Re: [algogeeks] Re: searching all matching words in a Trie with a given filter.
@Rahul, The problem is to get all the words matching to a given filter
like *...n..p *from already built trie.
@Don, I'm storing words at the ending node, now I'm trying to modify this
function so that it returns only one random word from all words matching
that filter with equal probability for each word, I'm trying like this
char* randomWord(trie *root, char *k)
{
int m;
if(!root || flag) return;
if(*k == '\0') {
if(root-word) {
strcpy(mWord,root-word); // storing the random word
in mWord.
flag = 1;
}
return;
}
else if(*k == '.') {
for(m=0;m26;m++) {
if(root-children[mark[m]]){//
where mark[26] is an array containing numbers [0,1,2,.,25] on random
indices.
randomWord(root-children[mark[m]],k+1);
}
}
randomize(); //
randomize function shuffles mark[].
}
else
randomWord(root-children[*k-'a'],k+1);
}
But this is not giving words with equal probability. It is returning words
more frequently which are having consecutive same letters.
On Thu, May 30, 2013 at 2:34 AM, rahul sharma rahul23111...@gmail.comwrote:
wat exaclty the question is.
We have to make a tire with filter or we have a trie(whole dictionary) and
we have to check filter out the elements.
plz explain question
On Wed, May 29, 2013 at 7:55 PM, Don dondod...@gmail.com wrote:
There has to be some way to know that a node is the end of a word, and
to know what that word is. This might be done by using a parent
pointer which lets you traverse the trie back to the root, rebuilding
the word, or you could keep track of the word as you traverse down the
trie. Putting the whole word in the node where the word ends would be
the most simple and time-efficient approach, if you have the memory to
support it.
Here is a different way to do it, if the trie has a wordEnd flag and
does not store the word in the node.
void findWords(trie *root, char *filter, String word=)
{
if (!root) return;
if (*filter == 0) // When you reach the end of the filter at the
end of a valid word, add the word.
{
if (root-wordEnd) words.add(word);
}
else if (*filter == '.') // Search for words with any letter
{
for(int i = 'a'; i = 'z' ; ++i)
findWords(root-link[i], filter+1, word+i);
}
else // Search for words with the required letter
{
findWords(root-link[*filter], filter+1, word+*filter);
}
}
On May 28, 11:36 pm, avinesh saini avinesh.sa...@gmail.com wrote:
Thank you Don, I was also trying in similar way. But here I'm confused
how
you are storing the traversed words. Are you adding whole words at the
node
on which word is ending during insertion.
On Wed, May 29, 2013 at 12:36 AM, Don dondod...@gmail.com wrote:
void findWords(trie *root, char *filter)
{
if (!root) return;
if (*filter == 0) // When you reach the end of the filter at the
end of a valid word, add the word.
{
if (root-words) words.add(root-word);
}
else if (*filter == '.') // Search for words with any letter
{
for(int i = 'a'; i = 'z' ; ++i)
findWords(root-link[i], filter+1);
}
else // Search for words with the required letter
{
findWords(root-link[*filter], filter+1);
}
}
On May 28, 4:47 am, avinesh saini avinesh.sa...@gmail.com wrote:
How to search all the matching words for a filter in a trie.
e.g.
searching by filter ...r..m will find all the words(of length =
7) in
trie in which 4th character is 'r' and 7th character is 'm'.
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Re: [algogeeks] Re: Array Problem
I was going through this problem on stackoverflow, and I found this classic article on this very topic http://www.americanscientist.org/issues/pub/2002/3/the-easiest-hard-problem Definitely, worth a read. -- * * *thanks regards,* *Avinesh Kumar National Institute of Technology, Calicut.* *Kerala- 673601* *+91 7849080702* -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To unsubscribe from this group and stop receiving emails from it, send an email to algogeeks+unsubscr...@googlegroups.com.
Re: [algogeeks] searching all matching words in a Trie with a given filter.
