@sunny why we look at all k number which are greater then x , correct ?
Lets think in this way
we wants to check if kth smallest element in heap thats =x isn't it ? so
if root of mean heap is greater then x then none other elements will less
then x so we terminate .
else our algorithm will
@anubhaw ..its like spoon feeding , never mind u could have googled it :D
http://www.google.co.in/search?q=3+sumie=utf-8oe=utf-8aq=trls=org.mozilla:en-US:officialclient=firefox-a
Thanks
Shashank Mani
CSE, BIT Mesra
http://shashank7s.blogspot.com/
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hey ,i got the bug and got AC ALSO..there was
several bugs ...sry ,for silly dbt
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@Lucifer- thnks got ur logic...:)
On Thu, Dec 15, 2011 at 12:07 PM, Lucifer sourabhd2...@gmail.com wrote:
Correction:
for NAN :
N(IT)A + TI + N = NITATIN
On Dec 15, 11:33 am, Lucifer sourabhd2...@gmail.com wrote:
@topcoder..
String: NITAN
RevStr: NATIN
LCS ( NITAN, NATIN) =
Twisted linked list problem: Two linked lists merge at some node p,both the
head pointers are given find the merging point under
the following restrictions.
1. You should not traverse to the end any of the linked list.
2. Merge node p should be detected by the time you reach at most most k
nodes
I like this offer http://www.chrisbeck.de/inf.php It could be
interesting for you, too
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in the link list take another variable suppose toggle and initialize all
with 0,,, at the time of declaring the link list( so no traversal of entire
link list).. now starting from both head traverse a single element at a
time but in alternating way... for one linklist the toggling value is set
to
seem like very twisted problem :P :P . lot of restriction.
On Thu, Dec 15, 2011 at 7:30 PM, Ankur Garg ankurga...@gmail.com wrote:
Twisted linked list problem: Two linked lists merge at some node p,both
the head pointers are given find the merging point under
the following restrictions.
1.
Sorry for replying late...ya posted the reply in hurry.But it appears that
finally it has been resolved what i meant to say in my original post.
On Thu, Dec 15, 2011 at 6:38 PM, Mohit kumar lal kumarmohit...@gmail.comwrote:
@Lucifer- thnks got ur logic...:)
On Thu, Dec 15, 2011 at 12:07 PM,
saurabh is right nd or U can also download lecture 2 of series
Introduction to Algorithm mit lecture . It is jst awesome .
On 12/15/11, saurabh singh saurab...@gmail.com wrote:
Introduction to Algorithms Coreman
On Wed, Dec 14, 2011 at 8:21 PM, Shashank Jain shashan...@gmail.com wrote:
I
http://www.geeksforgeeks.org/archives/14943 is a very similar problem.
On Thu, Dec 15, 2011 at 1:12 PM, WgpShashank shashank7andr...@gmail.comwrote:
@Azhar , also for 1st question u r trying Array DS will suffices. Its
Standard Coin Change Problem , u need to solve subproblem 1st , store it in
@Ravi : your solution will not work for all the cases because of the
restriction
*
*
*1)You should not traverse to the end any of the linked list.*
*
*
now consider one linked list to be very large while other is very small say
1/5 of the large one.
now because second linked list is short and it
@All
I don't think the algo given above is entirely correct.. Or may be i
didn't it properly...
So basically say a preorder traversal of the heap is done based on
whether the current root value X. As the algo says that at any point
if k0 and we hit a node value which is =X , then we are done.
@Ravi : if your solution is acceptable then there is no need ok modifying
data structure.
we can use hashMap which table address and count as an input.
then we can move in same fashion as you have mentioned , if count is of
given address is 1 then it means that address is the intersection point.
@Lucifer : yes even i found flaw in the above algo when i gave it a second
thought but didnt get time to post it.
bcoz min heap has property that the parent node is less than its both
child(subtree to be more precise ) but it does not confirm that left child
is always smaller than right child of
Hi Topcoder.
First of all you posted the wrong array .
