Write the code to find lexicographic minimum in a circular array, e.g.
for the array
BCABDADAB, the lexicographic mininum is ABBCABDAD.
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I don't think the inner loop is executing only once . Kindly check it for
this test case {1,3,5,8,9,2,6,7,6,8,9} . And try to print i in inner loop
you will find that for same values of i(Outer index) inner loop is called.
Its an O(n2) solution .
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x is even number probability
0.9x + x = 1
x = 1/1.9
P(2) = P(4) = P(6) = (1 / 1.9) / 3
P(4|5|6) = (P(4) + P(6)) / 0.75 = 2 / 1.9 / 3 / 0.75 = 0.4678362573099415
-weiq
On Thu, Jan 13, 2011 at 11:29 PM, snehal jain learner@gmail.com wrote:
An unbalanced dice (with 6 faces, numbered from 1 to
if the root of T2 is duplicated in T1,this code may give a wrong
answer(as there is no provision of backtracking in case there is a mis
match after some matchings).given above KMP is the best possible answer
i think..
On Wed, Jan 12, 2011 at 11:28 AM, pacific pacific
wow
This s a spoj problem
http://www.spoj.pl/problems/MINMOVE/
On Fri, Jan 14, 2011 at 1:40 PM, snehal jain learner@gmail.com wrote:
Write the code to find lexicographic minimum in a circular array, e.g.
for the array
BCABDADAB, the lexicographic mininum is ABBCABDAD.
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There s O(n) solution for this :)
On Fri, Jan 14, 2011 at 2:13 PM, radha krishnan
radhakrishnance...@gmail.com wrote:
append the string to original string and
index=answer of that spoj problem
now u can ouput the string from index to index+strlen(originalstring)-1
On Fri, Jan 14, 2011 at
FindStartIndex(char[] a)
{
int start = 0;
int current = 1;
while(current a.Length)
{
if(a[current] a[start])
{
start = current;
++current;
}else if(a[current] a[start])
{
++current;
}else //a[current] ==
how to find if two arrays of size n are disjoint or not in O(n)
time ?? You can use only O(n) space
The elements are +ve in the range 0 to n power 100..
Regards
Shashank Mani
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are they sorted?
On Fri, Jan 14, 2011 at 3:35 PM, bittu shashank7andr...@gmail.com wrote:
how to find if two arrays of size n are disjoint or not in O(n)
time ?? You can use only O(n) space
The elements are +ve in the range 0 to n power 100..
Regards
Shashank Mani
I guess u got confused with the comment I wrote, I have added 2 print
statements and now I guess it should be clear to you as to why the code is
O(n). The comment means that each element of the min_steps_dp will be
ACCESSED only ONCE over the execution of the entire program. Hence the outer
loop
ideally, a hashMap would be preferred
walk through one array and set the corresponding entry, and then through
another array, if any entry found, then they are not disjoint.
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
On Fri, Jan 14, 2011 at
You have N computers and [Ca, Cb] means a is connected to b and this
connectivity is symmetric and transitive. then write a program which
checks this
Regards
Shashank
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Floyd Warshall ?
On Fri, Jan 14, 2011 at 6:45 PM, bittu shashank7andr...@gmail.com wrote:
You have N computers and [Ca, Cb] means a is connected to b and this
connectivity is symmetric and transitive. then write a program which
checks this
Regards
Shashank
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You have N computers and [Ca, Cb] means a is connected to b and this
connectivity is symmetric and transitive. then write a program which
checks that all computers are interconnected and talk two each other?
Regards
Shashank
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Two different binary trees can have same set of Leaves/Inner Nodes and same
Preorder traversal
5
/ \
3 10
/ \
1 9
\
7
5
/ \
3 9
/ / \
1
If it is a BST...then having a pre-order traversal can give us the unique
binary tree.
Also, as per the problem statement,
every node can have 0 or at most 2 nodes.
that means every node can have 0 or two childs, which is not the case below.
On Fri, Jan 14, 2011 at 6:49 PM, Balaji Ramani
@qw
I think you solution will timed out?!
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@shehal
Concatenate string to itself and then find smallest suffix of length = n
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@Gene: Didn't you mean (2n)! / (n!)^2 ?
On Jan 13, 10:22 pm, Gene gene.ress...@gmail.com wrote:
The number of ways that n things can be positioned within 2n things is 2n
choose n. This is equal to (2n!) / (n!)^2.
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Bayes' theorem: http://en.wikipedia.org/wiki/Bayes'_theorem
P(x=even|x3) = P(x3|x=even)*P(x=even)/P(x3)===B
On Jan 14, 2:29 am, snehal jain learner@gmail.com wrote:
An unbalanced dice (with 6 faces, numbered from 1 to 6) is thrown. The
probability that the face value is odd is 90% of the
It's irrelevant but Building Special Tree has the same acronyms as
Binary Search Tree...lame joke I know
On Jan 14, 8:44 am, vaibhav agrawal agrvaib...@gmail.com wrote:
If it is a BST...then having a pre-order traversal can give us the unique
binary tree.
