I can only think of one way off the top of my head:
class IndexedDocument {
HashTableString, int index;
String contents;
/// Build an index of words and their position in the document
void buildIndex() {
for each word in document {
add word, position to index
use segment tree
http://en.wikipedia.org/wiki/Segment_tree
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use suffix tree it's much faster than simple trie
with ukkonen's method you can build it in O(size of document) and then
searching in it is practically O(1)
http://en.wikipedia.org/wiki/Suffix_tree
http://www.cs.helsinki.fi/u/ukkonen/SuffixT1withFigs.pdf
On Fri, Dec 10, 2010 at 7:54 PM, ADITYA
I think the answer is (2^m)^(2^n) (or 2^(m*2^n))
For m=1, the answer is 2^2^n.
We can express the function using a truth table with 2^n entries, one
entry for each possible input set.
And each entry has (n+m) fields, represent the n inputs and m outputs.
The m*2^n output fields can be filled
each value in A and B is 0 or 1 so the sum of all elements in A (or B) is n
so the sum of all elements in C which is the sum of differences between
values in A and B is between -n and n
now we want to maximize j-i for which C[i]+C[i+1]+...+C[j] = 0
suppose that sum(i) = C[1]+C[2]+...+C[i] which is
No. You made the same mistake as I.
Try this case: {1, 2, 2, 5, 5}.
Actually, this case defeats the solution of Manmeet's, yours, and mine.
(same min/max, same sum, same xor result)
I think the key point is that the N variable cannot be determined by 1
or 2 equation constraint.
On 2010-12-10
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@amir nice solution
On Sun, Dec 12, 2010 at 11:40 AM, Amir hossein Shahriari
amir.hossein.shahri...@gmail.com wrote:
each value in A and B is 0 or 1 so the sum of all elements in A (or B) is n
so the sum of all elements in C which is the sum of differences between
values in A and B is
visit this blog
http://code-forum.blogspot.com/
and http://coders-stop.blogspot.com/
for programming puzzles.
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Given that you have one string of length N and M small strings of
length L . How do you efficiently find the occurrence of each small
string in the larger one
I heard the best solution for this problem is to use generalized
suffix tree. Can any one explain the solution for this problem in
detail
KMP and RabinKarp
On Dec 12, 7:07 pm, Prims topcode...@gmail.com wrote:
Given that you have one string of length N and M small strings of
length L . How do you efficiently find the occurrence of each small
string in the larger one
I heard the best solution for this problem is to use
@Dave :but it will take more time. any other solution O(n^2).?
On Sat, Dec 11, 2010 at 7:48 PM, Dave dave_and_da...@juno.com wrote:
@Naresh: The sequence of numbers generated by this rule for any given
starting number is called a Collatz Sequence. Try googling it.
Here is a list of the
ans for 1st one is 24
On Sat, Nov 27, 2010 at 3:12 PM, vamsi achyuth vamsiachy...@gmail.comwrote:
these were the problems asked in tcs written exam.
On 27 November 2010 15:08, srinivas reddy srinivaseev...@gmail.comwrote:
@youngboy
you have to choose one value for variable only
It is NOT possible to multiply two matrices in less than O(n^2) simply
because there are n^2 elements, and you got to touch all of them at
least once!
Rakib
On 12/10/10, Luciano Junior luciano@gmail.com wrote:
Dave, thank You very much for yours information. Really, I want to
know a
@awesomeandroid
how you will use KMP in ths case, can you plz explain ?
Mohit
On Sun, Dec 12, 2010 at 8:01 PM, awesomeandroid
priyaranjan@gmail.comwrote:
KMP and RabinKarp
On Dec 12, 7:07 pm, Prims topcode...@gmail.com wrote:
Given that you have one string of length N and
i am facing problem in file handling in C
can any one suggest me how to implement them.
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