LRU implementation :
-Maintain a double linked list which stores LRU items(say 50).
-Keep one head and tail pointers of the linked list.
-Maintain a hash table which has value of the linked list as key and memory
location of that node as value.
- When ever hash table doesn't conta
@Ujjwal: Here's an algorithm for problem 2:
Given a number n, form the prime number factorization of n:
n = p1^k1 * p2^k2 * ... * pm^km, where the pi are distinct primes and ^
represents exponentiation.
If any ki = 1, terminate the factorization and report that n is not a
Trojan number.
T
plz mail it to me too... piyushdur...@gmail.com
On Tuesday, July 31, 2012 7:37:03 PM UTC+5:30, deepikaanand wrote:
>
> can anyone tell me the pattern (selection procedure )followed by directi
> this year
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"Algorithm G
sory fr prev wrng cmnt
On Mon, Jul 30, 2012 at 4:04 PM, SHOBHIT GUPTA
wrote:
> wat abt 32
>
>
> On Mon, Jul 30, 2012 at 4:02 PM, Lucifer wrote:
>
>> @ ruru
>> I think we can do it in log(n)..
>> I am not jolting down the code but giving out the idea that would lead
>> to log(n) time..
>>
>> If w
wat abt 32
On Mon, Jul 30, 2012 at 4:02 PM, Lucifer wrote:
> @ ruru
> I think we can do it in log(n)..
> I am not jolting down the code but giving out the idea that would lead
> to log(n) time..
>
> If we write down all the nos. in the binary format one below the
> other.. we will observe the fo
@vivek..
u can't sort.. its a stable merge...
The question is to stably merge the two arrays..the stable merge of
the 2 arrays can take place in (2n C n) ways.. 1 of these arrangements
will lead to the maximum sum..
We are required to find that sequence/maxsum..
A stable merge in this case is not
On Mon, Jul 30, 2012 at 3:56 PM, Lucifer wrote:
> @small correction:
> In the initialization condition the loop index i shall start from 2
> and not 0..
> // for(int i =2; i <=n; i+=2)
>
>
> On 30 July, 15:23, Lucifer wrote:
> > @Prakhar Jain
> >
> > I believe that the following recurrence shall
@ ruru
I think we can do it in log(n)..
I am not jolting down the code but giving out the idea that would lead
to log(n) time..
If we write down all the nos. in the binary format one below the
other.. we will observe the following pattern :-
at bit index '1' the bit value toggles in group of '01'
@small correction:
In the initialization condition the loop index i shall start from 2
and not 0..
// for(int i =2; i <=n; i+=2)
On 30 July, 15:23, Lucifer wrote:
> @Prakhar Jain
>
> I believe that the following recurrence shall solve it..
>
> Take an array C[n+1][n+1]...
>
> Now, we only need
int func(int start,int end)
{
int count=0;
for(int i=start;i<=end;i++)
{
int tmp=i;
while(tmp!=0)
{
tmp=tmp&(tmp-1);
count++;
}
}
return count;
}
Worst Case complexity : O((b-a)*32)
Please let me know if there is another gud way to d
@Prakhar Jain
I believe that the following recurrence shall solve it..
Take an array C[n+1][n+1]...
Now, we only need to calculate those C[i][j]'s where i+j is even..
// Assuming 1-based index...
Initialization condition...
C[0][0]=0;
for(int i =0; i <=n; i+=2)
{
C[0][i] = C[0][i-2] + B[i
Yes Manish, but how did you get to the answer?
On Jul 24, 10:00 pm, manish untwal wrote:
> I think the question is in the written round!!!
>
>
>
>
>
>
>
>
>
> On Tue, Jul 24, 2012 at 9:11 PM, algo bard wrote:
> > #include
> > #define RANGE_START 0
> > #define RANGE_END 100
>
> > int main()
> > {
@dave sir-your algorithm is having a complexity of n(2) but the solution
that i have given is of log(n) i guess.
