@jaspreet : i dont find much difference between using BFS or
backtracking...which is doing similar to DFS.
On Tue, Oct 16, 2012 at 4:28 AM, Jaspreet Singh
wrote:
> BFS
>
>
> On Sun, Oct 14, 2012 at 4:21 PM, Rahul Kumar Patle <
> patlerahulku...@gmail.com> wrote:
>
>> response awaited!!!
>> an
its not the best i think and also not a dp solution but can be done by this.
On Tue, Oct 16, 2012 at 4:28 AM, Jaspreet Singh
wrote:
> BFS
>
>
> On Sun, Oct 14, 2012 at 4:21 PM, Rahul Kumar Patle <
> patlerahulku...@gmail.com> wrote:
>
>> response awaited!!!
>> anyone??
>>
>> On Sat, Oct 13, 2012
BFS
On Sun, Oct 14, 2012 at 4:21 PM, Rahul Kumar Patle <
patlerahulku...@gmail.com> wrote:
> response awaited!!!
> anyone??
>
> On Sat, Oct 13, 2012 at 12:31 AM, Rahul Kumar Patle <
> patlerahulku...@gmail.com> wrote:
>
>> Pls help to solve this que.. does any one have DP solution for following
>
void findPaths(int x, int y, int depth)
{
if (isEnd(x,y)) showSolution(); // One path will be marked by
letters A,B,C...
else
{
maze[x][y] = 'A' + depth;
if (x && (maze[x-1][y]=='1')) findPaths(x-1,y,depth+1);
if (y && (maze[x][y-1]=='1')) findPaths(x,y-1,depth+
response awaited!!!
anyone??
On Sat, Oct 13, 2012 at 12:31 AM, Rahul Kumar Patle <
patlerahulku...@gmail.com> wrote:
> Pls help to solve this que.. does any one have DP solution for following
> que.
>
> http://www.geeksforgeeks.org/archives/24488
> section 5/question 2
>
> Write a program to find
@Atul
thanx for the code its working for the example you took...Please check
the same for i/p "abcmno","abcmnop"
Algo displays:- mno
It should display mno
mnop...
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@All *Here is the working code: test on input {-1,5,3,-8,4,-6,9} and
{1,-1,2}*
Algo:
increment current till first +ve number(p) and decerement end till last
-ve number(n)
now consider only array between [p..n]
If current is negetive, increment current
If current is positive, swap it with end and
@Navin: If I am correctly executing your algorithm on the data in the
original posting, {-1,5,3,-8,4,-6,9}, I get {-1,-6,-8,4,3,5,9}, when the
correct answer is {-1,-8,-6,5,3,4,9}. The array contains the correct
numbers, but the order of the positive numbers and the order of the negtive
numbers
@Dave :- a minor change
Initially, decrement the end pointer till it points to positive number,
Now end points to the last negative number.
Now,
If current is negative , increment current
If current is positive , swap it with the element at end and decrement
current and end both
If current >=
@Navin: Try this with {1,-1,2}. current points to the 1 and end points to
the 2. Since 1 is positive, the algorithm swaps the 1 and the 2, giving
{2,-1,1}. Then it decrements current to point outside the array and end to
point to the -1. How can this be right?
Dave
On Thursday, June 28, 2012
+1 naveen
On Thursday, June 28, 2012 8:29:26 PM UTC+5:30, Navin Gupta wrote:
>
> Keep two pointers - one at start of the array and other at end of the
> array
> Now current points to start of the array
> If current is negative , increment current
> If current is positive , swap it with the elem
Keep two pointers - one at start of the array and other at end of the array
Now current points to start of the array
If current is negative , increment current
If current is positive , swap it with the element at end and decrement
current and end both
If current >= end , then break.
Navin Kuma
keep swaping left most -ve and left most positive untill counter reaches at
the end of array, can be done in o(n) no extra space required..
3rd year
manit bhopal
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@wgp the ques is to maintain the order intact..
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i mean o(n) in single traversal .
On Fri, Jun 22, 2012 at 12:53 AM, sanjay pandey
wrote:
> single traversal n O(n) are 2 diff things...plz specify???
>
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Well they are the same you're going over an array once. As long as they are
not nested they are still counted as O(n) because leading constants are
dropped, at least that's what my acumen says. Need inputs on this guys!
On Friday, June 22, 2012 12:53:02 AM UTC+5:30, suzi wrote:
>
> single traver
single traversal n O(n) are 2 diff things...plz specify???
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can this be done in single pass in O(n) .
On Thu, Jun 21, 2012 at 8:10 PM, rusty wrote:
> guys this is my solution to the problem, it's a bit sloppy but as far as I
> checked it was working please have a go at it?
>
>
> #include
> #include
>
> int main() {
> int arr1[] = {0,-5,3,0,4,-6
guys this is my solution to the problem, it's a bit sloppy but as far as I
checked it was working please have a go at it?
