consider a case where tree is right skewed or left skewed , in dat case max
distance b/w two node found are root and leftmost or rightmost node(left
or right skewed) . so its not alwayzz true
On Sun, Jun 24, 2012 at 5:08 PM, Navin Kumar algorithm.i...@gmail.comwrote:
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Few Months back I found the problem
on Code Sprint
1/x + 1/y = 1/N! (N factorial). For large value of N
we have to find the par of (X,Y) which satisfy the equation
my sol was slow ,
can any pleas help me .
Thanks
Kumar Vishal
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please provide some good data structure to solve this problem:
http://www.careercup.com/question?id=14062676
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2! - x=y=4
3! - x=y=12
4! - x=y=48
5! - x=y=240
6! - x=y=1440
I don't have proof to prove x = y always.
But if x=y, then the answer should be x=y=2*n!
On Mon, Jun 25, 2012 at 5:04 PM, Roshan kumar...@gmail.com wrote:
Few Months back I found the problem
on Code Sprint
1/x + 1/y = 1/N! (N
sourabh singh, i think problem in your code may arise due to term (m/a
-n/a) ,instead it should be m/a -(n-1)/a .
there may be some other problem also .but this can be one of them
On Mon, Jun 25, 2012 at 5:10 PM, algogeeks@googlegroups.com wrote:
Today's Topic Summary
Group:
This is from interviewstreet named with equations
On Mon, Jun 25, 2012 at 11:19 AM, prakash y yprakash@gmail.com wrote:
2! - x=y=4
3! - x=y=12
4! - x=y=48
5! - x=y=240
6! - x=y=1440
I don't have proof to prove x = y always.
But if x=y, then the answer should be x=y=2*n!
On Mon, Jun
We have to find number of Pair(x,y)
which will satisfy the eq:
1/x + 1/y = 1/(n factorial)
for large 0 N 10 ^ 4 N is integer
Its problem from Code Sprint
I tried to solve it by taking LOG but the sol
was very slow , can any one please help
Thanks
Kumar Vishal
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What would be the diameter in case of a left skewed or right skewed tree?
On Mon, Jun 25, 2012 at 12:48 PM, atul anand atul.87fri...@gmail.comwrote:
consider a case where tree is right skewed or left skewed , in dat case
max distance b/w two node found are root and leftmost or rightmost
@Prakash
The Pattern given by u is because factorial (n) is always *even *so u
can always divide them
in two equal part .
what about
1/6= 1/8 + 1/24( 6 = factorial (3))
On Mon, Jun 25, 2012 at 11:24 PM, Kishore kkishoreya...@gmail.com wrote:
This is from interviewstreet
Sorry My Mistake *Number of pairs should be OUTPUT...*
On Mon, Jun 25, 2012 at 8:49 PM, prakash y yprakash@gmail.com wrote:
2! - x=y=4
3! - x=y=12
4! - x=y=48
5! - x=y=240
6! - x=y=1440
I don't have proof to prove x = y always.
But if x=y, then the answer should be x=y=2*n!
On
thiss is iterative inorder code i am using for printing all the paths of
the Btree .
but i am not able to manipulate the lengths properly when its perform pop
up so that i
can print the paths properly ..
Can any one help by modifying the changes in this code .
void inorder(struct node *root)
http://www.leetcode.com/2010/11/finding-minimum-window-in-s-which.html
On 6/25/12, Navin Kumar algorithm.i...@gmail.com wrote:
please provide some good data structure to solve this problem:
http://www.careercup.com/question?id=14062676
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@Vishal,
If the output should be the total no.of pairs, then i think there are
infinite no.of such pairs. but not sure.
Can someone provide me the link to the actual problem and some analysis of
solution?
Thanks,
~Prakash.
On Mon, Jun 25, 2012 at 9:45 PM, Kumar Vishal kumar...@gmail.com wrote:
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