values of n. Consider n = 3. Then
since rand() takes on 2^31 values, rand()%3 cannot take on the values 0, 1,
and 2 with equal probability since 2^31 is not divisible by 3.
Dave
On Saturday, August 25, 2012 1:44:03 PM UTC-5, Kunal Patil wrote:
How about using rand()%n ??
Like, calculate
How about using rand()%n ??
Like, calculate lucky_pos = rand()%n
Then print word at lucky_pos th position...
Am I missing anything? All words are still equiprobable to get printed
right?
On Aug 20, 2012 11:45 AM, Dave dave_and_da...@juno.com wrote:
@Navin: Okay. Here is a paraphrase. Assume
I am solving spoj problem Tiling a Grid With
Dominoes.(http://www.spoj.pl/problems/GNY07H/)..
I am not able to come up with a recurrence relation..
One of my friend said it has the recurrence relation as f(n) = f(n-1)
+ 5*f(n-2) + f(n-3) - f(n-4).
I am not convinced and have trouble deriving this
@shady: There were no specific constraints. Actually, they didn't expect
any best solution. People who wrote brute force also got shortlisted. Brute
force would be Just picking a number one by one from first row and then
checking other rows for existence of this number. I think it is a O(n^3)
Can you tell why (x * z * z) % MOD is different from (x * ( (z*z)%MOD) ) %
MOD
Again why ((x%MOD)*(z%MOD)*(z%MOD))%MOD is giving WA
I thought of a simple examples like
(100* 53*72) % 90
-- ((90+10)*53*72) % 90
-- (10*53*72)% 90 which is same as (100%90 * 53%90 * 72%90) % 90
Another
@Sunny: Thanks for the info !! That's gr8.. your logic also seems to be
working perfectly fine..
On Sat, Oct 15, 2011 at 12:16 PM, shady sinv...@gmail.com wrote:
u can always post the code.,... but before posting code, you must state
your algorithm
else code becomes useless for other users
Great work shady...+1
I was so bored of the company interview queries on this group..
Hope to see Real algorithmic discussion on the new group.
On Wed, Oct 5, 2011 at 11:27 PM, KARTHIKEYAN V.B. algo...@gmail.com wrote:
+1
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Why don't you post a code that compiles ?
On Fri, Sep 30, 2011 at 12:41 AM, Brijesh brijeshupadhyay...@gmail.comwrote:
void main()
{
void *ptr;
char *a='A';
char *b=TAN;
int i=50;
ptr=a;
ptr=(*char)malloc(sizeof(a));
printf(%c,*ptr);
ptr=i;
ptr=(*int)malloc(sizeof(i));
Nice question nice answer... :)
On Tue, Sep 27, 2011 at 6:25 AM, UTKARSH SRIVASTAV
usrivastav...@gmail.comwrote:
what's the algo of this question
On Mon, Sep 26, 2011 at 2:38 PM, vikas vikas.rastogi2...@gmail.comwrote:
simple graph question,
graph is given as list , just check the
@Kumar: +1
@Kumar Rajeshwar: ExitFailure is outputted because main is expected to
return something which is not done in your case. Just add return 0; at the
end of main to get expected output.
On Tue, Sep 13, 2011 at 11:33 PM, Ishan Aggarwal
ishan.aggarwal.1...@gmail.com wrote:
int main()
Allocate memory for pointer variables.
On Mon, Sep 12, 2011 at 11:03 AM, sukran dhawan sukrandha...@gmail.comwrote:
run time error
first int * a is not assigned some address... so it will be pointing to
some garbage value
second two pointers of different types without a typecast will result
@Brijesh: +1
On Sun, Sep 11, 2011 at 2:14 PM, Brijesh brijeshupadhyay...@gmail.comwrote:
This is the fastest way I can think of to do this, and it is linear to the
number of intervals there are.