#include
#include
int main() {
int arr1[] = {0,-5,3,0,4,-6,-9};
int arr2[7];
int j = 0;
for ( int i = 0 ; i < 7 ; i++ ) {
//loop for -ve numbers
@akshat ur code doesn't give intact output, so i modified ur code and
here is the code :
int j=0,k=0;
for (i = 0; i < n; ++i)
{
if(a[i]<0)
{
a[j] = a[i];
j++;
}
else
{
temp[k] = a[i];
k++;
}
}
k=0;
for (i = j; i < n; ++i)
{
@ayush goel couldnt really understand your algo , can you please explain
little bit more .
On Wednesday, 13 June 2012 21:49:49 UTC+5:30, Krishna Kishore wrote:
>
> Given a array of integers both positive and negative and you need to shift
> positive numbers to one side
> and negative numbers to
Root of a graph can be any node whose in-degree is zero. i.e. there are no
nodes pointing to that node.
It can be found by using O( |V| ) space and O( |E| ) time .
Now we can choose any node whose in-degree is zero if present.
or choose any random node
and itf DFS-tree is the desired tree.
Ti
@Partha
try with:
A = {2, 2, 9}
B= {1, 6, 6}
On Mon, May 21, 2012 at 7:08 PM, partha sarathi Mohanty <
partha.mohanty2...@gmail.com> wrote:
> a[] = [-1,-3,4,0,7,0,36,2,-3]
> b[] = [0,0,6,2,-1,9,28,1,6]
> b1[] = [0,7,0,36,4,-6,3,0,0]
> b2[] =[-1,-3,11,0,0,0,35,0,0]
>
> suma = 42 proda = -
a[] = [-1,-3,4,0,7,0,36,2,-3]
b[] = [0,0,6,2,-1,9,28,1,6]
b1[] = [0,7,0,36,4,-6,3,0,0]
b2[] =[-1,-3,11,0,0,0,35,0,0]
suma = 42 proda = -84*72*3
sumb = 51 prodb = -84*72*3
sumb1 = 44 prodb1 = -84*72*3
sumb2 = 42 prodb2 = 33*35
do the sum and prod operation w/o 0s and compare the values..
@ashish.. it wont be constant space then.. surely it will be o(n) though
On Mon, May 21, 2012 at 7:23 PM, Ashish Goel wrote:
> Dave,
>
> Cant we have a hash table with the item as key and its count as value
> (walk over array A and build HT).
> For permutation check, walk over second array and s
constant space vs no additional space and then O(n) time complexity not
possible..
Best Regards
Ashish Goel
"Think positive and find fuel in failure"
+919985813081
+919966006652
On Mon, May 21, 2012 at 8:01 PM, Dave wrote:
> @Ashish: Using a hash table violates the O(1) space requirement given
@Ashish: Using a hash table violates the O(1) space requirement given in
the original problem.
Dave
On Monday, May 21, 2012 8:53:44 AM UTC-5, ashgoel wrote:
> Dave,
>
> Cant we have a hash table with the item as key and its count as value
> (walk over array A and build HT).
> For permutation
in space
On May 21, 6:53 pm, Ashish Goel wrote:
> Dave,
>
> Cant we have a hash table with the item as key and its count as value (walk
> over array A and build HT).
> For permutation check, walk over second array and start reducing the count
> and remove when count becomes zero for that particul
^not an O(n)
On May 21, 6:53 pm, Ashish Goel wrote:
> Dave,
>
> Cant we have a hash table with the item as key and its count as value (walk
> over array A and build HT).
> For permutation check, walk over second array and start reducing the count
> and remove when count becomes zero for that part
Dave,
Cant we have a hash table with the item as key and its count as value (walk
over array A and build HT).
For permutation check, walk over second array and start reducing the count
and remove when count becomes zero for that particular key. If some char
not there in HT, return false, else retu
No way u can do it in O(1) space and O(n) time.sols above are not
gonna work..yeah, it is possible in O(n) space and O(n) time.
On May 20, 12:29 am, HARSHIT PAHUJA wrote:
> given 2 unsorted integer arrays a and b of equal size. Determine if b is a
> permutation of a. Can this be done in O
@Piyush: Did you even try this on any examples? If not, try a = {0,1,2,3}
and b = {0,2,2,2}.
Dave
On Sunday, May 20, 2012 1:39:25 AM UTC-5, Kalyan wrote:
> Piyush. I think we can use your logic. But You should check the product
> also.
> Have 4 variables, sum_a,sum_b , prod_a, prod_b
>
> Ca
@Gaurav: you are taking ia and ib as int so they will have 32 bits in
Java. So you can not set the bits for numbers in the array greater
than 32.
e.g if you have a[i]=59876 so you would want to set the 59876th bit in
ia : ia=ia | (1<<59876) but that is not possible. How do you handle
this?
