So, how do you do that sir?
Basically , we should identify all the paths between those two nodes and
keep track of maximum of sums.
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@ravi
yep, you're right.
But method is similar to finding diameter as given on geeksforgeeks as atul
suggested . Thanks.
On Mon, Aug 27, 2012 at 11:23 PM, Ravi Maggon maggonr...@gmail.com wrote:
@atul:
I think he is asking for max. sum of elements between 2 leaf nodes and not
the max
Guys Something good to share here..
Query :- Tomorrow I am Planning to take my car for a ride of 1 ( Ten
Thousands ) Km. As the journey will be rough so I am planning to pack 5
new tyres.. Now being very conservative, I want to use each of them such
that they all wear equally.. Now the
@Prem: You will travel 40,000 tire (American spelling) km. Therefore, each
tire needs to be on the ground for 8,000 km. One way to accomplish this is
to rotate the 5 tires each 2,000 km. Then, each tire will be on each wheel
for 2,000 km, and will be the spare for 2,000 km. Another way, with
If there are total 5 tyres, including the 4 attached ones...
you can change do a rotation, change tyre after every 2000 kms( or the
factors of 2000), in that case, the journey by each one would be 8000.
i think there can be several interpretations of the qn. IF for five extra
tyres, wud mean equal
@Dave.. Yeah that correct .. Its fairly simple to get in the
interviews. A bit trickier one for me is to get the corrupted link
list correction... I have seen but not concentrated enough during the
hours..
On Tue, Aug 28, 2012 at 10:02 PM, Aditya Virmani
virmanisadi...@gmail.comwrote:
Code:-
#includeiostream
#includevector
using namespace std;
void recursion(int sum,int i,vectorint v,int size)
{
vectorint v1=v;
int size1=size;
if(sum==0)
{
for(int k=0;ksize;k++)
coutv[k] ;
coutendl;
}
else
{
for(int j=i+1;j10;j++)
I guess it might be
finding maximum sum path for left subtree + max sum path for right subtree
+ root-data
As in case of finding the diameter which say
height(root-left)+height(root-right)+1
Please correct me!
On Mon, Aug 27, 2012 at 11:46 PM, kunal rustgi rustogi.ku...@gmail.comwrote:
@ravi
@atul-I think This Should work for n dimension-
complexity O(n^no.of dimesions)
:-
have N-dimension check for which Tile can contain which Tile.i.e (3,3,4)
can contain (2,3,4) .
Now
1.Take the titles which cannot contain any-other tile set no-of tile if it
is base =1;
2.now take tiles which can
Hope This will Work..
int Recursion(node *root)
{
if(root-left==Nullroot-right==Null)
{
return root-data;
}
int a=0,b=0;
a=recursion(root-left);
b=recursion(root-right);
if(a+b+root-datamax)
max=a+b+root-data;
return max(a,b)+root-data;
}
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@all-This is almost the same question asked in facebook..
Dave sir is correct!
Example-file contain only a b a c;
Then step1:
save a;
step2:
save b with probability of value 0 by rand()%2;
therefore pr[1]=1/2;pr[2]=1/2;
step3;
save a with probability of value 0 by rand()%3;
therefore
-- Forwarded message --
From: sachin singh sachin...@gmail.com
Date: Tue, Aug 28, 2012 at 10:23 PM
Subject: Snapdeal placement proceedure
To: algogeeks@googlegroups.com
Can anyone tell about the recruitment process of snapdeal?
I mean how to prepare for the written test?What
plz provide me algo for this,thnx
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@vaibhav-yup :-)
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I think the path between two leaves always passes thru root.
Now we can keep track of root-to-leaf path whose sum is max and secondMax
among all sums.
Now between this two leaves path, root is repeated.
Hence actual sum is maxsum + secondMaxSum - root-val;
This can be done by iterative inorder
int maxLtoLSum(NODE *root,int *sum)
{
int ls=0,rs=0,lsum=0,rsum=0;
if(root==NULL)
{
*sum=0;
return 0;
}
lsum= maxLtoLSum (root-left,ls);
rsum= maxLtoLSum (root-right,rs);
*sum=max(ls,rs) + root-data;
return max(ls + rs + root-data, max(lsum,rsum));
}
On Tue, Aug 28,
@Kailash: I assumed a function random() which generates uniformly
distributed double precision random numbers between 0 and 1. Usually such a
generator would produce numbers with 53 bit precision, so I guess that is
between 15 and 16 digits after the decimal point. Just google random
number
@Pankaj: My only complaint is that rand()%n is not uniformly distributed
between 0 and n-1. To see this, suppose rand() produces positive 32-bit
integers, so they have the range 0 to 2^31 - 1 = 2,147,483,647. Now,
suppose there are 100,000 words in the file. When you are computing
@Rahul: Please tell us what you mean by a pivoted array.
Dave
On Tuesday, August 28, 2012 12:56:03 PM UTC-5, rahul sharma wrote:
plz provide me algo for this,thnx
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@Dave Sir-Got ur point..thnks!!
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For more
http://marathoncode.blogspot.com.br/2012/08/ano-turing.html
http://marathoncode.blogspot.com.br/2012/08/logaritmo-discreto.html
http://marathoncode.blogspot.com.br/2012/08/registrador-de-deslocamento.html
http://marathoncode.blogspot.com.br/2012/08/paradoxo-do-aniversario.html
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