On Tue, Jul 24, 2012 at 8:10 PM, Dave wrote:
> @Ruru:
>
> int i,j,sum=100*101/2;
> for( i = 1 ; i <= 100 ; ++i )
> {
> j = i;
> while( j )
> {
> j >> = 1;
>
@Ruru:
int i,j,sum=100*101/2;
for( i = 1 ; i <= 100 ; ++i )
{
j = i;
while( j )
{
j >> = 1;
sum -= j
}
}
printf("%i\n",sum);
Dave
On Tuesday, July 24, 2012 4:39:42 AM UTC-5, ruru wrote:
> find no. of 1's in binary format of numbers from 1 to 100. like for
> 1
//the following functions will count number of bits in the number
int countbits(int n)
{
int count=0;
while(n)
{
n/=2;
count++;
}
return count;
}
int countnumberof1(int number)
{
if(number==0)
return 0;
if(number==1)
return 1;
if(number==2)
return 2;
if(number==3)
return 4;
>
>
> http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_cs;action=display;num=1280183627
>
>
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@ anshu -
should it not be like this :
f( x(i, n), y(j, n) ,0) = max( { x[i] * x[i+1] + max ( f( x(i+2, n), y(j,
n), 0) , f( x(i+2, n), y(j, n), 1) ) }, { x[i] * y[j] + max ( f( x(i+1,
n), y(j+1, n), 1 ), f( x(i+1, n), y(j +1, n), 0 ) } );
On Mon, Jul 16, 2012 at 2:15 AM, Anshu Mishra wrot
Please see *stable merge *in question.
On Sun, Jul 15, 2012 at 2:04 AM, sengar.mahi wrote:
> @naveen : 3*7+2*9+1*3 =42 is not maximum..
> sum of the product would me maximum wen, i guess, most weighted elements
> are adjacent
> like in this case
> if c={1,2,3,3,7,9}
> 1*2 + 3*3 + 7*9=74 (maximum
Awesomely done :) +1
On Mon, Jul 16, 2012 at 2:15 AM, Anshu Mishra wrote:
> two arrays are suppose x[n], y[n];
>
> take a function
> f( x(i, n), y(j, n) , 0) --> taking x[i] as a first element of merged
> array then max sum;
> f( x(i, n), y(j, n), 1) --> taking y[j] as a first element of
> merge
two arrays are suppose x[n], y[n];
take a function
f( x(i, n), y(j, n) , 0) --> taking x[i] as a first element of merged array
then max sum;
f( x(i, n), y(j, n), 1) --> taking y[j] as a first element of merged array
then max sum;
f( x(i, n), y(j, n) ,0) = max( { x[i] * x[i+1] + f( x(i+1, n),
O(n2) Time and O(n2) space solution -
1. Total of (n2 + 2n - 2) products possible with given combinations - (n -
1) times (n + 1) and once n for the first array for the last element; total
of (n -1) products for 2nd array -therefore (n -1)(n +1) + n + (n-1) = n2
+ 2n - 2. Products to be classifie
@naveen : 3*7+2*9+1*3 =42 is not maximum..
sum of the product would me maximum wen, i guess, most weighted elements
are adjacent
like in this case
if c={1,2,3,3,7,9}
1*2 + 3*3 + 7*9=74 (maximum )
thus this question is just merging both strings such resultant (here C) is
in sorted order which can b
As the final array contains element in stable-order, at any time we have 3
options for choosing the elements from A & B.
1- A[i] & A[i+1]
2- A[i] & B[I]
3- B[i] & B[i+1]
we will choose that pair whose product is maximum.
For ex:-
A-2,1,3
B-3,7,9
C- 3,7,2,9,1,3
Its a linear time solution with cons
int no_of_steps[arr_length] = {MAX};
if ( (arr_length==0) || (arr[0] == 0) ) //if there are no elements or
the very first element is 0 -> we can't jump anywhere
return MAX;
no_of_steps[0] = 0; //no jumps required to jump from element 0 to itself.
for (int i=1; i= (i - j) )//if it is po
^ cout<< no_of_steps[arr_length -1];
On Mon, Jul 9, 2012 at 8:44 PM, algo bard wrote:
> int no_of_steps[arr_length] = {MAX};
>
> if ( (arr_length==0) || (arr[0] == 0) ) //if there are no elements
> or the very first element is 0 -> we can't jump anywhere
> return MAX;
>
> no_of_steps[0] =
There is a greedy solution discussion about this approach. I don't have a
formal proof for this.
