Re: [algogeeks] Linked List question
Thank you for the code snippet.
What I don't understand is
if( (double)rand() / RAND_MAX 1 / k )
Here, Why do we use less than inequality?
Why won't Udit's solution of just using the minimum random number work?
On Tuesday, 1 January 2013 20:57:47 UTC+5:30, Dave wrote:
@Doom: Yes. It is necessary to go through the entire list. The code could
look like this:
int* p = head;
int k = 1;
while( head )
{
head = head - next;
k++;
if( k * (double)rand() RAND_MAX )
p = head;
}
Dave
On Sunday, December 30, 2012 12:40:28 PM UTC-6, Doom wrote:
Hi Dave
Please help me with some additional clarification regarding the
implementation of this algo. How do we make use of random number generator?
And is it necessary to traverse the list until the end?
On Friday, 28 December 2012 19:35:07 UTC+5:30, Dave wrote:
@Sanjeev: Set a pointer p to the first node. Traverse the list. When at
the kth node, set p to the kth node with probability 1/k. When you reach
the end of the list, return p, which will point to a random node with
uniform probability.
Dave
On Thursday, December 27, 2012 11:18:20 PM UTC-6, Sanjeev wrote:
i mean the length of the linked list is not known to us.
@udit how can we do this in single traversal ? i think we need to
traverse the linked list twice in your case.
Please reply if i am wrong ?
On Thu, Dec 27, 2012 at 10:48 PM, Udit Agarwal udit...@gmail.comwrote:
If the length of the linked is infinite then the above algo would do
the needful.
In case you have a finite length linked list, then you can normalize
the random value using following:
Suppose length of linked list is: l
random number is: r; and r l
then new random number would be: r1 = r%l;
now again use the above algo using new random number r1;
On Thu, Dec 27, 2012 at 4:12 PM, Vineeth vineet...@gmail.com wrote:
You said : Given a linked list of infinite length
On Thu, Dec 27, 2012 at 4:06 PM, naveen shukla
naveenshukla...@gmail.com wrote:
But suppose a random number generate a value 5 and your linked list
has only
four elements. In that case what would be the answer ???
On Thu, Dec 27, 2012 at 4:03 PM, Prem Krishna Chettri
hpre...@gmail.com
wrote:
Well my algo will be Something like this
1 Get a Random number. Perhaps You can have the function like
Randon(List
*head, int Randomnumber)
2 Use the function argument Randomnumber to loop the list.
i.e. for(int count=0;count=Randomnumber;count++ ){
head = head - next;
}
3 print (head-value);
4 return ;
Now as we are using byvalue when we return the value of head
remains the
same old head value. So everytime we call we are traversing the
same old
list.
The Random variable can be taken inside the function itself if
the user
is not taking the random value.
i.e. int Randomnumber = random(); and now the user can calll
Simple
Random(head);
On Thu, Dec 27, 2012 at 3:31 PM, naveen shukla
naveenshukla...@gmail.com wrote:
random node
--
--
With Best Wishes
From:
Naveen Shukla
IIIT Allahabad
B.Tech IT 4th year
Mob No: 07860896972
E-mail naveenshukla...@gmail.com
--
--
--
Udit Agarwal
B.Tech. ( Information Technology ) , 7th Semester,
Indian Institute of Information Technology
Allahabad - 211012, India
Email : udit...@gmail.com
Mobile: +91-9411656264
--
--
With Best Wishes
From:
Naveen Shukla
IIIT Allahabad
B.Tech IT 4th year
Mob No: 07860896972
E-mail naveenshukla...@gmail.com
--
Re: [algogeeks] Linked List question
Hi Dave
Please help me with some additional clarification regarding the
implementation of this algo. How do we make use of random number generator?
And is it necessary to traverse the list until the end?
On Friday, 28 December 2012 19:35:07 UTC+5:30, Dave wrote:
@Sanjeev: Set a pointer p to the first node. Traverse the list. When at
the kth node, set p to the kth node with probability 1/k. When you reach
the end of the list, return p, which will point to a random node with
uniform probability.
Dave
On Thursday, December 27, 2012 11:18:20 PM UTC-6, Sanjeev wrote:
i mean the length of the linked list is not known to us.
@udit how can we do this in single traversal ? i think we need to
traverse the linked list twice in your case.
Please reply if i am wrong ?
On Thu, Dec 27, 2012 at 10:48 PM, Udit Agarwal udit...@gmail.com wrote:
If the length of the linked is infinite then the above algo would do
the needful.
