- in general we use polynomial addition for the same.
If the numbers are carrying some additional information as ( particular
base or pattern) another mechanise can be designed
On Tuesday, 26 June 2012 15:40:39 UTC+5:30, ashgoel wrote:
Best Regards
Ashish Goel
Think positive and find fuel
In a stack, you can't access any element directly, except the top one.
On Mon, Jun 18, 2012 at 11:33 AM, Rituraj worstcod...@gmail.com wrote:
My iterative approach
/*code in c*/
#includestdio.h
int main()
{
int stack[]={1,2,3,4,5,6,7,8},top=7;//
int i,j,temp;
for(i=1;i=top;i++)
{
this is not a stack at all, u have just named it as a stack. for it to be a
stack u should access only the top most element at any point of time!!!
On Mon, Jun 18, 2012 at 11:33 AM, Rituraj worstcod...@gmail.com wrote:
My iterative approach
/*code in c*/
#includestdio.h
int main()
{
int
I think there is a problem in this solution.
U r accessing stack elements from 1 to n in the outer loop. It is not
possible. 1st element cannot be accessed without popping first n-1 elements
out.
On Mon, Jun 18, 2012 at 11:33 AM, Rituraj worstcod...@gmail.com wrote:
My iterative approach
1 search should in using KMP algo so that It can be seacrh in O(n) . let
function is int KMP(src,trget, searchDirection )
this kmpSearch funtion should be implemented is such a fashion that is
search in both direction.
3. assume that give 2d array name is array
const int row =1;
const int col
This can be done with a dfs to mark the path and a backtrack to
construct the path or the word itself.
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#include stdafx.h
#include iostream
using namespace std;
const int len = 20;
const int maxCount = 127;
int rle(char* pStr, int length, char* pNew) {
if (!pStr) return -1;
if (length 3) return -1;
int i = 0;
int k = 0;
char p1 = pStr[i++];
char p2 = pStr[i++];
char p3 = pStr[i++];
int pos=0;
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
On Fri, Jun 8, 2012 at 12:54 PM, Ashish Goel ashg...@gmail.com wrote:
#include stdafx.h
#include iostream
using namespace std;
const int len = 20;
const int maxCount = 127;
int rle(char* pStr,
The idea here is that there will be parts of the stream which actually
should not be compressed. For example abcdef as well as aa do not need any
compression. We need to compress only if 3 characters match because for
compressing two chars we will take up 2 chars so no compression benefit (:
So
Will fail for the sing having say 257characters all same
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
On Sat, May 26, 2012 at 12:26 PM, Navin Gupta navin.nit...@gmail.comwrote:
This is called Run-Length-Encoding (RLE) of a string.
Its purpose
u forgot to do inplace and you have wrong conversion of count
On Sat, May 26, 2012 at 11:31 AM, Anchal Gupta anchal92gu...@gmail.comwrote:
hey, here is the function that do the compression and store the output
in an array op.
void str_comp(char *str)
{
int count=0,j=0,i;
char
yeah i forgot inplace so to do that we simply add count and ch in str
input array instead of op.
btw whats wrong with count it give me right answer.
On May 26, 12:08 pm, Hassan Monfared hmonfa...@gmail.com wrote:
u forgot to do inplace and you have wrong conversion of count
On Sat, May 26,
1- try abb
On Sat, May 26, 2012 at 12:07 PM, Anchal Gupta anchal92gu...@gmail.comwrote:
yeah i forgot inplace so to do that we simply add count and ch in str
input array instead of op.
btw whats wrong with count it give me right answer.
On May 26, 12:08 pm,
http://michael.dipperstein.com/rle/index.html
and basic one is
http://www.fileformat.info/mirror/egff/ch09_03.htm
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
On Sat, May 26, 2012 at 1:10 PM, Hassan Monfared hmonfa...@gmail.comwrote:
1- try
This is called Run-Length-Encoding (RLE) of a string.
Its purpose is to save space.So in case of abcdef,I think the output needed
is abcdef (1 is implicit).
The added benefit is it makes the solution in-place.