Any counter example will be helpful.
at every place 'k' .. do the following.
--> find max ( a[k+i]+i ) where 1 <= i <= a[i]
for the given example:
A = {4 0 0 3 6 5 4 7 1 0 1 2}
initially a 4, the
@ashish it wont be...first we r finding one end from any node dat is "r" n
den frm dat end we r traversing to other deepest end...
it is possible dat r b d intermediate node...n distance from r to v2 is
smaller than from r to v1
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farthest from
2: Find a vertex v1 | the farthest form r.
3: Find a vertex v2 | the farthest form v1.
won't v2 be farthest from r? or we are talking about alternate pats also
Best Regards
Ashish Goel
"Think positive and find fuel in failure"
+919985813081
+919966006652
On Wed, Jun 20, 2012
As you traverse along and find the diameter of the tree, keep track of the
number of nodes thereby traversed. Let that be equal to n.
Now, centre is the node corresponding to floor((n+1)/2).
On Wed, Jun 20, 2012 at 5:19 PM, Nishant Pandey <
nishant.bits.me...@gmail.com> wrote:
> I am asking again
I am asking again .. can any one please suggest better method for getting
the median on the longest path of the tree ..
efficient method .
On Tue, Jun 19, 2012 at 5:08 PM, Nishant Pandey <
nishant.bits.me...@gmail.com> wrote:
>
> for getting diameter we can simply add the max height of left subtr
for getting diameter we can simply add the max height of left subtree and
max height of the right sub tree .
the main issue is how efficiently we find median on that longest path to
get the center of the tree .
can any bdy sugest optimized algo for that ?
On Mon, Jun 18, 2012 at 6:08 PM, rajesh p
I found it in some paper ;)
Diameter and center
De nition 4. The diameter of tree is the length of the longest path.
De nition 5. A center is a vertex v such that the longest path from v to a
leaf is minimal
over all vertices in the tree.Tree center(s) can be found using simple
algorithm.
Algor
I think this algorithm is used for calculating poset in graph.
On Sat, Jun 16, 2012 at 3:04 AM, Hemesh Singh wrote:
> + 1 for DK's solution. Is that a standard algorithm coz I feel like I have
> heard it somewhere ??
>
>
> On Mon, Aug 8, 2011 at 1:37 AM, DK wrote:
>
>> @KK: DFS and BFS are O(N)
@maddy: The students should be assigned consecutive books only. That is, u
CANNOT assign book 1, 2 and 5 to a single student. either assign book 1, 2
and 3 or 1 and 2 or any such combination of consecutive numbers.
On Thursday, June 14, 2012 12:26:09 PM UTC+5:30, algogeek wrote:
>
> In a li
On Jun 16, 2:1@shubham
couldnt understand following logic in you algo please explain
"when first k elemnts traversed then traverse again k elemnts of B
and
S[2*k-i]+=B[i])...again S[i-2*k]+=B[i][.so on till B is traversed
completely."
6 am, Shubham Sandeep wrote:
> wat constraints does dis bri
+ 1 for DK's solution. Is that a standard algorithm coz I feel like I have
heard it somewhere ??
On Mon, Aug 8, 2011 at 1:37 AM, DK wrote:
> @KK: DFS and BFS are O(N) and Floyd Warshall is O(N^3).
> Could you please state how you can use the traversals directly to get the
> center? (And prove yo
@Don : it seem that you are calculating diameter of a tree ??