In case you have a finite length linked list, then you can normalize
the random value using following:
Suppose length of linked list is: l
random number is: r; and r l
then new random number would be: r1 = r%l;
now again use the above algo using new random number r1;
On Thu, Dec 27, 2012 at 4:12 PM, Vineeth vineet...@gmail.com wrote:
You said : Given a linked list of infinite length
On Thu, Dec 27, 2012 at 4:06 PM, naveen shukla
naveenshukla...@gmail.com wrote:
But suppose a random number generate a value 5 and your linked list
has only
four elements. In that case what would be the answer ???
On Thu, Dec 27, 2012 at 4:03 PM, Prem Krishna Chettri
hpre...@gmail.com
wrote:
Well my algo will be Something like this
1 Get a Random number. Perhaps You can have the function like
Randon(List
*head, int Randomnumber)
2 Use the function argument Randomnumber to loop the list.
i.e. for(int count=0;count=Randomnumber;count++ ){
head = head - next;
}
3 print (head-value);
4 return ;
Now as we are using byvalue when we return the value of head remains
the
same old head value. So everytime we call we are traversing the same
old
list.
The Random variable can be taken inside the function itself if the
user
is not taking the random value.
i.e. int Randomnumber = random(); and now the user can calll Simple
Random(head);
On Thu, Dec 27, 2012 at 3:31 PM, naveen shukla
naveenshukla...@gmail.com wrote:
random node
--
--
With Best Wishes
From:
Naveen Shukla
IIIT Allahabad
B.Tech IT 4th year
Mob No: 07860896972
E-mail naveenshukla...@gmail.com
--
--
--
Udit Agarwal
B.Tech. ( Information Technology ) , 7th Semester,
Indian Institute of Information Technology
Allahabad - 211012, India
Email : udit...@gmail.com
Mobile: +91-9411656264
--
--
With Best Wishes
From:
Naveen Shukla
IIIT Allahabad
B.Tech IT 4th year
Mob No: 07860896972
E-mail naveenshukla...@gmail.com
--
Re: [algogeeks] expectation values..
If we expand it.. E(t) = E(t1) + E(t2) + E(t3) + ... + E(tn);
here I am able to derive E(t1) as N/1 using the expression E(t1) = 1/N +
((N-1)/N)(E(t1) + 1);
but how do I proceed?
How do I derive the E(t2) and so on??
What do these values mean??
Does it mean like E(t2) is the no. of expected throws to get value 2??
Any help on this?
On Sunday, 17 June 2012 00:09:13 UTC+5:30, amitesh wrote:
This problem is similar to Coupan collector problem.
http://en.wikipedia.org/wiki/Coupon_collector%27s_problem
In your case the answer is
[image: For N-Dice ; \newline \sum_{i=1}^{N} N/i \newline for\; N =~2 ;
\newline \sum_{i=1}^{2} 2/i = 2/1 + 2/2 = 3 \newline]
Hope it helps!
--
Amitesh
On Sat, Jun 16, 2012 at 5:18 PM, Gaurav Popli gpgaurav.n...@gmail.comwrote:
What is the expected number of throws of his die while it has N sides
so that each number is rolled at least once?
e.g
for n=2 ans 3.00
n=12 ans is 37.24...
i refrd to expectation tutuorial at
http://www.codechef.com/wiki/tutorial-expectation but still couldnt
get the logic...
any help?
--
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[algogeeks] Re: Find border of a binary tree.
75 is omitted because its the border. Think of border like putting an elastic rubber band around the tree. Print the nodes being touched by the rubber. On Monday, 9 April 2012 08:12:48 UTC+5:30, Gene wrote: Good question. The problem is not well-defined. It's possible that 75 should be omitted because there are deeper subtrees to the left and right. But we'll never know for sure because examples don't make a good definition. On Apr 8, 2:29 pm, atul anand atul.87fri...@gmail.com wrote: i guess in the given link 1st example should inculde 75 ?? correect me if i am wrong. On Fri, Apr 6, 2012 at 10:53 PM, Doom duman...@gmail.com wrote: Here is the reference: http://stackoverflow.com/questions/3753928/finding-border-of-a-binary... None of the proposed solutions is effective enough. Any ideas? -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To view this discussion on the web visit https://groups.google.com/d/msg/algogeeks/-/xjchdh2I_7MJ. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To view this discussion on the web visit https://groups.google.com/d/msg/algogeeks/-/vCCkW93pMCgJ. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Find border of a binary tree.