Approach:- (In-place and Linear Time)
Start from the left of string and
struct node
{
int data;
struct node *link;
};
node* CreateNode(int val)
{
node* root = (node*)malloc(sizeof(struct node));
root-data = val;
root-link = NULL;
return root;
}
node* createList(int *arr, int n)
{
node * root =
@Ashish : seems exactly similar to Lucifer code or you modified something
in his code ?? ...
On Tue, Jan 24, 2012 at 2:02 PM, Ashish Goel ashg...@gmail.com wrote:
struct node
{
int data;
struct node *link;
};
node* CreateNode(int val)
{
node* root =
oh, a possible mistake from my side, ignore my mail please...
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
On Tue, Jan 24, 2012 at 3:08 PM, atul anand atul.87fri...@gmail.com wrote:
@Ashish : seems exactly similar to Lucifer code or you
Steps:
1)Reverse the list ...
2)Now do the swap two nodes... consecutively...
PRAVEEN RAJ
DELHI COLLEGE OF ENGINEERING
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I just wrote a code which would work for any given size K.
I tested it with K = 1 till 7.
[ in the question asked above K=2]
Also, tested for corner cases..
If you guys are interested, then have a look at the code..
I have added few helper functions so that you can directly run the
code and use
@above
attaching the file..
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@rahul sharma, i ran this code, it is producing wrong answer :|
check it, http://codepad.org/THv1hJq1
anyone with correct solution?
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search archives :-/
On Mon, Oct 3, 2011 at 11:47 AM, pranav agrawal
pranav.is.cool.agra...@gmail.com wrote:
@rahul sharma, i ran this code, it is producing wrong answer :|
check it, http://codepad.org/THv1hJq1
anyone with correct solution?
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yeah it is wrong..i have a solution but uses 0(n+m) space.i need it
in 0(n*m) tymand o(1) space
On Mon, Oct 3, 2011 at 11:55 AM, shady sinv...@gmail.com wrote:
search archives :-/
On Mon, Oct 3, 2011 at 11:47 AM, pranav agrawal
pranav.is.cool.agra...@gmail.com wrote:
@rahul
keep two var row0 and col0 for checking if there is any 0 in row0flag
/col0flag
now walk over elements from 1,1 to n,m and set corresponding entry in 0th
row /column if you hit a zero.
now walk over zeroth column and rwo and set the complete row/col if a 0 is
there in 0th row/col.
after this
@ashish can u give an xample.plz...i have read a lot archives ...but
cant find in 0(1) spaceu using 2 var only...plz give xample...nended
urgent.thnx
On Tue, Oct 4, 2011 at 7:26 AM, Ashish Goel ashg...@gmail.com wrote:
keep two var row0 and col0 for checking if there is any 0 in
1 1 0 1
0 1 1 1
1 1 1 1
1 1 1 0
row0 is true
col0 is true
for (int i=1; in;i++)
for (int j=1;jm;j++)
if (a[i][j] == 0) {a[i][0]=0; a[0][j]=0;}
now after this
1 1 0 0
0 1 1 1
1 1 1 1
0 1 1 0
for (int i=1; in;i++)
if (a[i][0] ==0) for (int j=1; jm;j++) a[i][j]=0;
for (int j=1; im;j++)
if
row0 and col0 initilayy true coz we have 0 in 0 row???or these r default
values?