On Mon, Mar 26, 2012 at 7:18 PM, Don wrote:
> If the longest path passes through the root of the tree, then the
> length of the path is the depth of the left subtree plus the depth of
> the right subtree. If the longest path does not
@arunachalam you have misunderstood the problem.
On Tue, Mar 27, 2012 at 8:10 AM, Arunachalam wrote:
> This algorithm is almost right, but not exactly correct.
>
> Say for example you have a binary tree like this
>
>1
> 2
>3 4
> 5 6
>
> The
This algorithm is almost right, but not exactly correct.
Say for example you have a binary tree like this
1
2
3 4
5 6
The length of the longest path is 4, and this algorithm would return 5. The
algorithm can be slightly modified to find the m
If the longest path passes through the root of the tree, then the
length of the path is the depth of the left subtree plus the depth of
the right subtree. If the longest path does not pass through the root,
then it is the max of the longest path in the left subtree or the
right subtree.
int longes
I think we can tweak the standard "find the height of the tree"
program to get the result..
As we know that the 2 extremes of the longest path are nothing but
leaves. Hence, all we need to do is figure out is for which set of
leaves would the path be maximum..
[ Special Case : it need to be always
@Mohit
Agreed. The answer is O(n).
On Fri, Oct 28, 2011 at 6:48 PM, mohit verma wrote:
> @ankur - Ans-9 how can it be log n. The heap given is Max heap. I think
> it should be O(n) using array or tree traversal (as heap is implemented)
> keeping current min at hand. Correct me if m wrong.
>
>
@ankur - Ans-9 how can it be log n. The heap given is Max heap. I think it
should be O(n) using array or tree traversal (as heap is implemented)
keeping current min at hand. Correct me if m wrong.
On Sat, Oct 15, 2011 at 12:14 PM, shady wrote:
> already been answered... :-/
> but have to say
already been answered... :-/
but have to say you are damn quick...
On Sat, Oct 15, 2011 at 12:03 PM, Bittu Sarkar wrote:
> Q7. Correct answer is 12km west and 12km south for sure!!
>
>
> On 21 September 2011 13:28, Nitin Garg wrote:
>
>> Ohh i totally missed that line.
>> Thanx a lot :)
>>
>>
>
Q7. Correct answer is 12km west and 12km south for sure!!
On 21 September 2011 13:28, Nitin Garg wrote:
> Ohh i totally missed that line.
> Thanx a lot :)
>
>
> On Wed, Sep 21, 2011 at 10:46 AM, pankaj agarwal <
> agarwal.pankaj.1...@gmail.com> wrote:
>
>> @Nitin Garg
>>
>> Question 6 -
>>
>> i
Ohh i totally missed that line.
Thanx a lot :)
On Wed, Sep 21, 2011 at 10:46 AM, pankaj agarwal <
agarwal.pankaj.1...@gmail.com> wrote:
> @Nitin Garg
>
> Question 6 -
>
> i agree that greater the sum is and greater the probability to getting it.
> but in given question if sum>100 then rolling is
@Nitin Garg
Question 6 -
i agree that greater the sum is and greater the probability to getting it.
but in given question if sum>100 then rolling is stopped
so for
P(106)=P(100)*1/6
P(105)=P(100)*1/6+P(99)*1/6
.
.
.
P(101)=P(100)*1/6+P(99)*(1/6)+P(98)*(1/6)+P(97)*(1/6)+..+P(95)*(1/6)
now P(101)
Ans 7:
Why am I getting such an answer!!