Here is the reference: http://stackoverflow.com/questions/3753928/finding-border-of-a-binary-tree None of the proposed solutions is effective enough. Any ideas? -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To view this discussion on the web visit https://groups.google.com/d/msg/algogeeks/-/xjchdh2I_7MJ. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Path along the diameter of the binary tree
We have seen the question of computing the diameter of binary tree. Any ideas to compute the path of nodes along that diameter? No parent pointer exists. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To view this discussion on the web visit https://groups.google.com/d/msg/algogeeks/-/CMNxVKMrfasJ. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] how to code a deterministic or Non-Deterministic finite state automata model?
Any pointers to a code using NFA / DFA computation model? -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To view this discussion on the web visit https://groups.google.com/d/msg/algogeeks/-/UO5Em7j89scJ. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Pruning number of a binary tree
Can any one help me in understanding the concept? Here is the reference link: http://stackoverflow.com/questions/8288820/pruning-number-of-a-binary-tree -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To view this discussion on the web visit https://groups.google.com/d/msg/algogeeks/-/mheOlnSKgGYJ. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: 答复: [algogeeks] Given an array A[1..n] of log n
in-place is O(1) auxiliary space. Could you please of something else? On Thursday, 5 April 2012 10:24:09 UTC+5:30, Rujin Cao wrote: =?utf-8?Q?_bit_integers,_sort_them_in-place_in_O(n)_time?= MIME-Version: 1.0 Content-Type: multipart/alternative; boundary==_Part_1591_7508851.1333541114758 --=_Part_1591_7508851.1333541114758 Content-Type: text/plain; charset=utf-8 Content-Transfer-Encoding: quoted-printable using counting sort with an extra O(log n) space to store the appearance count of each number in the range[0,log n]. And then iterate from begin to end to set each number with its count times. I'm not sure whether it is appropriate to call it in-place, but at least it doesn't use O(n) space. Sent from my Windows Phone =E5=8F=91=E4=BB=B6=E4=BA=BA: Doom =E5=8F=91=E9=80=81=E6=97=B6=E9=97=B4: 2012/4/4 20:05 =E6=94=B6=E4=BB=B6=E4=BA=BA: algogeeks@googlegroups.com =E4=B8=BB=E9=A2=98: Re: [algogeeks] Given an array A[1..n] of log n bit int= egers, sort them in-place in O(n) time how are you going to do it in-place? On Wednesday, 4 April 2012 13:24:01 UTC+5:30, Siddhartha wrote: if the length of the binary representation of elements is logn, then the= =20 elements themselves are of size less than 2^log(n)=3Dn. as all the elements are less than n, use counting sort!!! --=20 You received this message because you are subscribed to the Google Groups = Algorithm Geeks group. To view this discussion on the web visit https://groups.google.com/d/msg/al= gogeeks/-/glXPUeyoh8IJ. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscribe@googleg= roups.com. For more options, visit this group at http://groups.google.com/group/algoge= eks?hl=3Den. --=_Part_1591_7508851.1333541114758 Content-Type: text/html; charset=utf-8 Content-Transfer-Encoding: quoted-printable htmlheadmeta content=3Dtext/html; charset=3Dutf-8 http-equiv=3DCont= ent-Type/headbodydivdiv style=3Dfont-family: Calibri,sans-serif; = font-size: 11pt;using counting sort with an extra O(log n) space to store= the appearance count of each number in the range[0,log n].brAnd then ite= rate from begin to end to set each number with its count times.brI'm not = sure whether it is appropriate to call it in-place, but at least it doesn= 't use O(n) space.brbrSent from my Windows Phonebr/div/divhrsp= an style=3Dfont-family: Tahoma,sans-serif; font-size: 10pt; font-weight: b= old;=E5=8F=91=E4=BB=B6=E4=BA=BA: /spanspan style=3Dfont-family: Tahom= a,sans-serif; font-size: 10pt;Doom/spanbrspan style=3Dfont-family: = Tahoma,sans-serif; font-size: 10pt; font-weight: bold;=E5=8F=91=E9=80=81= =E6=97=B6=E9=97=B4: /spanspan style=3Dfont-family: Tahoma,sans-serif; f= ont-size: 10pt;2012/4/4 20:05/spanbrspan style=3Dfont-family: Tahom= a,sans-serif; font-size: 10pt; font-weight: bold;=E6=94=B6=E4=BB=B6=E4=BA= =BA: /spanspan style=3Dfont-family: Tahoma,sans-serif; font-size: 10pt;= algogeeks@googlegroups.com/spanbrspan style=3Dfont-family: Tahoma,s= ans-serif; font-size: 10pt; font-weight: bold;=E4=B8=BB=E9=A2=98: /span= span style=3Dfont-family: Tahoma,sans-serif; font-size: 10pt;Re: [algog= eeks] Given an array A[1..n] of log n bit integers, sort them in-place in O= (n) time/spanbrbr/body/htmlhow are you going to do it in-place?