Also, as per the problem statement,
http://www.cs.ubc.ca/labs/beta/Projects/SATzilla/
http://www.csail.mit.edu/events/eventcalendar/calendar.php?show=eventid=2784
Automated Algorithm Configuration and Selection: Enabling Technologies
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Speaker: Frank Hutter, University of British Columbia, Vancouver
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In recent years, Artificial Intelligence researchers have largely
focused their efforts on solving specific problems, with less emphasis
on 'the big picture' - automating large scale tasks which require
human-level intelligence to
Here is general dicsussion on this topic:
http://forums.topcoder.com/;jsessionid=A92760DB520220B76858A0675626EBAC?module=ThreadthreadID=641215start=0mc=4#1100409
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@balaji_ramani
That is why there is additional info (L or N). So it is solvable during dfs.
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Simple BFS or DFS solves this problem.
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For
http://www.springer.com/series/7373?changeHeader is the source of the
following.
Researchers and practitioners alike are increasingly turning to
search, optimization, and machine-learning procedures based on natural
selection and genetics to solve problems across the spectrum of human
endeavor.
http://www.genetic-programming.com/johnkoza.htm
Scientific Research Interests—John R. Koza
Our main research interest is automatic programming (also called
program synthesis or program induction)—that is, getting computers to
solve problems without explicitly programming them.
This goal can be
http://www.genetic-programming.com/johnkoza.htm
Scientific Research Interests—John R. Koza
Our main research interest is automatic programming (also called
program synthesis or program induction)—that is, getting computers to
solve problems without explicitly programming them.
This goal can be
http://zobayer.blogspot.com/2009/12/cse-102-practice-recursions.html
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@juver++, can you give me a test case that will time out?
On Fri, Jan 14, 2011 at 6:44 AM, juver++ avpostni...@gmail.com wrote:
@qw
I think you solution will timed out?!
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@qw
Please ignore my post. I've already removed it. Cause I found it seems to be
ok.
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If I am getting the question right, I believe both the above trees represent
this input:
5(N) 3(N) 1(L) 9(N) 7(L) 10(L)
5
/ \
3 10
/ \
1 9
\
7
5
/ \
3 9
Here is compact linear solution (it is based on some non-trivial
observations):
int min = 0, p = 1, l = 0;
while (p n min + l + 1 n) {
int cmp = buffer[min + l] - buffer[(p + l) % n];
if (cmp == 0)
++l;
else if (cmp 0) {
p += l + 1;
l = 0;
}
else {
I thought that node has 0 or 2 childs. If so, than supplied information is
enough.
If node may has 1 child, then it is failed.
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To
@Avi Greedy approach doesn't work since you can't ensure the choice is
locally optimum. Consider 3,9,2,1,8,3. Using greedy alg. would give
you 3,1,8,3 while otherwise DP would give you 3,9,3.
On Jan 14, 6:11 am, Avi Dullu avi.du...@gmail.com wrote:
I guess u got confused with the comment I
@jammy Even I felt the same, but the greedy 'algo' u suggest is actually
IMHO not a greedy approach. You just take each arr[i] and jump *without
deciding a locally optimal policy* . SO, if u were to *see* arr[i] and
*decide* on the optimal policy I feel one would follow d same steps as in a
DP
Integers larger than which fit in 32/64 bits.
On Thu, Jan 13, 2011 at 8:58 PM, bittu shashank7andr...@gmail.com wrote:
1. 10 test cases for entering 3 values representing sides of a
triangle and the program giving output as scalene, isosceles or
equilateral--Means At Least 10
On Wed, Jan 12, 2011 at 2:44 PM, snehal jain learner@gmail.com wrote:
1. Quick-sort is run on two inputs shown below to sort in ascending
order
(i) 1,2,3, ….,n
(ii) n, n - 1, n - 2, …., 2, 1
Let C1 and C2 be the number of comparisons made for the inputs (i) and
(ii) respectively. Then,
At each location if the value is k ,
find the largest value in the next k elements and jump there.
This greedy approach works in 0(n^2) and i believe it works. If not can
someone give me a counter example ?
On Sat, Jan 15, 2011 at 3:30 AM, Avi Dullu avi.du...@gmail.com wrote:
@jammy Even I
@pacific..
the approach needs a little bit of tuning...
for instance, for the set 9 8 7 6 5 4 3 2 1 1 2, per ur approach, u
will pick 9, 8, 7, 6 etc...
minimum jumps in reality is from 9 to 1 to 2.
On Jan 14, 8:19 pm, pacific pacific pacific4...@gmail.com wrote:
At each location if the value
@balaji - In the first tree , 9 has one only child which is not possible
for this question.
@juver++ - Can you give some algo for this problem
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Sorry.. u need 2 sort only one array. whichever u likei was thinking to
sort 1 array and wrote to sort all 3 anyway ... sort only one
On Tue, Jan 11, 2011 at 1:18 PM, MOHIT mohit...@gmail.com wrote:
@ puneet then we have to sort only one array, on which we doing binary
search
@ snehal I have a doubt. Are you sure this will terminate and find the
looping node?
slow1=head;
while( slow1!=slow)
{
prev=slow;
slow=slow-next;
slow1=slow1-next;
}
Consider this example list with nodes 1-12 looping node 5.
1 2 3 4 5 6 7 8 -
|
@Decipher -
something like this:
void build_tree(Node* node, int index) {
node = new Node(value[index++]); // create an empty node
if (label[index] == 'L') return; // we are at leaf, there is no need to
process further current subtree
build_tree(node-left, index); // build left and right
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