is the number of points in the input will be distinct or there may
be some duplicate point . like suppose the input is like this...
9
0 1
0 2
1 2
1 3
2 3
3 3
3 0
1 0
1 1
now the output of this program will
(0
well... i dont check this grp that frequently.
but well... if u know hungerian then, Hopcroft is similar to hungarian except u need BFS here, and in initial push all the unmatched vertex of Left in Queue, thats all.
On 8/21/06, phoenixinter [EMAIL PROTECTED] wrote:
Hi guysI'm recently trying to
hi SPX2,
can u explain your algorithm in a simple language?
thank you
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anybody can answer in detain??
thanks
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Isn't third problem solved by sorting where comparison function is
replaced by the equivalence tester? After sorting we just run through
the array to see if there n/2 repetitions.
ajay mishra wrote:
@stefan , ur idea seems correct to me.
On 3/8/06, SPX2 [EMAIL PROTECTED] wrote:
ajay
@pramod
not quite ... the equivalence tester does not give a less than
operation which is needed for sorting. It only tells whether they are
equal or not. You cannot order them on the basis of that.
-Dhyanesh
On 3/9/06, pramod [EMAIL PROTECTED] wrote:
Isn't third problem solved by sorting where
That's true Dhyanesh, thanks. Any prose explanation for SPX2's pseudo
code?
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@stefan , ur idea seems correct to me.On 3/8/06, SPX2 [EMAIL PROTECTED] wrote:
-- Ajay kr. Mishrahttp://ajay.mishra19.googlepages.comIIT KGP
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i believe for the second problem it is enough to sort them using
quicksort method in O(nlgn) but retain a vector with initial
positions(wich after quicksort is applied will be a permutation vector
of the original one).
Now you have to find out number of pairs (ai,aj) with ij and
ai2aj.
Now this
Finding ah...isnt that simple... you have to do a binary search for
it... that means
that if you're on step j in the for above...you will have to do
(n-j)lg(n-j)computations
well...I think it would bring the algorithm to a total of O(sum from
j=0 to n of (n-j)lg(n-j)
O(n^2lgn)
if anyone can
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