Hi,
I think you have misread the problem. You are allowed to take away letters from
the middle of the input string. So, for ADAM, you have to check
AAM (D is removed)
ADM (A is removed)
in addition to the usual substrings (i.e. strings formed by taking away letters
from the beginning or the
Are you really sure that your assumption is correct?
I am not sure about the fact that you can take sub strings from the word by
removing characters from between them...
AAM is certainly not valid sub string for this case as far as my
understanding of the problem stands.
On 3/30/07, Muntasir Azam
We have set named S. We assume we have an algorithm that specifies if
there is a subset in S that sum of it's elements equals to K in O(n)
and returns TRUE or FALSE.
That assumption is either wrong, or you have specified the problem
incorrectly. Is subset the right word? Do you mean
And we have to find all the elements of the set S using the sum method...
hence we don't have the members of the set S.
The method hence should more likely be like SUM(K) which returns TRUE or
FALSE if there's a subset, sum of whose elements is equal to k.
It is a very interesting problem...
On
Yes, i read the problem again. My interpretation was incorrect.
Generally one tends to interpret a problem in a way it's a bit easier to
understand :D. That happened with me on this occasion.
It's a pretty complex problem from where i see it. and time limit:10
seconds!!! Seems rather too less...
Oh I think i wasn't thinking straight...
I think the problem can be solved by simply removing 'i' letters from the
string and check if that is a palendrome. where i would go from 1 to lenght.
But when i 1...will i need to select all possible 2 length strings from
the word and remove them one by
- Original Message -
From: Mahdi [EMAIL PROTECTED]
To: Algorithm Geeks algogeeks@googlegroups.com
Sent: Friday, March 30, 2007 1:27 PM
Subject: [algogeeks] Sum of subsets
We have set named S. We assume we have an algorithm that specifies if
there is a subset in S that sum of it's
On Mar 30, 10:07 pm, Muntasir Azam Khan [EMAIL PROTECTED]
wrote:
- Original Message -
From: Mahdi [EMAIL PROTECTED]
To: Algorithm Geeks algogeeks@googlegroups.com
Sent: Friday, March 30, 2007 1:27 PM
Subject: [algogeeks] Sum of subsets
We have set named S. We assume we have an
Hello everyone!
We have n points in a chart that none of them have the same length(I
mean their 'x' parameter). We want to find a cycle with MINIMUM length
with these assumptions :
We start from the point with minimum length, meeting all of the points
and finally return to the starting point. We
OK. that is the assumption of the problem that this is done in O(n).
Perhaps it is imaginary.
The Question is :
assuming this algorithm and using that, now we want to find the exact
members that their sum equals to K, with the best order.
Thank you.
On Mar 30, 7:38 pm, Muntasir Azam Khan [EMAIL
Hello!
We have a graph that is not directional. We want an algorithm to find
out if this graph is divided to two parts or not.
mofid
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I second that. I believe O(n^2) is a correct call.
Consider a function LONGEST_PALINDROME (left, right), where it accepts the
substring definition based on left and right, and will return the maximum
length of palindrome which can be generated from that substring:
LONGEST_PALINDROME (left,
there are existing algorithms to find connected components which wud do it.
On 3/30/07, Mofid [EMAIL PROTECTED] wrote:
Hello!
We have a graph that is not directional. We want an algorithm to find
out if this graph is divided to two parts or not.
mofid
- Original Message -
From: Mofid [EMAIL PROTECTED]
To: Algorithm Geeks algogeeks@googlegroups.com
Sent: Saturday, March 31, 2007 12:28 AM
Subject: [algogeeks] A graph problem
Hello!
We have a graph that is not directional. We want an algorithm to find
out if this graph is divided
DFS
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For more
Ok, so you have an oracle that gives you the answer for any SUBSET-SUM
(the decision problem) instance in O(n), and you want to use it to
actually construct the solution. It's obvious that you can construct
the solution in O(n^2) (at most n queries to the oracle).
Start with the initial set S.
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