The array should be
4, 5, 10, 7, 12, 13
a+b a+c a+d b+cb+d c+d
Now first calculate a+b+c+d which will be (sumofarray)/N-1
So here a+b+c+d = 17
Now take a[1] is a+c
and a[N-1] = b+c
subtracting them
on above algo , there is no need to calculate sum of array so we can do it
in O(N) and not O(n).
On Fri, Dec 16, 2011 at 12:59 AM, Ankur Garg ankurga...@gmail.com wrote:
Hi Topcoder.
First of all you posted the wrong array .
The array should be
4, 5, 10, 7, 12, 13
oops...
wanted to write the same but yeah its meaning turns out to be totally
different :(
anyways very well explained by Lucifier
@shashank
i think now u will be able to get why there will be only 2k comparisons in
the worst case
On Thu, Dec 15, 2011 at 10:51 PM, atul anand
I saw this term non-decreasing order
So we need to sort the numbers first..it will take nlogn .
On Fri, Dec 16, 2011 at 1:12 AM, Ankur Garg ankurga...@gmail.com wrote:
on above algo , there is no need to calculate sum of array so we can do it
in O(N) and not O(n).
On Fri, Dec 16, 2011 at
i got my mistake..
den how to fulfill all the conditions??
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shashank.dude the algorithm whch u suggested in the morning for
tsum ...hv u tried it wid urself bcoz wid the wiki link method
if we wl sort the array den wat b the delimiter so that ,i wl get get all
combination of 3 .in dat method dat wz given for sum of three
we have to submit it in 120 byte cn ne 1 tl me dat whr z the chances of
further byte reduction in this code.
#includestdio.h
#define s(n) scanf(%d,n)
#define I int
I a[14][14];I d(I m,I n){I s=a[m][n];I k;if(!m||!n)k=1;else
if(s)k=s;else{k=d(m-1,n)+d(m,n-1);s=k;}return k;}main(){I
Post the formatted code too.(With proper indents)Then it would be easier
for others to work on it,
On Thu, Dec 15, 2011 at 11:47 PM, anubhav raj anubhaw@gmail.com wrote:
we have to submit it in 120 byte cn ne 1 tl me dat whr z the chances of
further byte reduction in this code.
Problem Statement
A sequence of numbers is called a zig-zag sequence if the differences
between successive numbers strictly alternate between positive and
negative. The first difference (if one exists) may be either positive
or negative. A sequence with fewer than two elements is trivially a
It is dynamic programming.
Search in topcoder algo tutorials
On Dec 16, 2011 10:33 AM, Sangeeta sangeeta15...@gmail.com wrote:
Problem Statement
A sequence of numbers is called a zig-zag sequence if the differences
between successive numbers strictly alternate between positive and
negative.
Given a list of N coins, their values (V1, V2, ... , VN), and the
total sum S. Find the minimum number of coins the sum of which is S
(we can use as many coins of one type as we want), or report that it's
not possible to select coins in such a way that they sum up to S.
For a better understanding
use dynamic programming to solve this problem :-
here is the algo:-
result[S];
result[0]=1;
index;
for i=0 to len(coins[])
c=coins[i];
for k=c to S
index=k-c;
result[k]= result[k] + result[index];
end
end
print result[S];
On Fri, Dec 16, 2011 at 10:52 AM,
give explaination
On Fri, Dec 16, 2011 at 11:14 AM, atul anand atul.87fri...@gmail.comwrote:
use dynamic programming to solve this problem :-
here is the algo:-
result[S];
result[0]=1;
index;
for i=0 to len(coins[])
c=coins[i];
for k=c to S
index=k-c;
int LIDS ( vectorint a , int n ){
int s[51][2];
for(int i = 0 ; i n ; i++)
for(int j = 0 ; j 2 ; j++)
s[i][j] = 1;
for(int i = 0 ; i n ; i++){
for(int j = 0 ; j i ; j++){
if( a[i] != a[j] ){
s[i][a[i]a[j]] =
You are given a stream of numbers which can be positive or negative.
You are
required to provide an operation FIND MEDIAN..which when invoked
should be
able return the median of the numbers in stream (in sorted order) in
O(1)
time.
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