On Tue, Oct 4, 2011 at 8:07 AM, Ashish Goel ashg...@gmail.com wrote:
1 1 0 1
0 1 1 1
1 1 1 1
1 1 1 0
row0 is true
col0 is true
for (int i=1; in;i++)
for (int j=1;jm;j++)
if (a[i][j] == 0) {a[i][0]=0;
0 in 0th row as well as 0 in 0th col and hence true
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
On Tue, Oct 4, 2011 at 8:28 AM, rahul sharma rahul23111...@gmail.comwrote:
row0 and col0 initilayy true coz we have 0 in 0 row???or these r default
so we shoul d aslo add loop at the top to find only for firrst row and
column the initial values
On Tue, Oct 4, 2011 at 8:30 AM, Ashish Goel ashg...@gmail.com wrote:
0 in 0th row as well as 0 in 0th col and hence true
Best Regards
Ashish Goel
Think positive and find fuel in failure
got it..thnx yr
On Tue, Oct 4, 2011 at 8:34 AM, rahul sharma rahul23111...@gmail.comwrote:
so we shoul d aslo add loop at the top to find only for firrst row and
column the initial values
On Tue, Oct 4, 2011 at 8:30 AM, Ashish Goel ashg...@gmail.com wrote:
0 in 0th row as well as 0 in 0th
*@all to median of BST time O(n) space O(1) (modified code of nitin to
get median)
medianBST*(node, n)
int x = 0;
*while* hasleftchild(node) *do*
node = node.left
*do*
x++;
if (x == n/2) return node-val;
*if* (hasrightchild(node)) *then*
node = node.right
If you're given that it's a sparse matrix, then you must assume
storage is in a sparse matrix data structure to get time less than
O(mn).
In fact, if you assume the right data structure, then the operation
can take O(1) time.
For example if you say the structure is an array of sets of indices of
Suppose matrix is
1 0 0 1
1 0 1 0
0 0 0 0
then we traverse the matrix for each 1 we found at a[i][j] , we will check
for i=i to irow and j=j to jcol if that contains any more 1
if it contains 1 in row then we don't make the whole row as 1..we ignore the
row and same will be for column
And you have to use the pointer-reversing trick to traverse the tree
because you don't have space for a stack.
On Sep 27, 4:52 am, anshu mishra anshumishra6...@gmail.com wrote:
do the inorder traversal as soon as reached at n/2th element that will be
median.
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This requires O(n) extra space.
On Sep 27, 7:43 am, anshu mishra anshumishra6...@gmail.com wrote:
int bstMedian(node *root, int n, int x)
{
if (node-left) return bstMedian(root-left, n, x);
x++;
if (x == n/2) return node-val;
if (node-right) return bstMedian(root-right, n,
its not o(n) it is O(max height of tree) :P
i have not seen the constraint.
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a simple one is rabit-tortoise method, and using stackless traversal,
facing a lot of corner cases in coding this, can someone check this as
well?
On Sep 27, 6:41 pm, anshu mishra anshumishra6...@gmail.com wrote:
its not o(n) it is O(max height of tree) :P
i have not seen the constraint.
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@anshu
can middle element can be found if the no. of nodes are not given...
On Tue, Sep 27, 2011 at 8:34 PM, vikas vikas.rastogi2...@gmail.com wrote:
a simple one is rabit-tortoise method, and using stackless traversal,
facing a lot of corner cases in coding this, can someone check this as
Recursion also requires space, so the problem is how to traverse without
extra space.
Once this is done, nothing is left in the problem.
Sanju
:)
On Tue, Sep 27, 2011 at 8:35 AM, Dheeraj Sharma dheerajsharma1...@gmail.com
wrote:
@anshu
can middle element can be found if the no. of nodes
Do inorder traversal, to find out the total no. of nodes.
Next time, do the inorder traversal but keeping the count of nodes visited
and stop when you visit n/2 nodes.
Non recursive In-order Traversal -
*inorder*(node)
*while* hasleftchild(node) *do*
node = node.left
*do*
Since we are given pointer to root node, we can easily find the minimum
element in the tree.
This will be the first node in the inorder traversal, now use method to find
the inorder successor of a each node. Do it iteratively.
Complexity will be O(n log n) and O(n) if tree is skewed.
Correct me
morris Inorder traversal can do the task i think
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I assume we don't want to use extra storage.
So one way is this: Go over the matrix and mark the first row with a 1
and the first column with a 1 for each 1 you find. Because row and
column 1 are used for temporary storage in this manner, you must first
remember whether they contained a 1, then
Guys an Update ,
This has been asked in MS by me.. I suggested O(m*n) but they were looking
for a solution in nlogn ( n*n Sparse Matrix ) ..Any idea ...