Don, I'm trying to get all the words from trie iteratively, because I'm
creating trie of whole dictionary (more than 200k words) and searching
recursively will consume a lot of stack space.
Thanks for your help!
On Wed, May 29, 2013 at 9:44 AM, rahul sharma rahul23111...@gmail.comwrote:
@don u r searching in a previously built trie with the given filter...then
wat is this add fxn doing?correct me if m getting u wrng
On Wednesday, May 29, 2013, avinesh saini avinesh.sa...@gmail.com wrote:
Thank you Don, I was also trying in similar way. But here I'm confused
how you are storing the traversed words. Are you adding whole words at the
node on which word is ending during insertion.
On Wed, May 29, 2013 at 12:36 AM, Don dondod...@gmail.com wrote:
void findWords(trie *root, char *filter)
{
if (!root) return;
if (*filter == 0) // When you reach the end of the filter at the
end of a valid word, add the word.
{
if (root-words) words.add(root-word);
}
else if (*filter == '.') // Search for words with any letter
{
for(int i = 'a'; i = 'z' ; ++i)
findWords(root-link[i], filter+1);
}
else // Search for words with the required letter
{
findWords(root-link[*filter], filter+1);
}
}
On May 28, 4:47 am, avinesh saini avinesh.sa...@gmail.com wrote:
How to search all the matching words for a filter in a trie.
e.g.
searching by filter ...r..m will find all the words(of length = 7)
in
trie in which 4th character is 'r' and 7th character is 'm'.
--
*
*
*thanks regards,*
*Avinesh
*
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thanks regards,
Avinesh
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*Kerala- 673601*
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[algogeeks] searching all matching words in a Trie with a given filter.
How to search all the matching words for a filter in a trie. e.g. searching by filter ...r..m will find all the words(of length = 7) in trie in which 4th character is 'r' and 7th character is 'm'. -- * * *thanks regards,* *Avinesh * -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To unsubscribe from this group and stop receiving emails from it, send an email to algogeeks+unsubscr...@googlegroups.com.
Re: [algogeeks] Re: searching all matching words in a Trie with a given filter.
Thank you Don, I was also trying in similar way. But here I'm confused how
you are storing the traversed words. Are you adding whole words at the node
on which word is ending during insertion.
On Wed, May 29, 2013 at 12:36 AM, Don dondod...@gmail.com wrote:
void findWords(trie *root, char *filter)
{
if (!root) return;
if (*filter == 0) // When you reach the end of the filter at the
end of a valid word, add the word.
{
if (root-words) words.add(root-word);
}
else if (*filter == '.') // Search for words with any letter
{
for(int i = 'a'; i = 'z' ; ++i)
findWords(root-link[i], filter+1);
}
else // Search for words with the required letter
{
findWords(root-link[*filter], filter+1);
}
}
On May 28, 4:47 am, avinesh saini avinesh.sa...@gmail.com wrote:
How to search all the matching words for a filter in a trie.
e.g.
searching by filter ...r..m will find all the words(of length = 7) in
trie in which 4th character is 'r' and 7th character is 'm'.
--
*
*
*thanks regards,*
*Avinesh
*
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*
*thanks regards,*
*Avinesh
*
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Re: [algogeeks] least common ancestore bst
If one node is parent of other then Parent Node is lowest common ancestor.
Source- http://en.wikipedia.org/wiki/Lowest_common_ancestor (just read it)
On Mon, May 13, 2013 at 1:42 AM, rahul sharma rahul23111...@gmail.comwrote:
[image: BST_LCA]
what should be ancestor of 12 and 14.it should be 12 or
14...if 12 then for 20 and 22 also it is 20...if 12 ans 14 has
ancestor as 8 then for 20 and 22 it is NULL.
@allplz give me clear one line definition in case if one node is
parent of other then what is LCA
On Sun, Apr 21, 2013 at 10:32 PM, Tushar Patil tushar01pa...@gmail.comwrote:
@rahul : It's fine solution, but can we check the root-data == n1 || n2
before calling function recursively, I think if we check this condition 1st
it will reduce unnecessary function calls.