[co-ordinates] next_step_size
[-2, -2] 5
[-4, -4] 9
[-6, -6] 13
[-8, -8] 17
[-10, -10] 21
[-12, -12] 25
[-14, -14] 29
[-16, -16] 33
[-18, -18] 37
[-20, -20] 41
[-22, -22] 45
[-24, -24] 49
[-26, -26] 53
[-28, -28] 57
[-30, -30] 61
[-32, -32] 65
[-34, -34] 69
Q9 - > 1 *logn for getting the minimum element ..Normal heap Sort procedure
Q3 - > n+logn-2 comparisions so 51 -2 + log 51
Regards
Ankur
On Mon, Sep 19, 2011 at 7:59 PM, Ashima . wrote:
> m getting result in 95 matches
>
> Ashima
> M.Sc.(Tech)Information Systems
> 4th year
> BITS Pilani
> R
m getting result in 95 matches
Ashima
M.Sc.(Tech)Information Systems
4th year
BITS Pilani
Rajasthan
On Mon, Sep 19, 2011 at 7:07 PM, malay chakrabarti wrote:
> i have explained :)
>
> On Sun, Sep 18, 2011 at 11:53 PM, Ashima . wrote:
>
>> @malay: how cm n+logn-2?
>> cn u explain the logic ?
i have explained :)
On Sun, Sep 18, 2011 at 11:53 PM, Ashima . wrote:
> @malay: how cm n+logn-2?
> cn u explain the logic ?
>
> Ashima
> M.Sc.(Tech)Information Systems
> 4th year
> BITS Pilani
> Rajasthan
>
>
>
>
> On Sun, Sep 18, 2011 at 11:07 AM, Ashima . wrote:
>
>> rite! 62.5%
>>
>> Ashima
For question 5 reentrant
On 19-Sep-2011 1:43 PM, "Nitin Garg" wrote:
> Can someone tell answers to question 2 and 5 with explanation??
>
>
>
>
> On Mon, Sep 19, 2011 at 1:40 PM, Nitin Garg wrote:
>
>> In Question 4 i just kept counting new processes that are being added in
>> every iteration
Given in the question
.
On 19-Sep-2011 2:57 PM, "Bhanu Chowdary" wrote:
> @Nithin: Sorry I did not understand your logic!! If a person looses a
match
> he should be knocked out of the tournament. Could you please explain why 2
> matches to knock out a person??
>
> On Mon, Sep 19, 2011 at 2:47 PM,
@Nithin: Sorry I did not understand your logic!! If a person looses a match
he should be knocked out of the tournament. Could you please explain why 2
matches to knock out a person??
On Mon, Sep 19, 2011 at 2:47 PM, praveen raj wrote:
> For question 2 see ashima link.
>
> On 19-Sep-2011 1:43 PM,
For question 2 see ashima link.
On 19-Sep-2011 1:43 PM, "Nitin Garg" wrote:
>
> Can someone tell answers to question 2 and 5 with explanation??
>
>
>
>
> On Mon, Sep 19, 2011 at 1:40 PM, Nitin Garg
wrote:
>>
>> In Question 4 i just kept counting new processes that are being added in
every iterati
@VIHARRI
i think q.5 is *Which is not thread safe ??* :D :D
On Mon, Sep 19, 2011 at 1:43 PM, Nitin Garg wrote:
> Can someone tell answers to question 2 and 5 with explanation??
>
>
>
>
> On Mon, Sep 19, 2011 at 1:40 PM, Nitin Garg wrote:
>
>> In Question 4 i just kept counting new processes that
Can someone tell answers to question 2 and 5 with explanation??
On Mon, Sep 19, 2011 at 1:40 PM, Nitin Garg wrote:
> In Question 4 i just kept counting new processes that are being added in
> every iteration.
> No. of new processes being created is equal to the already running no. of
> even pi
In Question 4 i just kept counting new processes that are being added in
every iteration.
No. of new processes being created is equal to the already running no. of
even pid processes.
Time - PId
0 - 0 1
1 - 0,12
2, - 0,1,23
3, -
Question 6 -
Intuitively you can see that the greater the sum is, the greater the
favorable events in sample space.
e.g. - sum = 1 .. cases {(1)} Pr = 1/6
sum = 2 cases {(2),(1,1)} Pr = 1/6 + 1/36
sum = 3cases {(3),(2,1)(1,2)(1,1,1)} Pr = 1/6 + 1/36 +1/36 +
1/216
for
Question 3 -
To eliminate one player, you need to host atleast 2 matches and make him
loose in both 2. These 2 matches can not contribute to elimination of any
other player.