This post was discussed earlier but every1 came with O(m*n) solution so
instead of cluttering it ..opened a new One ...
On Tue, Sep 27, 2011
yep, trie needs to be built
On Aug 24, 10:49 pm, Ankur Garg ankurga...@gmail.com wrote:
It means when u call that func u get the next word in the document
Regards
Ankur
On Wed, Aug 24, 2011 at 6:59 PM, vikas vikas.rastogi2...@gmail.com wrote:
what do you mean by a function for
wat abt doing wid hashing?
On Thu, Aug 25, 2011 at 3:55 PM, vikas vikas.rastogi2...@gmail.com wrote:
yep, trie needs to be built
On Aug 24, 10:49 pm, Ankur Garg ankurga...@gmail.com wrote:
It means when u call that func u get the next word in the document
Regards
Ankur
ya why not hashing ?
On Thu, Aug 25, 2011 at 3:31 PM, SANDEEP CHUGH sandeep.aa...@gmail.comwrote:
wat abt doing wid hashing?
On Thu, Aug 25, 2011 at 3:55 PM, vikas vikas.rastogi2...@gmail.comwrote:
yep, trie needs to be built
On Aug 24, 10:49 pm, Ankur Garg ankurga...@gmail.com wrote:
how will we exactly implement hashtables in this? What will be
appropriate keys? ??
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for every word in the document , apply hash funtion.. store the string n
its frequency too..
if we get the same word , then increment the frequency.
after storing.
whenver we want to search the word , searching in O(1) time , jst applying
hash function again on the word to be searched ..
But it says finding the next word and also it's position of
occurence. If we use frequency..won't position of occurence overlap?
Am not sure whether I got the question right or not!
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no ..
question says that there is a function that gives the next word in the
document..
this means after hashing one word , we hav to go to another word ... for
this going to next word in the document , we hav already provided wid a
function..
nothing to do wid the occurence
On Thu, Aug 25,
design a code which efficiently search
the word and find occurrence of it in given document
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ohh sry my mistake .. got it
On Thu, Aug 25, 2011 at 7:36 PM, Shrey Choudhary choudharyshre...@gmail.com
wrote:
design a code which efficiently search
the word and find occurrence of it in given document
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@shrey but its not given in question that we have to return the position
or the occurence in the document.. we hav to only the tell whether the word
is in document..
read
.design a code which efficiently search the word and find occurrence of it
in given document .
its nt given that return
i made a boo boo
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what do you mean by a function for finding the next word is given ?
On Aug 22, 1:56 am, Ankur Garg ankurga...@gmail.com wrote:
Question-- Given a document containing some words ...and a function
for finding the next word is given .design a code which efficiently
search
the word
It means when u call that func u get the next word in the document
Regards
Ankur
On Wed, Aug 24, 2011 at 6:59 PM, vikas vikas.rastogi2...@gmail.com wrote:
what do you mean by a function for finding the next word is given ?
On Aug 22, 1:56 am, Ankur Garg ankurga...@gmail.com wrote:
Anyone??
Regards,
Priyanshu Gupta
On Fri, Aug 5, 2011 at 6:09 PM, priyanshu priyanshuro...@gmail.com wrote:
Give an efficient algorithm to determine which part of the video
should be displayed as a thumbnail??
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I guess , it can be done using indexing , with time stamp as key , and frame
pointer as data ..
Please correct me if I am wrong.
On Tue, Aug 9, 2011 at 11:03 PM, Priyanshu priyanshuro...@gmail.com wrote:
Anyone??
Regards,
Priyanshu Gupta
On Fri, Aug 5, 2011 at 6:09 PM, priyanshu
i guess ur qn was how will u decide which frame shud b ur thumbnail...as
after tht, the frame cud be set as a thumbnail which is pretty much an eay
task in windows programming.i think most, dun hav much info abt it though,
thumbnails shud contain more of data; i mean shudnt be unicolour(blank
@gopi.. didnt got you...
@adi.. ya thats what i am talking about...