Let L be your original list of numbers and A be a hash of empty arrays
where initially A[0] =
, 2011 at 4:20 PM, Ankur Garg ankurga...@gmail.com wrote:
This solution is not in O(n) time :(
Unfortunately interviewer wants O(n) .
On Sun, Sep 11, 2011 at 4:01 PM, Kunal Patil kp101...@gmail.com wrote:
@Brijesh: +1
On Sun, Sep 11, 2011 at 2:14 PM, Brijesh brijeshupadhyay
@Piyush: +1
On Sat, Sep 10, 2011 at 1:07 AM, Piyush Grover piyush4u.iit...@gmail.comwrote:
pseudo algo:
=array idx[0...k-1] indicates the current pointer position in the ith
stream(initialized to 0).
=heap tree of size k where each node stores value of the data and value of
stream
@Dave: TC in your first case will be O(klogn + n).
Transforming array into heap would be O(n).
Correct me If i am wrong.
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To
. :)
:)
On Sat, Sep 10, 2011 at 5:10 PM, Kunal Patil kp101...@gmail.com wrote:
@Piyush: In 2 ways I will prove you wrong.
1)
Lets take 2 innermost loops.
for(j=0;ji*i;j++)
{
for(k=0;kj;k++)
}
Let i*i be m. Thus, It becomes.
for(j=0;jm;j++)
{
for(k=0;kj
Cool.. If you can share/ask algorithm related programming questions.. :)
In fact it is the main cause of this group. :) :) :)
There are sufficient SPOJ problem discussions over here..
You can continue asking questions if you think, that particular problem
needs special data handling techniques to
sizeof never executes whatever expression is given as parameter.
It needs only size of parameter.
In this case, parameter main() returns an int type.
So sizeof returns sizeof int. It doesnt need to call main() to get its task,
of finding size, done.
On Tue, Sep 6, 2011 at 11:41 PM, siddharam
If I understand question correctly, then just read as many characters as you
can using standard string functions. (like fgets n all related)
Output these all characters in a file.
Iterate until you have read all the characters and go on appending what you
have read to file.
You intended to ask
@siddharam suresh: sizeof(void) is an prohibited operation. So your code
would result in compile time error.
On Tue, Sep 6, 2011 at 11:47 PM, siddharam suresh
siddharam@gmail.comwrote:
my guess,
sizeof() takes only the declaration properties not the definition
properties(that means it
@kartik:
typedef struct
{
int bit1:29;
int bit2:4;
}bit;
This makes total number of bits 33, which is not on int boundry. (multiple
of 32 bits)
So to make it aligned on int boundry, 31 extra bits are padded at the end
of bit2.
This makes size of struct 29 + 4 + 31 = 64 bits -- 8 bytes.
When bit1
@Dave: +1 for nice solution... :)
On Sat, Aug 27, 2011 at 10:12 PM, Dave dave_and_da...@juno.com wrote:
@Avinash: What do you mean by exceeding limit(already overflowed)?
The number maxint is the limit. Every positive number is = maxint.
Dave
On Aug 27, 10:31 am, Avinash LetsUncomplicate..
@Dave: Still your approach to solve the problem remains correct.
(subtracting a number from max possible value then comparing this
difference with another number). So, no need to think that you were brain
dead (If you were, you would have posted a movie story here)..[?]
Mathematically it is
@Abhishek:
Its not always that you reach a leaf through the node.
But still your logic seems correct.
There would be 3 candidates for minimum:
--predecessor
--current node
--successor.
On Sat, Aug 20, 2011 at 1:13 PM, Abhishek Yadav
algowithabhis...@gmail.comwrote:
No your solution is correct
+1 Yasir.
On Wed, Aug 17, 2011 at 11:39 PM, Yasir yasir@gmail.com wrote:
For questions specifically asking about test cases, I would suggest
following 3 step approach:
First think of a* basic flow that MUST work for the application* (what is
expected with the application. Firstly make
Yes..i agree with Dave..Here is what i think.
As you have integers upto n^3 in your input, it would need [3*lg(n) + 1]
bits to represent each integer.