So, min 2 matches for every player who is to be eliminated, hence 100.
On Mon, Sep 19, 2011 at 11:54 AM, Bhanu Chowdary wro
@Nitin: Answer to question 3 is 50.
On Mon, Sep 19, 2011 at 11:44 AM, praveen raj wrote:
> @nitin Plz explain how u have reached answer of question no. 4 and 6
>
> On 19-Sep-2011 12:26 AM, "Nitin Garg" wrote:
> >
> > Answer 3 - 100
> > Answer 6 - 103
> > Answer 4 - 194 total processes includin
@nitin Plz explain how u have reached answer of question no. 4 and 6
On 19-Sep-2011 12:26 AM, "Nitin Garg" wrote:
>
> Answer 3 - 100
> Answer 6 - 103
> Answer 4 - 194 total processes including the parent
> Answer 7 - 12 km south, 12 km east
>
>
> On Sun, Sep 18, 2011 at 11:53 PM, Ashima . wrote:
@Nithin Garg how 194 for the 4 question?
On Mon, Sep 19, 2011 at 12:26 AM, Nitin Garg wrote:
> Answer 3 - 100
> Answer 6 - 103
> Answer 4 - 194 total processes including the parent
> Answer 7 - 12 km south, 12 km east
>
>
> On Sun, Sep 18, 2011 at 11:53 PM, Ashima . wrote:
>
>> @malay: how cm n
Answer 3 - 100
Answer 6 - 103
Answer 4 - 194 total processes including the parent
Answer 7 - 12 km south, 12 km east
On Sun, Sep 18, 2011 at 11:53 PM, Ashima . wrote:
> @malay: how cm n+logn-2?
> cn u explain the logic ?
>
> Ashima
> M.Sc.(Tech)Information Systems
> 4th year
> BITS Pilani
> Raj
@malay: how cm n+logn-2?
cn u explain the logic ?
Ashima
M.Sc.(Tech)Information Systems
4th year
BITS Pilani
Rajasthan
On Sun, Sep 18, 2011 at 11:07 AM, Ashima . wrote:
> rite! 62.5%
>
> Ashima
> M.Sc.(Tech)Information Systems
> 4th year
> BITS Pilani
> Rajasthan
>
>
>
>
> On Sat, Sep 17, 201
rite! 62.5%
Ashima
M.Sc.(Tech)Information Systems
4th year
BITS Pilani
Rajasthan
On Sat, Sep 17, 2011 at 9:04 PM, malay chakrabarti wrote:
> create a tournament tree.in each round one value is eliminated to obtain
> in the process the winner or the highest value in n-1 comparisons. Then
> chec
create a tournament tree.in each round one value is eliminated to obtain in
the process the winner or the highest value in n-1 comparisons. Then check
the queue of the winner which contains log(n) entries of the values beaten
by the winner which implicitly will contain the runners up.Then log(n)-1
Q3. 101 matches
..
On Sun, Sep 18, 2011 at 8:02 AM, VIHARRI wrote:
> hey i'm also thinking n + logn -2.. but couldnt able to figure out
> how??? can you please explain the logic
>
> --
> You received this message because you are subscribed to the Google Groups
> "Algorithm Geeks" group.
> To po
hey i'm also thinking n + logn -2.. but couldnt able to figure out
how??? can you please explain the logic
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some more questions apart from them
1. area of the largest square such that it lies only on the black square of
chess such that the side of the chess square is 2 cm.
2. there is a rectangle area such that in it m number of roads are moving
from east to west and n number of roads are moving from no
k thanxgot it
On Aug 19, 9:07 pm, sagar pareek wrote:
> search the archives ...
> direct i's questions already discussed many times
>
>
>
> On Fri, Aug 19, 2011 at 9:33 PM, Agyat wrote:
> > Hi, friends
> > tom i m having directi written test followed by coding(ol) and written
> > test is eli
In written test
all the basics
like apti(major), OS, networking(major), shell programming(major) were there
i m not sure about C programming in written
and in coding i too faced that same log ACCESS program...i was unable to
fetch system's current time stamp
but still my logic was correct so i w
@Kunal: in the written test wer the ques technical or aptitude??
