Can anybody mail me this.
Thanks,
Sourav
On Thu, Jun 23, 2011 at 9:56 PM, Swathi chukka.swa...@gmail.com wrote:
Can someone send it to me please...
Thanks,
Swathi
On Thu, Jun 23, 2011 at 12:13 PM, pullasunil pullasu...@gmail.com wrote:
Can you mail me the book for Algorithms
My implementation tries to create a doubly linked list, each node of
which will hold the value of sum of all nodes in a particular
vertical.If there is no requirement for the final output to state
vertical number against the sum (and indeed there is no such
requirement in the question ), then my
:07 am, yq Zhang zhangyunq...@gmail.com wrote:
@sourav,
As Juvier++ pointed out, it's an **amortized** O(n) algorithm. That's
because each element can be at most popped twice.
Thanks
Yq
On Mon, Jan 3, 2011 at 11:20 AM, sourav souravs...@gmail.com wrote:
@yq Zhang,
To pop if you
are constant time.
Thanks,
Sourav
On Jan 3, 9:44 pm, yq Zhang zhangyunq...@gmail.com wrote:
Push into one stack. When pop first pop all from the first stack and push
into the second stack. Then pop from the second stack
On Jan 3, 2011 7:42 AM, MOHIT mohit...@gmail.com wrote:
if only two
Wish you and your family a Bright and Prosperous Diwali.
Regards,
Sourav
[image:
diwali-diyas.jpeg]?ui=2ik=4f1ebe5da5view=attth=12c1a980b57e423fattid=0.1disp=inlinerealattid=f_gg4neyjf0zw
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complexity, where you use a matrix of side (length of array
1)X(length of array 2) is known to me.
I am looking for a better order solution.
Thanks for your time.
Sourav
On Oct 23, 10:53 pm, nishaanth nishaant...@gmail.com wrote:
It is nothing but a common subsequence problem...isnt it ?
On Wed, Oct
you are given 2 arrays sorted in decreasing order of size m and n
respectively.
Input: a number k = m*n and = 1
Output: the kth largest sum(a+b) possible. where
a (any element from array 1)
b (any element from array 2)
The Brute force approach will take O(n*n). can anyone find a better
logic.
You are given an array of positive numbers in which one number is
repeated. Rest all are present only once. Find the duplicate number in
linear or sub linear time.
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@Anuj
You will find the implementation in the book Algorithms on Strings,
Trees and Sequences: Computer Science and Computational Biology by
Dan Gusfield
Chapter 3 Exact Matching: A Deeper Look at Classical Methods Section
3.5.3
Sourav
On Sep 18, 11:51 am, ANUJ KUMAR kumar.anuj...@gmail.com
based on the fact that min heap can be constructed
from an array in linear time.
Thanks,
Sourav
On Sep 27, 1:32 pm, rahul rahulr...@gmail.com wrote:
If their is no constrain on assumptions.
1.Sort the array.
2. check the dif between 2 elements.
{ 99,35,45,33,88,9098,112,33455,678,3}
sorted
a for loop into code above (given
by @Dave) to find if majority element has 2n/3 occurance.
Thanks,
Sourav
On Sep 22, 9:02 am, Navin Naidu navinmna...@gmail.com wrote:
Use majority vote algorithm:
http://userweb.cs.utexas.edu/~moore/best-ideas/mjrty/index.html
On Wed, Sep 22, 2010 at 9:12 AM, pre
O(n) solution.
Please provide your comments in any
Thanks
Sourav
On Sep 27, 11:09 am, prodigy 1abhishekshu...@gmail.com wrote:
Actual complexity of above algorithm = O(n^1.6)
On Sep 27, 8:19 am, Gene gene.ress...@gmail.com wrote:
If the array is m by n, pick the element a[m/2][n/2], i.e
@Dave
I cannot see any code at the links you have provided. I only see the
prime numbers. am I missing something here..?