= brbrOn Wednesday, 4 April 2012 13:24:01 UTC+5:30, Siddhartha wrote:bl= ockquote class=3Dgmail_quote style=3Dmargin: 0;margin-left: 0.8ex;border= -left: 1px #ccc solid;padding-left: 1ex;if the length of the binary repre= sentation of elements is logn, then the elements themselves are of size les= s than 2^log(n)=3Dn.divas all the elements are less than n, use counting = sort!!!/div /blockquote p/p -- br / You received this message because you are subscribed to the Google Groups = Algorithm Geeks group.br / To view this discussion on the web visit a href=3D https://groups.google.c= om/d/msg/algogeeks/-/glXPUeyoh8IJ https://groups.google.com/d/msg/algogeek= s/-/glXPUeyoh8IJ/a.br /=20 To post to this group, send email to algogeeks@googlegroups.com.br / To unsubscribe from this group, send email to algogeeks+unsubscribe@googleg= roups.com.br / For more options, visit this group at http://groups.google.com/group/algoge= eks?hl=3Den.br / --=_Part_1591_7508851.1333541114758-- -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To view this discussion on the web visit https://groups.google.com/d/msg/algogeeks/-/LGcbPhxQFrsJ. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Find the maximum boxes which can fit each other?
@Don: Any ideas which oppose the above proposed solution? On Saturday, 24 March 2012 21:52:49 UTC+5:30, Don wrote: Build a graph in which each box is a vertex and there is an edge from A to B if B can fit inside A. Then use the longest path algorithm to find the solution. Don On Mar 24, 1:55 am, Ratan success.rata...@gmail.com wrote: You are given a lot of cuboid boxes with different length, breadth and height. You need to find the maximum subset which can fit into each other. For example: If Box A has LBH as 7 8 9 If Box B has LBH as 5 6 8 If Box C has LBH as 5 8 7 If Box D has LBH as 4 4 4 then answer is A,B,D A box can fit into another only and only if all dimensions of that is less than the bigger box. Also Rotation of boxes is not possible... can ny1 suggest a good algo for this? -- -- Ratan | 3rd Year | Information Technology | NIT ALLAHABAD -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To view this discussion on the web visit https://groups.google.com/d/msg/algogeeks/-/C1yQ1v_QAboJ. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: water-jug problem
@Don: Could you please explain ur tree approach with an example? Thanks Doom On Tuesday, 3 April 2012 02:52:17 UTC+5:30, Don wrote: I did that by building a tree in which each node stores the configuration, including number of steps to that point, amount of water in each bucket, and a pointer to the parent, sibling, and leftmost child. Start out with one node with stepCount set to zero and all the buckets empty. You want the solution with the least number of steps, so use a breadth-first search. For each node, create a list of children based on the following: 1. For each bucket which is not full, fill it 2. For each bucket which is not empty, empty it 3. For each pair of buckets A and B where A is not empty and B is not full, pour A into B When you find a bucket with the desired amount of water, you can trace back through the parent pointers to get the series of moves required to get there. If you just need the minimum number of steps, the problem is much easier. Just use a 99x99x99 table D[99][99][99]. If the bucket capacities are a,b,c, start with all table values set to a large value except D[a][0][0] = D[0][b][0] = D[0][0][c] = 1 Then iteratively start with all states with depth d and set all states which can be reached from that state by a single step to d+1 if their current depth is greater than d+1. You'll end up with a table of all states reachable with the set of buckets, and the minimum number of steps to reach each state. Don On Apr 1, 12:57 pm, bharat b bagana.bharatku...@gmail.com wrote: A common puzzle asked during interviews is how to measure 4 liters of water with just two uneven shaped buckets, one of 3 liter another of 5 liter, in minimum number of steps. To extend and generalize this, write a program to evaluate the minimum number of steps to measure “X” liters of water, with “Y” buckets each having a separate denomination. The assumptions and sample input/output as given below: *Assumptions:* * * 1. Each bucket used for measuring water should be unique in denomination and the number of buckets will be = 3 2. The target amount to be reached has to finally reside in a single bucket (at the end of the measuring activity). 3. The bucket capacities and target amount will be = 99 4. If there are multiple ways to measure the same amount, only one way, having the minimum number of steps, has to be shown in the output. If there are multiple minimum steps solutions present, displaying one solution should be enough. 5. The console input as well as the output should be in single line and in the format shown below. 