On Tue, Aug 9, 2011 at 4:42 PM, NIKHIL JAIN
wrote:
> thnx
>
> can you please tell some of the written questions so that i will have an
> idea what kind of questions are being asked by them
>
> --
> You received this message because
thnx
can you please tell some of the written questions so that i will have an
idea what kind of questions are being asked by them
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The procedure was like this:-
First they had a written test of 45 questions of 90 mins. Depending on the
performance in the test they divided selected students into two buckets. one
for software dev and other for sys ops.
Then there was coding round.
For sys ops one has to extract some information
can any dce student give an idea whats the procedure and what are the
questions asked by directi recruitment group
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Yeah..right..! u have to select continuous 10 petrol pumps , so just
traverse through every set of 10 pumps with complexity of o(n). e.g. (1,10)
(2,11) (3,12)(91 ,100)
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I can think of this solution
Put first elements in a max heap ..O(10)
Now take 11 th element ..compare it with root of this max heap ..If less
than root ignore ..else remove root and call max-heapify and put this
element in right place
Similarly do this for rest
Complexity will be n+nlogk wher
@DK can you please explain your solution, i couldn't understand...
@all any alternate solution ?
On Sun, Aug 7, 2011 at 11:02 PM, DK wrote:
> This is a simple problem:
>
> 1. Understand that if you are selecting 2 petrol pumps, say P and Q with
> say another petrol pump R between them, then it
@KK: DFS and BFS are O(N) and Floyd Warshall is O(N^3).
Could you please state how you can use the traversals directly to get the
center? (And prove your correctness too?)
The solution given by Wladimir (& expanded upon by me) is O(N) and uses
(somewhat) the inverse of a BFS as a traversal.
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@kunal patil:
U were proceeding in correct way...in next series u cud hav seen a
formation of arithmatico geometric series...
It doesnt matter what the value of no of faces in a dice is..ans will
be always 2...:)
My simplified soln: o+e+1
o=probability of odd number coming in 1
throw of dice
I
Codechef Ques(Initiative of Directi)
use DFS, BFS or Floyd Warshall... :)
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algogeek
This is a simple problem:
1. Understand that if you are selecting 2 petrol pumps, say P and Q with say
another petrol pump R between them, then it is optimal to include R into the
set of selected petrol pumps as the max-distance for the chosen pumps
remains PQ but number of included petrol pump
starting from each petrol pump binary search for the value of distance
On Sun, Aug 7, 2011 at 9:15 PM, subramania jeeva
wrote:
> No.. They are not uniform.. They will give you distance between two
> adjacent stations..
>
>
>
>
>
>
>
> Cheers
> ~ Jeeva ~
>
>
>
> On Sun, Aug 7, 2011 at 9:
No.. They are not uniform.. They will give you distance between two adjacent
stations..
Cheers
~ Jeeva ~
On Sun, Aug 7, 2011 at 9:13 PM, iama wrote:
> I think you are missing some part of the question
> Are the petrol pumps placed uniformly?
> How many do we have to select exac
I think you are missing some part of the question
Are the petrol pumps placed uniformly?
How many do we have to select exactly? (you have added a 'say' in the
statement above. Is it upto the us to choose?)
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@Everyone: Wladimir has posted the correct solution to the problem and it is
an O(N) bottom up solution.
*The original solution:*
On Wednesday, 6 October 2010 21:10:40 UTC+5:30, wladimirufc wrote:
>
> Find the leaf of tree then put all leaf in a queue.
>
> While queue not empty:
> remove u fr
@kunal patil
expected number of rolls required to get even sum is inverse of that
i.e. 2 """
can we say like this all the time ?
i have understood the alternative method, thanks :D
On Sun, Aug 7, 2011 at 1:33 PM, Kunal Patil wrote:
> Probability of getting an even sum in one roll is 1/2..