Also how
@Piyush: Try this i/p 8,0,0 ; 2,6,0-- Ur algo aint adequate..
On Sat, May 19, 2012 at 11:24 PM, Piyush Khandelwal <
piyushkhandelwal...@gmail.com> wrote:
> Hiii!! I have some idea about the solution. Please notify me if i am
> wrong
>
> a= [ 4,3,5 ] and b= [ 3,5,4 ]
> diff=0;
> for (i=0; i
Piyush. I think we can use your logic. But You should check the product also.
Have 4 variables, sum_a,sum_b , prod_a, prod_b
Calculate Sum and product of array 'a' and store it in sum_a,prod_a
Calculate Sum and product of array 'b' and store it in sum_b,prod_b
if sum_a=sum_b && prod_a==prod_b the
@piyush :
your solution will fail for the case
a={5,1,1}
b={3,3,1}
On Sun, May 20, 2012 at 11:54 AM, Piyush Khandelwal <
piyushkhandelwal...@gmail.com> wrote:
> Hiii!! I have some idea about the solution. Please notify me if i am
> wrong
>
> a= [ 4,3,5 ] and b= [ 3,5,4 ]
> diff=0;
> for (i=0;
U are checking if the sum is same or not.. which can be same even if the
elements are different.
On Sun, May 20, 2012 at 11:54 AM, Piyush Khandelwal <
piyushkhandelwal...@gmail.com> wrote:
> Hiii!! I have some idea about the solution. Please notify me if i am
> wrong
>
> a= [ 4,3,5 ] and b= [
Hiii!! I have some idea about the solution. Please notify me if i am
wrong
a= [ 4,3,5 ] and b= [ 3,5,4 ]
diff=0;
for (i=0; i wrote:
> @Harshit: These are a few unanswered questions that came to mind when I
> read your solution attempt: What do you do with negative elements? What is
> the -12t
@Harshit: These are a few unanswered questions that came to mind when I
read your solution attempt: What do you do with negative elements? What is
the -12th prime number? How do you deal with overflow in the cases where
you have a lot of large prime numbers and the product exceeds your native
d
Well, the interviewer gave a hint to use hash-table.
The key of hash-table will be memory address of original node and value
will be the memory address of the new node.
On Wed, Mar 14, 2012 at 9:43 PM, atul anand wrote:
> @umer : did interviewer told any one of the solution provided in
> the g
@umer : did interviewer told any one of the solution provided in the given
link below or different?
http://www.geeksforgeeks.org/archives/1155
On Tue, Mar 13, 2012 at 11:31 AM, Umer Farooq wrote:
> Yes that is exactly what they wanted. I proposed BFS for this solution.
> Anyway, there was ano
Sorry, a small test showed a loop quitting one iteration too soon.
Here is the fix.
import java.util.Random;
public class ListCopy {
class Node {
int val;
Node next, other;
Node() { }
Node(int val, Node next, Node other) {
this.val = val;
Copying a full graph requires a BFS or DFS. But here we have a big
advantage. We know the nodes are linked in a single chain. So we
don't need to search to find all nodes. We can just use .next
pointers.
No matter how we do things, we will need Map(A) that returns the copy
of node A if it alre
This problem is close to copying a graph. For that as you said, just
do DFS or BFS and maintain a map from original nodes to copies. Use
the copy to set pointers whenever it exists. When it doesn't exist,
make a new node and add it to the map.
You can implement the map in various ways. A hash t
http://www.geeksforgeeks.org/archives/1155
On Tue, Mar 13, 2012 at 11:31 AM, Umer Farooq wrote:
> Yes that is exactly what they wanted. I proposed BFS for this solution.
> Anyway, there was another problem that I was able to solve; but the
> interviewer brought up a much more efficient approach.
Yes that is exactly what they wanted. I proposed BFS for this solution.
Anyway, there was another problem that I was able to solve; but the
interviewer brought up a much more efficient approach.
The problem was:
- Given a linked a linked list with one pointer pointing to next,
another poin
Yes that is exactly what they wanted. I proposed BFS for this solution.
Anyway, there was another problem that I was able to solve; but the
interviewer brought up a much more efficient approach.
The problem was:
Given a linked l
On Mon, Mar 12, 2012 at 11:31 PM, Gene wrote:
> Since there is no
Since there is no mention of weights, you are looking for any spanning
tree. Primm and Kruskal are for _minimum_ spanning tree. They are
overkill for this problem.
You can construct a spanning tree in any graph with DFS. Just record
every edge you find that reaches a vertex that has never been vi
@Teja Bala
U dont need the last line for a[0][0]
else code will be wrong
conside
0 0 0 0 1
0 0 0 0 0
0 1 0 0 0
0 0 0 1 0
Regards
On Sun, Sep 11, 2011 at 11:56 PM, teja bala wrote:
> //pseudo code dis will work for sure.