Thanks,
Sourav
On Sep 23, 10:59 pm, Rishi Agrawal rishi.b.agra...@gmail.com wrote:
Apart from 1, 2 and 3, all the prime numbers are either of the form (6*n -
1) or (6*n
Very Nice Simple approach @Dave
On Sep 24, 12:56 am, Dave dave_and_da...@juno.com wrote:
Do a single-elimination tournament of the numbers, where the larger
wins each game. This will take n/2 + n/4 + ... + 1 = n-1
comparisons. The second largest will be among the numbers that lost to
the
most child)
InsertNode(tree-left,temp);
Do share your views.
Thanks,
Sourav
On Sep 27, 7:58 am, prodigy 1abhishekshu...@gmail.com wrote:
15
/ \
8 25
/ \
20 22
On Sep 26, 10:45 am
All of them are correct.
You need to verify that the probability of getting each of the numbers
from 0 to 7 are same.
In the example given by Gene, here is the explanation below:
rand04()rand04()5*rand04() Sum Sum/3
0 0 0 0 0
1 0 0
You are given a random no. generator function rand04() that generates
random numbers between 0 and 4 (i.e., 0,1,2,3,4) with equal
probability. You have to design a random no. generator function
rand07() that generates numbers between 0 to 7 (0,1,2,3,4,5,6,7) using
rand04() such that rand07()
this is O(n) will be of great help.
Thanks in Advance.
Sourav
On Jul 3, 1:35 am, Abhishek Sharma jkabhishe...@gmail.com wrote:
@Anand: good one
On Sat, Jul 3, 2010 at 2:02 AM, Anand anandut2...@gmail.com wrote:
This is an example of bitonic sequence if we reverse the bottom half of the
array
Let A[1..n] be an array such that the first (n − √n) elements are
already sorted
(though we know nothing about the remaining elements). Give an
algorithm that
sorts A in substantially better than (n log n) steps.
This question is from chapter 4 : Algorithm Design Manual by S Skiena
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You
@Rahul
Agree with your approach. When you say merge the last root(n)
elements with the starting elements, it means you are doing something
like merge sort using an additional O(n) space. Correct me if I am
wrong. This should give O(n) overall time complexity.
How about an in - place approach?
The median of a set of n values is the \lceil n/2 \rceilth smallest
value.
1. Suppose quicksort always pivoted on the median of the current
sub-array. How many comparisons would Quicksort make then in the worst
case?
2. Suppose quicksort were always to pivot on the n/3 th smallest
value of
consider a link list l which can contain nodes representing the bits
GenerateBitStrings(int length)
{
if (length == 0)
{
//print all values in link list l
}
Add 1 to l
call GenerateBitString(length-1)
Remove the 1 that was added.
Add 0 to l
call GenerateBitString(length-1)
f(n) = sqrt(n) [squate root of n]
g(n) = log(^2) [log of (n square)]
For the above pair of functions is f(n) = Ω(g(n))? i.e., is there some
c 0, such that f(n) = g(n) for all n? Give proof in case answer is
yes or no.
A small correction.
You need to prove if f(n) = O(g(n)).
My Proff (under Note) is for f(n) = Ω(g(n))
On Sat, Jul 31, 2010 at 12:08 AM, sourav souravs...@gmail.com wrote:
f(n) = sqrt(n) [squate root of n]
g(n) = log(^2) [log of (n square)]
For the above pair of functions is f(n) = Ω(g(n
Suppose we start with n companies that eventually merge into one big
company. How many different ways are there for them to merge?
With three companies {a,b,c}, we need to find the number of ways that
the three companies can become two companies, and for every one of
those possibilities, the two
@Shiv
Collision is itself a wel known issue in hashing and much need to be
done to avoid collision. When you say your appraoch is hash the
numbers, how do u make sure without knowing the nature of the numbers
that you can hash them without bringing ing collision of inequal items
of the array?
A sorting algorithm takes 1 second to sort 1,000 items on your local
machine. How long will it take to sort 10,000 items ...
if you believe that the algorithm takes time roughly proportional to
nlogn?
[Show your calculations / logic to arive at an answer]
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@akshay
Does the array contain numbers in the range 1 to n?
On Jul 28, 8:16 pm, akshay akshayrastogi2...@gmail.com wrote:
An array of unsorted numbers n is given with one no.repeated once ie
n-1 distinct nos to find duplicate no. in o(n) complexity
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