>
@kunal patil:
how can u say "Probability of getting an even sum in one roll is 1/2."...
if tht is the case..can u find the ans in one go if the dice is having *5
faces*??
i mean the numbers on dice are 1 2 3 4 5 ...and it is unbiased...
what will be the *Probability of getting an even sum in one r
Probability of getting an even sum in one roll is 1/2..
Thus, expected number of rolls required to get even sum is inverse of that
i.e. 2.
Alternatively, Going by basics...
Let P(x) be probability of getting Even sum in x rolls.
P(1) = 1/2(Even)
P(2) = (1/2) * (1/2) (Odd + Odd)
P(3) = (1/2)
Traverse the tree inorder. Store the values in an array. Find the element in
the middle of the array.
On Sun, Aug 7, 2011 at 8:57 AM, subramania jeeva
wrote:
> 5
> /\
> 6 7
>/
> 8
>
> Here the centre is both 5 and 6 . we need to find both of them..:)
>
>
>
>
>
>
>
>
>
> Cheers
Dave shouldn't it be :: E(x) = Summation( xp(x) ) thus it should
be 1*1/2 + 2*(1/2 * 1/2) + 3*(1/2 * 1/2 * 1/2) .
On Sun, Aug 7, 2011 at 11:55 AM, Nitish Garg wrote:
> @Dave - Can you please explain how did you generate the series? Shouldn't
> it be : 1/2 + 2(1/4) + 3(1/8) and so
@Dave - Can you please explain how did you generate the series? Shouldn't it
be : 1/2 + 2(1/4) + 3(1/8) and so on?
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5
/\
6 7
/
8
Here the centre is both 5 and 6 . we need to find both of them..:)
Cheers
~ Jeeva ~
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@Shady: There is a 1/2 probability of getting an even number on the
first roll. In the 1/2 of the cases where the first roll is odd, roll
again and note that there is a 1/2 chance of getting an odd number.
Etc. The probability of needing n rolls is 1/2 to the nth power.
The series is 1 + 2 * (1/2)
>From any node, find the farthest node by DFS, save its location(say
A). Now start DFS from A, and reach the farthest node, say B. AB will
be diameter of tree and longest path. The central node will be the
centre of tree. O(n) solution.
On Jul 27, 1:53 am, sivaviknesh s wrote:
> we can do like cr
13.75 lacs p.a
On Sat, Aug 6, 2011 at 6:41 PM, Anuj Kumar wrote:
> how much does de shaw offer?
>
>
> On Sat, Aug 6, 2011 at 6:40 PM, SANDEEP CHUGH wrote:
>
>>
>> can any one tell me the same about D.e Shaw ??
>>
>> On Sat, Aug 6, 2011 at 6:27 PM, SANDEEP CHUGH wrote:
>>
>>> can any one tell me
how much does de shaw offer?
On Sat, Aug 6, 2011 at 6:40 PM, SANDEEP CHUGH wrote:
>
> can any one tell me the same about D.e Shaw ??
>
> On Sat, Aug 6, 2011 at 6:27 PM, SANDEEP CHUGH wrote:
>
>> can any one tell me the same about D.e Shaw ??
>>
>>
>> On Sat, Aug 6, 2011 at 5:35 PM, raj kumar wrot
can any one tell me the same about D.e Shaw ??
On Sat, Aug 6, 2011 at 6:27 PM, SANDEEP CHUGH wrote:
> can any one tell me the same about D.e Shaw ??
>
>
> On Sat, Aug 6, 2011 at 5:35 PM, raj kumar wrote:
>
>> Really helpful info thanks guys..
>>
>> --
>> You received this message because you are
can any one tell me the same about D.e Shaw ??
On Sat, Aug 6, 2011 at 5:35 PM, raj kumar wrote:
> Really helpful info thanks guys..
>
> --
> You received this message because you are subscribed to the Google Groups
> "Algorithm Geeks" group.
> To post to this group, send email to algogeeks@googl
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