Correct me if i am wrong?
Thanks,
Tushar Patil.
On Sun, Apr 21, 2013 at 10:26 PM, rahul sharma
rahul23111...@gmail.comwrote:
int leastCommanAncestor(struct node* root, int n1, int n2)
{
if(root==NULL)
return -1;
if(root-datan1 root-datan2)
return leastCommanAncestor(root-left,n1,n2);
else if(root-datan1 root-datan2)
return leastCommanAncestor(root-right,n1,n2);
return root-data;
}
Does this code miss any case?N suppose if we have to find LCA of n1 and
n2 and suppose n1 is parent of n2 ..then this will return n1..is this fyn
or if n1 is parent of n2 then we should return -1??
Plz. comment
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*Avinesh Kumar
National Institute of Technology, Calicut.*
*Kerala- 673601*
*+91 7849080702*
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[algogeeks] No of nodes in B-Tree
*Suppose that we insert the keys (**1,2,...,n) **into an empty B-tree with minimum **degree **2. How many nodes does the final B-tree have?* explain your answer.! -- *Avinesh Kumar, * -- 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.
[algogeeks] Hello World
Explain this code-
It is supposed to print- Hello World
#include iostream
using namespace std;
int main()
{
long long int
l1l[]={72,-11037827,917043223,-47519989,1450408591,-194718605,2037206149,-8912843,279667,-26713,-3617,1571,-79};
longlongintll1[]={1,9240,277200,13440,725760,290304,14515200,483840,193536,483840,14515200,15966720,31933440};
long double lll=1;
long long int l1='\0'+1, ll='\0'+1;
for (int i='\0'; i'\r'; i=i+1)
{
for (int j='\r'-1; j='\0'; j=j-1)
{
for (int k='\0'; kj; k=k+1)
{
ll=l1=l1*(!!i);
for (int l='\0'; li-1; l=l+1)
{
l1=l1+ll;
}
}
lll+=(long double)l1/(*(ll1+j))*(*(l1l+j));
l1='\0'+1;
}
std::cout(char)(lll+0.5);
lll=!lll;
}
std::coutstd::endl;
return 0;
}
--
*Avinesh Kumar,
MCA
NIT Calicut.
*
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Re: [algogeeks] Reverse the string word by word . Can someone tell what is wrong in my code ?
Try this one...
#includestdio.h
#includestring.h
void reverse(char *p,char*q)
{
char c;
while(pq)
{
c=*p;*p=*q;*q=c;
p++;
q--;
}
}
int main()
{
char A[50];
printf(\n Enter a String:\n\n);
gets(A);
int len=strlen(A);
reverse(A,A[len-1]);
printf(%s\n,A);
return 0;
}
On Sun, Sep 25, 2011 at 10:51 AM, Dheeraj Sharma
dheerajsharma1...@gmail.com wrote:
u shud do TWO things in..your reverseword function..
first is str[i]=='\0' and not str[i]='\0'
second is while(i=len) and not while(ilen)
On Sun, Sep 25, 2011 at 6:49 AM, Deoki Nandan deok...@gmail.com wrote:
//Reverse String word by word
// if string is :- I am a good boy
//output string should be :- boy good a am I
#includestdio.h
#includestring.h
void reverse(char *p,char*q)
{
int i;char c;
while(pq)
{
c=*p;*p=*q;*q=c;
p++;
q--;
}
}
void reverseWordByWord(char str[],int len)
{
int i=0,j=0;
while(ilen)
{
if((str[i]==' ')||(str[i]=='\t')||(str[i]='\0'))
{
reverse(str[j],str[i-1]);
j=i+1;
}
i++;
}
}
int main()
{
char A[]=Ram is a good person;
int i;
int len=strlen(A);
reverse(A[0],A[len-1]);
printf(%s\n,A);
reverseWordByWord(A,len);
printf(%s\n,A);
return 0;
}
--
**With Regards
Deoki Nandan Vishwakarma
*
*
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