So take each group of r = ceil(lg(n)) bits at a time.
So this becomes number of bits needed to represent single digit.
Each digit thus can take 2^r
:
isn't it a simple question of applying radix sort from most significant
to
least signigicant digit and concatenating all the sorted numbers to get
the
largest number..
On Sat, Aug 13, 2011 at 11:13 PM, Kunal Patil kp101...@gmail.com
wrote:
Let me clarify.
Lets take example
@Yasir: Yups...I also have the same algo...
Just to improvise your solution, you need not do binary search on both sides
of the pivot.
Just check End points (min-max) of the both sub-array to decide which side
to do a binary search..This works in the case of duplicate elements too.
for e.g.
2 5
Following approach should work:
1) Count max number of digit in any integer of input. Let it be m. (Thanks
to dave..)
2) For each int having less than m digits:
Convert it to string of length m where you append circularly.
For e.g. if m=5
53 -- 53535
100 -- 10010
@rohit: Cast pointer to an integer into an int to get what you are
expecting.
for e.g.
printf(%d,(int)a[4]-(int)a[0]);
This will give 16.
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PM, aditi garg aditi.garg.6...@gmail.comwrote:
@kunal: what is the best way to implement step 2?
On Sat, Aug 13, 2011 at 7:33 PM, Ashish Sachdeva
ashish.asachd...@gmail.com wrote:
@kunal: seems fine.. tried it on some cases...
On Sat, Aug 13, 2011 at 5:17 PM, Kunal Patil kp101
equal to
the max length of any number...
whr r v doing dat chking in dis algo?
On Sat, Aug 13, 2011 at 10:25 PM, Kunal Patil kp101...@gmail.com wrote:
I dont know whether this is best approach to do step 2 or not. But it's
certainly good.
//I will show for two strings s1 and s2
len1 = s1
@Ankit Sambyal: Agree with ankuj...TC of your solution is O(nlogn) and not
O(n^2)...
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@Ankuj: +1 for different approach. (Though selection algo is more efficient
than this.)
On Wed, Aug 10, 2011 at 1:44 PM, nick tarunguptaa...@gmail.com wrote:
nice logic :)
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@Ankit: Ohh Sorry..I didnt actually read the question properly..
I didnt see we have to check for sum which must be another element in the
array not some user provided constant value..I mis-understood it with sum
upto k problem which can be solved on sorted array in O(n)...
thats why gave a wrong
@Saurabh: +1 for your second code.
On Sun, Aug 7, 2011 at 12:17 PM, Amol Sharma amolsharm...@gmail.com wrote:
nice :)
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On Sun, Aug 7, 2011 at 8:16 AM, saurabh singh saurab...@gmail.com wrote:
Yup it
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)
@Shady :
No...we can say this only at the time when following constraints are
satisfied:
1) *Outcome* of event *should be* *binary*. (In above example Sum can have
binary outcomes only i.e. EVEN or ODD)
2) Random variable x in P(x) should be supported on set {1,2,3,4,} i.e.
It *should start
I agree with Ved that Apti of GS is really really tough and technical is
relatively easy..They ask CAT (Quant) like questions and you have to solve
them in little time too...Also there will be sectional cut-off for this
written test.
So dont be afraid if you can solve only 10-15 questions from
Sort both the lists...
(Keep track of their original indices as we need to return to original list)
Modify the merge process of merge_sort:
// Assume size of list1 is m and that of list2 is n
curr_list1=0;curr_list2=0;curr_output=0;
while( curr_list1 m curr_list2 n )
{
If
@Shady: +1
But I cant stop laughing.[?] Hacking some1's account and doing such hilarious
posts...[?]
On Wed, Aug 3, 2011 at 9:55 PM, Anil Arya anilarya...@gmail.com wrote:
@shadythank you very much
.Good job
On Wed, Aug 3, 2011 at 9:48 PM, shady sinv...@gmail.com wrote:
utkarsh -
@Kumar:Thanks for the very good question and
@Dave:Thanks for very very good solution...