Regards,
Priyanshu Gupta
On Tue, Aug 9, 2011 at 11:18 AM, Aditya Virmani virmanisadi...@gmail.comwrote:
i guess ur qn was how will u decide which frame shud b ur thumbnail...as
after tht, the frame cud be set as a
Given : ddbbccae
O/P : 2d4a2b2c1a1e
What is happening? What algo?
Thanks and regards
Monish
On Aug 9, 5:59 pm, ankit sambyal ankitsamb...@gmail.com wrote:
Given an array of characters, change the array to something as shown in the
following example.
Given : ddbbccae
O/P :
Below is a solution -
#include stdio.h
int main(int argc, char* argv[]){
int i = 0;
char *array = argv[1];
char prev = array[0];
while(array[i]){
int count = 0;
while(prev == array[i]){
i +=1;
what has palindrome to do with this?
On Mon, Jul 18, 2011 at 1:11 AM, swetha rahul swetharahu...@gmail.comwrote:
Thanks everyone!!
On Sun, Jul 17, 2011 at 10:56 PM, ankit sambyal ankitsamb...@gmail.comwrote:
Check for case senstivity also:
eg: Madam and madam both are palindromes.
Test cases for MSN Search Engine
1 : Check for auto completion feature (The auto completion based on the
prefix should provide suggestions for the most searched keyword with that
prefix)
.
2 : The spelling correction feature which shows as to what the user
actually meant. This should be highly
1.If string is NULL then it should return 1 i.e. string is palindrome
2.If there is only one character in string then it is palindrome
3.If reverse of given string is same as string
On Jul 17, 8:18 pm, swetha rahul swetharahu...@gmail.com wrote:
ya got it..thanks...
how abt test cases for
is this ans sufficient..? any other possible test cases for palindrome ?
On Sun, Jul 17, 2011 at 8:53 PM, Nishant mittal.nishan...@gmail.com wrote:
1.If string is NULL then it should return 1 i.e. string is palindrome
2.If there is only one character in string then it is palindrome
3.If
Well, for the third case, you can elaborate on the implementation. We
need to compare the first half with the later half. So, in case of
even no. of characters it splits perfectly and in case of odd number
of characters, just ignore the middle character.
On Jul 17, 8:28 pm, swetha rahul
check for query injection,string like (spaces,nothing) entered.
On Sun, Jul 17, 2011 at 9:04 PM, Dumanshu duman...@gmail.com wrote:
Well, for the third case, you can elaborate on the implementation. We
need to compare the first half with the later half. So, in case of
even no. of characters
Check for case senstivity also:
eg: Madam and madam both are palindromes.
Also for clarify with the interviewer whether whitespace and
punctuation can be ignored.
eg: is the following string a palindrome or not:
A man, a plan, a canal, Panama
And check for this condition accordingly
--
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Thanks everyone!!
On Sun, Jul 17, 2011 at 10:56 PM, ankit sambyal ankitsamb...@gmail.comwrote:
Check for case senstivity also:
eg: Madam and madam both are palindromes.
Also for clarify with the interviewer whether whitespace and
punctuation can be ignored.
eg: is the following string a
@Dave: Think of it again counting sort won't work, If you are just
considering the last character as
the even though key is just last character, the value is the whole string..
On Tue, Jun 28, 2011 at 1:53 PM, juver++ avpostni...@gmail.com wrote:
List[letter] - linked list of all words with
@Rizwan: I don't see the problem. Initialize an array of counters, one
for each possible last character, to zero. If there are no
restrictions on the last character, use an array of 256 counters.
Then, for each string, increment the counter corresponding to the last
character of the string by the
@Dave:can u please explain the second method.
On Mon, Jun 27, 2011 at 11:24 PM, Dave dave_and_da...@juno.com wrote:
@Nishant: Here are a couple of O(n) alternatives, given that there are
a limited number of last characters and no requirement to break ties
in any particular way:
1. A
List[letter] - linked list of all words with the last character as letter.
Then iterate over all letters and concatenate lists.
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Do we really need to reverse? Can't we apply any normal sorting algo
like merge sort or quick sort and for the comparison part of it, use
the last characters of each string only?
