if integers are positive,then go on a cycle... like a[2]goes to its final
position, the element in a[2]'s final position goes to its final position,
and so on... each time on visiting an element, put some marker on it...
like make it negative... finally after an element comes to position of
@Wladimir, yeah I have heard about that. Another way of populating primitive
pythagoreans is, for any natural number m 1 (m^2 - 1, 2m, m^2 + 1) forms
a pythagorean triplet. This is useful in populating pythagorean tiplets but
here the problem is to search such triplets from a given int array.
@
You can take advantage of a basic property of triagle that
sum of largest side of triangle sum of other two sides,
After sorting you could easily deside the range in which possible solution
could be found for a chosen hypotenuse
On Fri, Oct 14, 2011 at 11:10 AM, ravindra patel
can u please take an example and explain itcould not understand
the statement a[2]'s final position goes to its final position
On Oct 14, 10:40 am, Siddhartha Banerjee thefourrup...@gmail.com
wrote:
if integers are positive,then go on a cycle... like a[2]goes to its final
position, the
@Rahul
Pls elaborate with an example ...
On Fri, Oct 14, 2011 at 2:35 PM, rahul patil
rahul.deshmukhpa...@gmail.comwrote:
You can take advantage of a basic property of triagle that
sum of largest side of triangle sum of other two sides,
After sorting you could easily deside the range in
using properties of tiangle wont help i guess. the will give the range
of VALUES you want to restrict yourself to. not the range of INDEX's
of the array...
On Fri, Oct 14, 2011 at 3:14 PM, Ankur Garg ankurga...@gmail.com wrote:
@Rahul
Pls elaborate with an example ...
On Fri, Oct 14, 2011 at
suppose sorted array is
1,2,3,5,10,12,13,17,19,25
so if u want to find possible combinations, with 25 as hypotenuse, then only
range 10 ... 19 could have answer
as 19 + 10 25
On Fri, Oct 14, 2011 at 3:14 PM, Ankur Garg ankurga...@gmail.com wrote:
@Rahul
Pls elaborate with an example ...
@rahul It still will be O(n^2) time complexity
On 14 October 2011 15:14, Ankur Garg ankurga...@gmail.com wrote:
@Rahul
Pls elaborate with an example ...
On Fri, Oct 14, 2011 at 2:35 PM, rahul patil
rahul.deshmukhpa...@gmail.com wrote:
You can take advantage of a basic property of
http://stackoverflow.com/questions/575117/pythagorean-triplets
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
On Fri, Oct 14, 2011 at 3:14 PM, Ankur Garg ankurga...@gmail.com wrote:
@Rahul
Pls elaborate with an example ...
On Fri, Oct 14,
@siddharth what is the complexity?
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#includeiostream#includestdlib.husing namespace std;char *
remove_spaces(char *s){
char *result;
char *temp;
temp=s;
result=s;
while(*s!='\0')
{
if(*s==' ')
{
s++;
}
else
{
*temp = *s;
temp++;
s++;
@Utkarsh As efficient as possible..
On Fri, Oct 14, 2011 at 6:25 AM, UTKARSH SRIVASTAV
usrivastav...@gmail.comwrote:
@siddharth what is the complexity?
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Interview-Street Group until Whole Thing is Done , You Can Search
The
@Gaurav how will u do for
a1a2a3a4a5b1b2b3b4b5c1c2c3c4c5
On Fri, Oct 14, 2011 at 6:49 AM, gaurav yadav gauravyadav1...@gmail.comwrote:
consider following example...
suppose initailly we have a1a2a3b1b2b3c1c2c3
then do the following-
a1a2a3 b1b2b3 c1c2c3 (look for b1 in the remaining
Its Out Shuffle not In shuffle, although both are similar and you can read
both
On Fri, Oct 14, 2011 at 8:26 PM, sunny agrawal sunny816.i...@gmail.comwrote:
this problem is like Card shuffling problem(search for In-shuffle)
i think solution is
if indexing is zero based
each i will go to -
this problem is like Card shuffling problem(search for In-shuffle)
i think solution is
if indexing is zero based
each i will go to - k*i % (N-1)
k = 3 and N = 3*n -1
n = no of cards in one pile Or No of a's
On Fri, Oct 14, 2011 at 7:52 PM, shiva@Algo shiv.jays...@gmail.com wrote:
i dnt knw why still these posts are allowed in algogeeks
On Fri, Oct 14, 2011 at 7:37 PM, WgpShashank shashank7andr...@gmail.comwrote:
@All Join Interview-Street Group For Detailed Discussion About Interview
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What should be the output in this case : a1a2a3a4b1b2b3b4c1c2c3c4 ??
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@shiva...keep swapping the underline elements
a1*a2*a3a4a5*b1*b2b3b4b5c1c2c3c4c5
a1b1*a3*a4a5a2b2b3b4b5*c1*c2c3c4c5
a1b1c1*a4*a5*a2*b2b3b4b5a3c2c3c4c5
a1b1c1a2*a5*a4*b2*b3b4b5a3c2c3c4c5
a1b1c1a2b2*a4*a5b3b4b5a3*c2*c3c4c5
a1b1c1a2b2c2*a5*b3b4b5*a3*a4c3c4c5
a1b1c1a2b2c2a3b3*b4*b5a5a4*c3*c4c5
Pick a language to work in.
Open the file in whatever way your language of choice allows.
Read the data.
And... you are done.
A better answer will probably require a better question.
Dan ;-)
On Oct 13, 10:44 am, karthikeya s karthikeya.a...@gmail.com wrote:
Does anyone have an idea about
i have the code solution to this.. if m allowed to post it here then i can i
post it. m i allowed to post code here?
On Fri, Oct 14, 2011 at 9:40 PM, gaurav yadav gauravyadav1...@gmail.comwrote:
@shiva...keep swapping the underline elements
a1*a2*a3a4a5*b1*b2b3b4b5c1c2c3c4c5
Is there a pattern in the indices of the numbers we are swapping. some
formula which may tell the next two indices to swap.
Thanks,
- Ravindra
On Fri, Oct 14, 2011 at 9:40 PM, gaurav yadav gauravyadav1...@gmail.comwrote:
@shiva...keep swapping the underline elements
On Oct 13, 7:52 pm, shiva@Algo shiv.jays...@gmail.com wrote:
Convert an array a1 a2 a3...an b1 b2 b3...bn c1 c2 c3...cn to a1b1c1
a2b2c2...anbncn, inplace
See the algorithm for memory efficient rearrangement of array elements
in one of the books by Robert Sedgewick such as Algorithms in C++ or
for instance, you'll need some kind of 'magic-key verification' or any
other 'pattern matching' with a big dictionary, a wide and complete
one; such as the one implemented by command 'file' in unix.
On Thu, Oct 13, 2011 at 3:44 PM, karthikeya s karthikeya.a...@gmail.com wrote:
Does anyone have
+1 Dan.
I've been seeing lots of vague questions around. I thought of doing
the same but for the sake of peace I decided not to troll this one.
I'm not saying its OP's case but many people either think the job of
answering questions is costless or maybe think it's our duty to do so.
If you need
Given a n vertex polygon, find in how many ways k non intersecting diagonals
can be drawn ?
or
in How many ways it can be divided into k+1 regions such that no 2 diagonals
intersect ?
Limits:1 = k = N = 10^9
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Sunny Aggrawal
B.Tech. V year,CSI
Indian Institute Of Technology,Roorkee
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