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x^(x^x) - (x^x)^x = 0
Thus, x^(x^x) = (x^x)^x
Let's open it up by taking log on both sides...
(x^x)*log(x) = x* log(x^x)
(x^x)*log(x) = x*x*log(x)
If x==1 equation is satisfied as log(x) becomes 0..
so x=1 is definitely a solution. what if when x != 1
cancelling log(x) on both the sides..
x^x =
Insufficient data to calculate what you need to find out !!!
On Thu, Jul 28, 2011 at 9:39 PM, shubham shubh2...@gmail.com wrote:
A man leaves his office daily at 07:00 PM. His driver arrives with the car
from home to his office at sharp 07:00 PM. So he doesn't need to wait for
any transport
@Ankur Garg: Nice explanation at the link given by u...
On Tue, Jul 26, 2011 at 10:35 PM, Ankur Garg ankurga...@gmail.com wrote:
Check this
http://codesam.blogspot.com/2011/03/find-all-subsets-of-given-set.html
On Tue, Jul 26, 2011 at 9:41 PM, Vishal Thanki vishaltha...@gmail.comwrote:
@Sunny: Excellent explanation ( solution) !!
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@surender: I assume you want to give general solution to Josephus problem in
which we shoot every kth person...In that case formula for pos must be:
pos = ( x*k )%n + (k-1)
In current context, k=2...thus the formula which you gave also holds
true..but not when k != 2
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I agree with skript: Number of ways of doing this is n!
One in the first row can be placed in n ways.
After one in first row has been placed,
we can place One in second row in n-1 ways and so on.
So total num of ways is n*(n-1)*...*1 = n!
One possible solution to this problem can be coded as
@Antony: To post in this group..Just login to your gmail account. Compose
new mail containing your question and send it to algogeeks@googlegroups.com
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I had one in my MS apti...
Given a randomly generated 2 d matrix find an element which occurs in all 3
rows...
On Sat, Jul 9, 2011 at 2:47 PM, Nitish Garg nitishgarg1...@gmail.comwrote:
I think Strassen Algorithm is used to multiply Matrices efficiently. Read
Cormen.
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@My post: matrix was randomly generated 3X3 matrix...(Not any 2D matrix)
On Sat, Jul 9, 2011 at 9:08 PM, Kunal Patil kp101...@gmail.com wrote:
I had one in my MS apti...
Given a randomly generated 2 d matrix find an element which occurs in all 3
rows...
On Sat, Jul 9, 2011 at 2:47 PM
@Sunny : Excellent !!! Keep posting such nice solutions !! :)
On Sat, Jul 2, 2011 at 1:47 AM, Anika Jain anika.jai...@gmail.com wrote:
ohh ok i got it.. thanx :)
On Sat, Jul 2, 2011 at 12:48 AM, sunny agrawal sunny816.i...@gmail.comwrote:
it is (2,4] not [2,4].
open interval, close
@Sandeep: Without storing explicit pointer to last element, how would you be
able to access last element in a Singly Linked List in O(1) ??? Is there
any parallel data structure that needs to be maintained ?? and if it is
larger than size of 2 explicit pointers to last and first elements then 2
Assuming everything is unbiased:
probability of the next slot to contain a bullet (given, first was empty)
would be (1/4) = 0.25
After spinning: prob(bullet) = (2/6) = 0.334...
We want to minimize the probability...
thus answer should be just to pull the trigger again..
On Mon, Jun 13, 2011 at
@Divye Kapoor : It was interesting...use of inheritance concept to print
that...