6. You can assume (not mandatory) minimum 2 steps will be required. You can ignore the cases where solution can be achieved in single step. 7. You can assume that maximum 10 steps will be required to reach the target amount. If any solution requires more than 10 steps to reach the target amount, you can stop the calculation and ignore that test case. *Input Format:* The input will be of the following comma separated format – No. of buckets, capacity of bucket 1, capacity of bucket 2, …, Target amount desired. The following examples will provide more clarity *Sl. No.* *Input string* *Explanation* 1 2,3,5,4 Implies that there are 2 buckets of capacity 3 and 5 lts each and the target amount desired is 4 lts. 2 3,4,6,9,7 Implies that there are 3 buckets of capacity 4, 6 and 9 lts each and the target amount desired is 7 lts. . *Output Format:* Each step should be a 11 bytes field left aligned right padded with spaces along with the right boundary as “|”. Together “|” character the length should be 12 characters. The following examples will provide more clarity *Sl. No.* *Input string* *Expected Output String* 1 2,3,5,4 Fill 2 |Move 2 to 1|Empty 1|Move 2 to 1|Fill 2 |Move 2 to 1| 2 3,4,6,9,7 Fill 1 |Move 1 to 2|Fill 3 |Move 3 to 2| * * *Let’s explain 1**st** output sample in more detail*: As explained earlier, effectively the input 2,3,5,4 means there are 2 buckets of capacity 3 liter and 5 liter and using these two buckets we have to measure 4 liter (which is the target). The answer is: Fill 2 |Move 2 to 1|Empty 1|Move 2 to 1|Fill 2 |Move 2 to 1| Let’s explain the answer also for test case 1: *Step #* *Step Description* *Status of 3 liter bucket (Bucket 1)* *Status of 5 liter bucket (Bucket 2)* *Explanation* Step 1 Fill
Re: [algogeeks] Given an array A[1..n] of log n bit integers, sort them in-place in O(n) time
how are you going to do it in-place? On Wednesday, 4 April 2012 13:24:01 UTC+5:30, Siddhartha wrote: if the length of the binary representation of elements is logn, then the elements themselves are of size less than 2^log(n)=n. as all the elements are less than n, use counting sort!!! -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To view this discussion on the web visit https://groups.google.com/d/msg/algogeeks/-/glXPUeyoh8IJ. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Find the maximum boxes which can fit each other?
What is wrong with atul's solution of longest decreasing subsequece on length after sorting on area? I think the relationship is transitive as well. e.g. box a fits inside b. box b fits inside c by transitivity, box a fits inside c. Isn't this fine? On Saturday, 24 March 2012 22:25:33 UTC+5:30, Don wrote: There is more to it than a longest increasing subsequence because the greater than relationship is not transitive. Don On Mar 24, 3:05 am, atul anand atul.87fri...@gmail.com wrote: ok you need to put box into a box... so first requirnment willl be to sort on the basis of area of box. after this bcoz you are adding one box into another...the box you are putting inside big box ..shud have base length less than a big box or it wont fit in...even if its area is smaller.. now we reduced this problem of finding longest inc subsequence on the basis of base length. On 24 Mar 2012 13:14, Ratan success.rata...@gmail.com wrote: @atul can u plzz describe in detail the algo of modified subsequence prob used here i m nt able to get it ... though tried a lot On Sat, Mar 24, 2012 at 1:05 PM, atul anand atul.87fri...@gmail.com wrote: it is modified longest increasing subsequence problem.. On 24 Mar 2012 12:26, Ratan success.rata...@gmail.com wrote: You are given a lot of cuboid boxes with different length, breadth and height. You need to find the maximum subset which can fit into each other. For example: If Box A has LBH as 7 8 9 If Box B has LBH as 5 6 8 If Box C has LBH as 5 8 7 If Box D has LBH as 4 4 4 then answer is A,B,D A box can fit into another only and only if all dimensions of that is less than the bigger box. Also Rotation of boxes is not possible... can ny1 suggest a good algo for this? -- -- Ratan | 3rd Year | Information Technology | NIT ALLAHABAD -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- -- Ratan | 3rd Year | Information Technology | NIT ALLAHABAD -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.- Hide quoted text - - Show quoted text - -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To view this discussion on the web visit https://groups.google.com/d/msg/algogeeks/-/a06lUGOcKvgJ. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Given an array A[1..n] of log n bit integers, sort them in-place in O(n) time
Any ideas? -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To view this discussion on the web visit https://groups.google.com/d/msg/algogeeks/-/aGzMcjTFcAYJ. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
