You are given many slabs each with a length and a breadth. A slab i can be
put on slab j if both dimensions of i are less than that of j. In this
similar manner, you can keep on putting slabs on each other. Find the
maximum stack possible which you can create out of the given slabs
and for
Q1 Solution: . we can use doubly linked list with hash to implement all the
operation in O(1).
By keeping track of head and tail pointer we can do enqueue and dequeue in
O(1) time.
In hash we will keep track of each element present in linked list(queue).
With node value as a hash key and address
its a LIS problem.
need to think for n-dimension...
On 8/26/12, Ravi Ranjan ravi.cool2...@gmail.com wrote:
You are given many slabs each with a length and a breadth. A slab i can be
put on slab j if both dimensions of i are less than that of j. In this
similar manner, you can keep on putting
Hi all,
Whether anyone appeared for CISCO this year? If so pls provide me the *written
test pattern* as well as the interview process.
Thanks in advance!!
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Swaminathan M
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see A(n)ie the average case will always be smaller or equal to the
worst case...
ie something like... A(n)= c. w(n) for some c as constt ... which the
definition of big O...
correct me if i'm wrong..
On Sat, Aug 25, 2012 at 7:11 PM, rahul sharma rahul23111...@gmail.comwrote:
*Let
Atrenta is visiting our campus on tuesday.Can anyone please mail me the
placement papers of atrenta..
Thanks in acknowledgement...
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Yeah Atul is right.
Here is my solution:--
1) first rearrange the dimensions of slabs i.e. put bigger dimension in y
and smaller dimenson in x (rotate the slab)
2) then arrange all slabs in increasing order of x dimension
3) and then find the longest increasing sub-sequence based on y
Hi,
Can anyone suggest the best approach for finding max sum b/w 2 leaf nodes
in a binary tree ( not BST ) ?
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its the diameter of tree.
you can find implementation on geeksforgeeks
On 8/25/12, kunal rustgi rustogi.ku...@gmail.com wrote:
Hi,
Can anyone suggest the best approach for finding max sum b/w 2 leaf nodes
in a binary tree ( not BST ) ?
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@kailash : you can simply find area of each slab area=x*y ,,,store it;
then just run LIS on this area.
On 8/26/12, Kailash Bagaria kbkailashbaga...@gmail.com wrote:
Yeah Atul is right.
Here is my solution:--
1) first rearrange the dimensions of slabs i.e. put bigger dimension in y
and
@Atul007: No, because a (1,4) tile will not fit on a (2,3) tile.
Dave
On Sunday, August 26, 2012 7:45:27 AM UTC-5, atul007 wrote:
@kailash : you can simply find area of each slab area=x*y ,,,store it;
then just run LIS on this area.
On 8/26/12, Kailash Bagaria kbkailas...@gmail.com
@dave : correct..
i guess this will work :-
sort in decreasing area.
then run LIS such that for i j , length( i ) length( j ) width(
i ) width( j )
On 8/26/12, Dave dave_and_da...@juno.com wrote:
@Atul007: No, because a (1,4) tile will not fit on a (2,3) tile.
Dave
On Sunday, August
Q2) get to n/2 nodes
Reverse link list after n/2 nodes
Now check from 1st node and n/2 node for equality
Can make the checking for equality function recursive
Sent from my iPhone 4
On 26-Aug-2012, at 12:41 AM, Ashish Goel ashg...@gmail.com wrote:
Q1. Design a data structure for the following
@ll...thnx a lot
On Sat, Aug 25, 2012 at 8:22 PM, GAURAV CHAWLA togauravcha...@gmail.comwrote:
see A(n)ie the average case will always be smaller or equal to the
worst case...
ie something like... A(n)= c. w(n) for some c as constt ... which the
definition of big O...
correct
written test will be (apti + tech) no negative marking
apti from time , speed and distance, probablity , mixtures , profit
and loss (easy)
2 qs to infer from the passage given(critical reasoning)
1 DI set (too simple)
and technical qs 2 simple C o/p qs
A large %age of qs was from digital logic
Okay..missed it..thnx fr info..
Your approach is nice..it took me some time to understand your code's
though.. Great answer!!
On Aug 26, 2012 3:55 AM, Dave dave_and_da...@juno.com wrote:
@Kunal: Yes. You are missing that you don't know the number of words in
the file in advance. So, to use your
In mathematics, *binomial coefficients* are a family of positive integers
that occur as coefficients in the binomial theorem. [image: \tbinom nk]denotes
the number of ways of choosing k objects from n different objects.
However when n and k are too large, we often save them after modulo
@Dave: Can you throw some light on random() function?? How it is generating
numbers between 0.0 and 1.0, and how many digits are there after the
point...because if there is only one digit then we will not be able to
print words after 10th place because 10*0.1(lowest number generated by
@Ankit: Apply Lucas' Theorem, which you can find written up in Wikipedia.
Dave
On Sunday, August 26, 2012 3:57:18 PM UTC-5, Ankit Singh wrote:
In mathematics, *binomial coefficients* are a family of positive integers
that occur as coefficients in the binomial theorem. [image: \tbinom
Der were many questions from electronics part also..20 apti 20 electronic
ques 10 c/netwrking ques..So computer science people will have a tough
luck.So revise basics of electronics. ques like half wave rectifier
efficiency were asked
On Sun, Aug 26, 2012 at 10:24 PM, deepikaanand
but how to solve this problem for with n-dimension ??
On Sun, Aug 26, 2012 at 6:47 PM, atul anand atul.87fri...@gmail.com wrote:
@dave : correct..
i guess this will work :-
sort in decreasing area.
then run LIS such that for i j , length( i ) length( j ) width(
i ) width( j )
On
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