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. the one in
the middle. There are now 3 possibilities:
1) The middle element is the one you're looking for, so you're done.
2)
how about this
n+(8-(n7))
On Sun, Sep 26, 2010 at 4:48 AM, Dave dave_and_da...@juno.com wrote:
@Ashwath: Thanks for the correction.
Dave
On Sep 26, 1:20 am, aswath G B aswat...@gmail.com wrote:
@Dave
Slight change u have to do
#includestdio.h
main()
{
int a = 24;
step1construct heap using siftdown. // time complexity wil be O(n) and
space complexity O(1).
step2take mindifference=a[1].
step3for i=1 ,i=n/2 do
{ find the difference of (root,root-left),(root,root-right)and
(root-left,root-right).and maintain mindifference is the smallest
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 arrary : { } would be something.
now update the min_diff.
another example : { 7,8,1,3,5,4}
sorted arrary : { 1,3,4,5,7,8}
min diff is 1.
Please
try x+8-(x7 )
On Sep 26, 4:47 am, Dave dave_and_da...@juno.com wrote:
@Shrevan: I mistyped what I intended. Try answer = (x | 7) + 1;
Dave
On Sep 26, 5:51 am, Shravan shravann1...@gmail.com wrote:
@Dave
Your answer will be always 2 irrespective of the value of 'x'.
BTW I too didn't
#includeiostream
using namespace std;
int main()
{
int n,temp,ans=1;
cin n;
temp=n/8;
temp++;
cout hi temp\n;
while(temp--)
{
temp=temp3;
}
cout temp;
return 0;
}
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Please correct.
ans = (x|7)+1.
Thanks Dave.
On Mon, Sep 27, 2010 at 5:42 PM, saurabh agrawal saurabh...@gmail.comwrote:
This problem is already solved.
ans=(x|1)+1
On Mon, Sep 27, 2010 at 5:15 PM, nishaanth nishaant...@gmail.com wrote:
try x+8-(x7 )
On Sep 26, 4:47 am, Dave
I feel the binary search approach as given by Saurabh Singh or like
moving from top right row, doing binary search in the row 0, find the
element being searched (say x) or one less than that (say y), followed
by binary search in the column below 'y' and proceeding in this way
can give a less than
@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
For solution to this problem see
http://groups.google.co.in/group/algogeeks/browse_thread/thread/adc95dcf96020a7/2943fd3c5a116fee?hl=enlnk=gstq=binary+tree+to+binary+serach+tree#
In that thread, I have a doubt regarding solution posted by @Algoose
Chase. The code posted is good as I feel except
@Prodigy
As per your example, 8 15 20 25 which is the is indeed the maximum binary
search tree in this binary tree is only a solution to smaller problem used
to solve a bigger problem.
The solution to smaller problem can be translated directly to the solution
of the bigger problem.
On Mon, Sep
@Sourav: What you are missing is that the algorithm is a manual one
that will answer the question asked in a few moments without writing
any computer code. Looking in the first file, you find that there are
1229 prime numbers less than 10,000. Looking in the second file, you
find that the sum of
These are called Twin Primes ...:)))
On Thu, Sep 23, 2010 at 11:29 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 + 1).
Can we use this property to generate the primes till we get 1 primes.
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Hi can any one suggest an efficient algorithm for iterative quick
sort
Thanks in advance.
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any type of replace would need at least one extra memory space.
recursion is the worst, depends how you implement recursion. one
iteration might depends on another, which depends one other, and so
on.. each iteration hold its own stack
On Sep 23, 1:59 pm, Albert alberttheb...@gmail.com
it's a compression problem. using hex instead of oct saves more space
00aaa0ss yyy = 50 2a 0 1s 3f 1\s ay
On Sep 15, 8:21 am, bittu shashank7andr...@gmail.com wrote:
A file is given with many 0s stored in continuous way , store it in
another file such that when you store try
Move-To-The-Front.
On Sep 27, 11:58 pm, Anand anandut2...@gmail.com wrote:
Given an array of integers, for each index i, you have to swap the value at
i with the first value smaller than A[ i ] that comes after index i
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Suppose you have N strings that can be generated on-the-fly, and you
wanted to discover if a hash function generates any collisions. Is
there a way to do this without O(N) storage?
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u can use log(n)+1 space to do that by using bit string
On Mon, Sep 27, 2010 at 10:37 PM, AdrianW atw...@gmail.com wrote:
Suppose you have N strings that can be generated on-the-fly, and you
wanted to discover if a hash function generates any collisions. Is
there a way to do this without
hey isn't it suppposed to be tournament problem..
On Fri, Sep 24, 2010 at 12:06 AM, Divesh Dixit
dixit.coolfrog.div...@gmail.com wrote:
finding largest and second largest elements from a set of n elements
by means of
Minimum comparison of n+celling(log n) +2
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Well, friend, we're both wrong.
The algorithm will find 6 just fine. It will choose 3 as the middle
element. Since 6 is bigger, it will throw away the subarray
1 2
2 3
and check the other 3 subarrays. When it checks
6 7
7 8
It will find the 6 on the first try. I just verified this by
Twin primes are a pair of prime numbers that differ by 2.
Dave
On Sep 27, 11:26 am, radha krishnan radhakrishnance...@gmail.com
wrote:
These are called Twin Primes ...:)))
On Thu, Sep 23, 2010 at 11:29 PM, Rishi Agrawal
rishi.b.agra...@gmail.com wrote:
Apart from 1, 2 and 3, all the
Thanks for pointing out. I was wrong because I assumed that numbers would be
in continuous increasing order when numbers in each row are written in a
line taking each row at a time.
On Mon, Sep 27, 2010 at 5:49 PM, Nikhil Jindal fundoon...@yahoo.co.inwrote:
@saurabh.nsit:
Consider the
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