On Mon, Jun 13, 2011 at 12:09 AM, Divye Kapoor divyekap...@gmail.comwrote:
This will probably be the best solution yet ;)
Compile time template metaprogramming:
templateint N
class X : public XN-1 {
public:
Yups...that seems best.. :)
On Fri, Jun 10, 2011 at 4:03 PM, sunny agrawal sunny816.i...@gmail.comwrote:
initialy cal size of queue then apply a for loop
On Fri, Jun 10, 2011 at 4:00 PM, sunny agrawal sunny816.i...@gmail.comwrote:
First algorithm taht comes in mind
deque a element
if
IF allowed ???
If yes...Use recursion..
On Fri, Jun 10, 2011 at 9:12 PM, Navneet Gupta navneetn...@gmail.comwrote:
Take n from user and print 1 to n. No loops like for/while/do-while are
allowed to use.
--Navneet
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How about this???
*
unsigned int flip_j_to_k_bits (unsigned int n,unsigned int j,unsigned int k)
{
unsigned int temp;
int num_of_on_bits = k-j+1;
temp = (1num_of_on_bits)-1;
temp = j;
return (n^temp);
}*
I dont know whether shift operation is O(1) or not !
But i think
@ross: seems logically correct..nice solution..
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@ Dumanshu:
With memory restriction also XOR method works.. :)
In this case difference is just that you will be working with 400/ X
number of files..where X is size of the RAM...just maintain a variable
Curr_XOR_value and go on XORing it with element read from the file.
When you are done
Automatic deletion will solve the trouble only for you not for the group as
such messages are not reported spam.. [?]
What i do is: When i login just type in search box 'panzer'...mark
all...report spam and then delete from spam box..That way they are
definitely reported as spams..[?]
We can't
Ohh...that was a hard one...[?]
On Tue, Jun 7, 2011 at 10:30 AM, shashankreddy509
shashankreddy...@gmail.com wrote:
http://mathworld.wolfram.com/TwinPrimes.html
http://mathworld.wolfram.com/TwinPrimes.html
have look at this link...
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Simple solution of order O(n^2), similar to bubble_sort, is obvious...
Any improvements ???
On Sun, Jun 5, 2011 at 7:03 PM, Arunachala Karthik
arunachalakart...@gmail.com wrote:
What is the order of time specified in the question?
On Thu, Jun 2, 2011 at 5:03 PM, siva viknesh
Manually calculated it to be 14.. [?]
Can't think of any general formula but i think a formula or at least
recursive function must be there to solve this.
On Sat, Jun 4, 2011 at 1:01 PM, siva viknesh sivavikne...@gmail.com wrote:
Stack A has the entries a,b,c ( with a on the TOp) Stack B is
If you are not going to allow extra space, you have to compromise on time
complexity..[?]
If you dont have your string already stored in a trie/hashmap usage of it
requires additional buffer.
Simple solution would be:
Sort given string using in-place sorting algorithm and then removal of
@Ashish: your solution is not O(N). I dont think you are taking advantage of
the statement ( A and B need not be sorted in the end)
@sravanreddy: excellent solution.
On Thu, Jun 2, 2011 at 7:46 PM, Ashish Goel ashg...@gmail.com wrote:
int i=lenA-1;
int j=lenB-1;
while (j=0)
{
if (A[i]
Go to hell u spammer..[?]
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For more options,
Hahaha..nice one.. :) :)
On Thu, May 26, 2011 at 9:43 PM, DeVaNsH gUpTa devanshgupta...@gmail.comwrote:
Mark as it reads ?(Question Mark) Crimson.
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@Gaurav: You might want to say circular doubly linked list, didn't you ?
coz without that, its not possible to reach last node if we are at first
node or vice-versa.
On Thu, Jun 2, 2011 at 9:31 AM, Gaurav Aggarwal 0007gau...@gmail.comwrote:
other solution might be to use doubly linked list,
@Piyush:
I don't think 1010 (and any even number) will be a binary palindrome (Unless
u allow single leading zero)...(Either you allow all or allow none)
If its not so what about 1001 then ? Whether it will be a palindrome or not
?[?]
Don't you think, it isn't possible to do this in less than
Not strictly an algorithmic question, rather a test-of-design-skills type of
question.