On Jun 27, 4:21 am, varun pahwa varunpahwa2...@gmail.com wrote:
reverse all strings and then sort.
On Mon, Jun 27,
@Nishant: Here are a couple of O(n) alternatives, given that there are
a limited number of last characters and no requirement to break ties
in any particular way:
1. A counting sort.
2. Form a linked list of entries corresponding to each last character,
and then merge these lists in collating
I don't think you need a temp string (though, this may be a language
dependent issue). Just use the string as given and keep track of
the character positions within the string. The length of the original
string is irrelevant as long as your system can handle it.
Dan :-)
On Jun 23, 2:44 pm,
Good one! That halved the memory requirement. :)
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@Sunny: what if the number is repeated multiple times? u seem to
consider only the duplicates(2 times).
On Jun 13, 7:36 pm, sunny agrawal sunny816.i...@gmail.com wrote:
no we can take care of duplicates without any extra memory
modify 2nd step of my previous solution as follows
if T[a[i]] is
i meant if N = { 1, 1, 1, 2, 12}
and M = { 1, 1, 3, 12}
then answer should be = {1, 1, 12}
On Mon, Jun 13, 2011 at 8:06 PM, sunny agrawal sunny816.i...@gmail.comwrote:
no we can take care of duplicates without any extra memory
modify 2nd step of my previous solution as follows
if T[a[i]] is
Nice Confusion... :)
Consider the following case
A[M] = {1,1,3,12};
B[N] = {1,2,12}
here again i think answer should be {1,1,12} , why are u binding one
occurrence of 1 in array A with one in B. Question is which elements of
first array is present in second array. so for this case A[0], A[1],
Use a hash table of size 1025 with bits per table entry = 2.
1. Go through the N sized array and set bit 0 of the hash table entry to 1
if it is present in the first array.
2. Go through the M sized array and set bit 1 of the hash table entry to 1
if the element belongs to 0 to 1024.
3. Go
why do we need 2 bits at all ??
i think single bit per table entry will do.
say table is T[1025], and array is A[M]
1. Go through the N sized array and set bit 0 of the hash table entry to 1
if it is present in the first array.
2. Go through the M sized array and if T[a[i]] is set then this
we can take care of the duplicate entries, but then that would cost more
space(int), as of now we are working with bool
On Mon, Jun 13, 2011 at 5:51 PM, sunny agrawal sunny816.i...@gmail.comwrote:
that will report duplicate entries multiple times :(
On Mon, Jun 13, 2011 at 5:38 PM, sunny
no we can take care of duplicates without any extra memory
modify 2nd step of my previous solution as follows
if T[a[i]] is set then this element is there in second array so report this
element and Reset T[a[i]].
now no duplicates will be reported. and only 1025 bits will be required.
any
You can really impress 'em by pointing out that this problem is
related to some common ways of representing numbers and arithmetic
operations in the lambda calculus.
On Aug 24, 7:16 pm, Ashish Goel ashg...@gmail.com wrote:
inline int decrease (int v) {
int decreasedVal = 0;
int counter = 0;
Given a file with a lot of words (10 million) find out the top 10%
most
frequently occuring words.
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*TRIE*: Using a trie we would get a linear time solution but i think the
memory requirement would be huge ..
On Mon, Dec 14, 2009 at 11:02 PM, vicky mehta...@gmail.com wrote:
Given a file with a lot of words (10 million) find out the top 10%
most
frequently occuring words.
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You
can u make use of binary index trees
On Tue, Dec 15, 2009 at 4:23 PM, abhijith reddy abhijith200...@gmail.comwrote:
*TRIE*: Using a trie we would get a linear time solution but i think the
memory requirement would be huge ..
On Mon, Dec 14, 2009 at 11:02 PM, vicky mehta...@gmail.com
On Mon, Dec 14, 2009 at 11:02 PM, vicky mehta...@gmail.com wrote:
Given a file with a lot of words (10 million) find out the top 10%
most
frequently occuring words.
http://www.cc.gatech.edu/classes/semantics/misc/pp2.pdf
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