You are asked to design a 3-in-1 remote control for TV, a DVD player, and a
cable box.
How will you go with design ?
In my approach it should have following buttons.
1) *Device Key* :
Will represent device
@ Piyush: Excellent Solution...It appears both Correct and O(n)...Good work
!![?]
Just a minor correction in your algo.[?]
* while(B[i]C[j]) *
* j++;
must also check for J's bound as:
**while ( j ( sizeof(A)/sizeof(A[0]) )* *B[i]C[j] )
j++;
Or it will crash when
@Amit: Ohh..Your test case is correct but not my solution..[?]
It only works if it is guaranteed that one end will be at the extreme of the
array ! (UseLess ! [?])
Sorry folks...
So can anybody prove that O(n) solution does not exist for this problem? [?]
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Hahaha...Nice answer Piyush !
On Wed, May 18, 2011 at 12:44 PM, Piyush Sinha ecstasy.piy...@gmail.comwrote:
*NOTHING*.
*Greater than GOD - NOTHING*
*Worse than EVIL - NOTHING*
*Poor have it and rich want it - NOTHING*
*if you eat it, you die - NOTHING
*
On Wed, May 18, 2011 at 12:35
Ohh..If it is so...Sorry !![?] I understood it the different way...[?]
But if the question is as mentioned in your 2nd case then also I believe
there is O(n) solution.[?]
Maintain
two pointers: *START* and *END*
two variables: max1 and max2
Assume arr[MAX_SIZE] to be the array containing
Nice explanation Dave..Thnx for the extra info !!
On Tue, May 17, 2011 at 11:00 AM, Piyush Sinha ecstasy.piy...@gmail.comwrote:
thanks Dave :)
This is a standard Google question
On 5/17/11, Dave dave_and_da...@juno.com wrote:
@Piyush. The simplest algorithm is to sort the array
@Anuj Piyush:
You didn't get the algo. It works on unsorted array also. You might have
missed the statement
*else // next element is smaller than or equal to current element
reset curr_max to 1;*
Here, the comment itself shows unsorted elements have been taken into
consideration.
If you
Each team plays a total of 14 matches. Top 50% teams(4 out of 8) qualify for
the semis.
Thus, u must win more than 50% matches to be sure to get into semis. Thus, 8
is the answer.
On Fri, May 13, 2011 at 12:14 AM, amit amitjaspal...@gmail.com wrote:
Consider a series in which 8 teams are
@ Anil: +1 dude...Nice answer...
On Wed, May 11, 2011 at 8:52 PM, anil chopra anil.chopra2...@gmail.comwrote:
i will stop imaging.
On Wed, May 11, 2011 at 7:38 PM, Dave dave_and_da...@juno.com wrote:
I was on a river boat in Europe, and the emergency drill was to go to
the upper deck and
OMG !!! Was it so easy
I feel lyk crying that i didnt think of this...Really bad of mine.. [?][?]
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Must admit it was nice !!!
Well use of English and sentence construction..
On Wed, Apr 27, 2011 at 1:59 PM, Lavesh Rawat lavesh.ra...@gmail.comwrote:
* Short Riddle
*
*How far can a dog run into the forest?
*
*Update Your Answers at* : Click
@Hary rathor: Your program also crash on my Dev-Cpp (Version 4.9.9.2)..Make
sure whether it runs on your PC !!
I don't know whether Literal can be type-casted...
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geeksforgeeks.org
BTW..
Thanks for creating this post...I have came across really nice sites that
interest me in above replies..
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@All : Please report it spam and then delete. Gmail will look into it if
reported spam by many people.
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Friends in my class appeared for it recently. They were asked OS n
Networking based questions.
On Thu, Apr 7, 2011 at 11:09 PM, nitish goyal nitishgoy...@gmail.comwrote:
If anyone had telephonic interview at cisco regarding summer internship,
then
please the experience asap. I am having this
Nice to See so many Kunal here !!! :P
On Wed, Mar 30, 2011 at 11:42 PM, Kunal Yadav kunalyada...@gmail.comwrote:
One is grandpa
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Regards
Kunal Yadav
(http://algoritmus.in/)
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Excellent answer Kunal !!
On Wed, Mar 30, 2011 at 1:55 PM, kunal srivastav kunal.shrivas...@gmail.com
wrote:
grandfather, father and son went for fishing... here we have two fathers
and two sons
On Wed, Mar 30, 2011 at 1:52 PM, Lavesh Rawat lavesh.ra...@gmail.comwrote:
*Fishing Problem *
@rk: thnx...
@raunak: https://www.spoj.pl/ITRIX11/problems/main
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I am able to view the solutions but not able to get the logic behind it...
Can you explain it...Or any1 who has solvd it can xplain !!
A link, if any, to an article explaining logic behind it would be awesome...
Also idea of writing an editorial explaining logic to solve problems in this
contest
@Bittu:
Can you elaborate more how Constructing BST (I hope it stands for Binary
Search Tree) would be o(n)...
I think It would also be O(nlog(n))...
My xplanation:
Single element can be inserted in BST in O(log n)
So inserting n elements would be n * O(log n) -- O(n log n)
So as per me
How to solve that Lucky Sequence Again problem...
i tried it using vectors...for small values it succeeded..
but it wasn't calculating for 10^10 or for large condition..
So what was the logic to solve the problem ???
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Yes...Kunal is right !!!
It doesn't matter whether getMax() removes element or not, because MAX value
is maintained in each node of stack which is pointing to the maximum of the
value present below or up to that element...
So if stack is
5 3 2 8 6
Then nodes would have following structure
5
@dave:
Very nice answer yaar..Excellent !!!
On Wed, Mar 23, 2011 at 7:08 PM, Dave dave_and_da...@juno.com wrote:
It is LCM(2,3,4,5,6,7,8,9,10) - 1 = 2^3 * 3^2 * 5 * 7 - 1 = 2519.
LCM = least common multiple.
Dave
On Mar 23, 2:41 am, Lavesh Rawat lavesh.ra...@gmail.com wrote:
*A
@cegprakash:
y dont u use formula...
log2(n) -- log(n) / log(2)
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Yes...
See first num is 1
second is 11 (there is 1 number of 1 in previous number i.e. in 1)
third is 21 (there is 2 number of 1 in previous number i.e. in 11)
fourth is 1211 (there is 1 number of 2 and 1 number of 1 in previous number
i.e. in 21)
ans so on...
Hope u got it
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Nopes...why precalculate and waste space...+ It gets bigger n bigger
afterwards
Look at this algorithm..
en.wikipedia.org/wiki/Run-length_encoding
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@Gunjan Sharma:
Why do you think 50 is not an acceptable answer???
You might be thinking last 5 zeroes not representing the number of zeroes
followed..thus you gave an answer as 10 in one of your previous
post...
But as per the problem is exemplified, *if it is the original precise
Nothing !!! :P :P
On Fri, Mar 18, 2011 at 1:02 PM, Lavesh Rawat lavesh.ra...@gmail.comwrote:
*A Riddle Problem Solution*
*
*What is one thing that all wise men, regardless of their religion or
politics, agree is between heaven and earth
Update Your Answers at : Click
Hey..
I also got into same trouble today...
I submitted it 6 times..then got bored and de moralised cause i cudnt find
flaw in code...
When i read your mail and corresponding sorry mail...it just struck me..I
also had to print YES and NOand i was printing NO as NoIt
got ac den...
:P
thnx
@ ashu:---Very interesting and nice solution...
On Tue, Mar 15, 2011 at 2:58 PM, Terence technic@gmail.com wrote:
To the east shop :)
One can not get his haircut by himself.
On 2011-3-15 15:55, Lavesh Rawat wrote:
* barbershop **Problem Solution*
*
*A traveller arrives in a small
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