>
> for(i=0;i for(j=0;j {
> if (a[i][j] == 1)
>
//pseudo code dis will work for sure.
for(i=0;ihttp://groups.google.com/group/algogeeks?hl=en.
for(i=0;ihttp://groups.google.com/group/algogeeks?hl=en.
whts the question??
With regards,
Praveen Raj
DCE-IT 3rd yr
735993
praveen0...@gmail.com
On Fri, Sep 9, 2011 at 12:17 PM, Amit Gupta wrote:
> Guys, why don't we do something like this :
>
> 1. If (arrayHasBeenTraversed, Goto 4).
>Else, Traverse the 2-D array [row,column] wise. In
Guys, why don't we do something like this :
1. If (arrayHasBeenTraversed, Goto 4).
Else, Traverse the 2-D array [row,column] wise. Inspect element
array[row][column]. Goto 2.
2. If you encounter a '1' (array[row][column]),
change all the 0's in the corresponding [row,column] to '-1'
Al
how abt this:
if(!a[0][0])
{
first traverse the 1st row till we find 1.
if dere is 1 do a[0][0]+=2;
then traverse the first column till 1..
if dere is 1... do a[0][0]+=3;
}
apply dave's method
if(a[0][0]==2)
make 1st row 1
else if(a[0][0]==3)
how abt this:
if(!a[0][0])
{
first traverse the 1st row till we find any 1...
On Sep 2, 12:32 pm, kranthi raj wrote:
> oops missed Space Complexity
>
>
>
>
>
> On Fri, Sep 2, 2011 at 12:55 PM, kranthi raj wrote:
> > for( i = 0 ; i < n ; ++i )
> > for( j = 0 ; j < m ; ++j )
> > if(
oops missed Space Complexity
On Fri, Sep 2, 2011 at 12:55 PM, kranthi raj wrote:
> for( i = 0 ; i < n ; ++i )
>for( j = 0 ; j < m ; ++j )
>if( a[i][j] != 0 )
> row[j]=col[i]=1;
>
>
>
> for( i = 0 ; i < n ; ++i )
>for( j = 0 ; j < m ; ++j )
>
>{
>if
for( i = 0 ; i < n ; ++i )
for( j = 0 ; j < m ; ++j )
if( a[i][j] != 0 )
row[j]=col[i]=1;
for( i = 0 ; i < n ; ++i )
for( j = 0 ; j < m ; ++j )
{
if (row[j]==1 || col[i]==1)
a[i][j]=1;
}
Does this work?
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You received this
@dave:ur algo s nice:)
On Thu, Sep 1, 2011 at 9:49 AM, Dave wrote:
> @Piyush: What does it do on
>
> 0 0 0
> 0 0 0
> 1 0 0
>
> The output should be
>
> 1 0 0
> 1 0 0
> 1 1 1
>
> Dave
>
> On Aug 31, 8:24 pm, Piyush Grover wrote:
> > What's wrong with this??
> >
> > for( i = 0 ; i < n ; ++i )
> >
@Piyush: What does it do on
0 0 0
0 0 0
1 0 0
The output should be
1 0 0
1 0 0
1 1 1
Dave
On Aug 31, 8:24 pm, Piyush Grover wrote:
> What's wrong with this??
>
> for( i = 0 ; i < n ; ++i )
> for( j = 0 ; j < m ; ++j )
> if( a[i][j] != 0 )
> a[i][0] = a[0][j] = 1;
> for( i
I think this code is perfect.
Sanju
:)
On Thu, Sep 1, 2011 at 6:54 AM, Piyush Grover wrote:
> What's wrong with this??
>
>
> for( i = 0 ; i < n ; ++i )
>for( j = 0 ; j < m ; ++j )
>if( a[i][j] != 0 )
>a[i][0] = a[0][j] = 1;
> for( i = 0 ; i < n ; ++i )
>for( j = 0 ;
What's wrong with this??
for( i = 0 ; i < n ; ++i )
for( j = 0 ; j < m ; ++j )
if( a[i][j] != 0 )
a[i][0] = a[0][j] = 1;
for( i = 0 ; i < n ; ++i )
for( j = 0 ; j < m ; ++j )
if( a[i][0] + a[0][j] != 0 )
a[i][j] = 1;
On Thu, Sep 1, 2011 at 5:45 AM, Dave
@Replying to my own posting: Propagating a[0][0] as in my most recent
post isn't correct. Gene is correct to have two flags that indicate
whether the first row and/or the first column are to be filled with
1s.
Dave
On Aug 31, 7:01 pm, Dave wrote:
> @Icy: I forgot about a[0][0]. So I need to add
@Icy: I forgot about a[0][0]. So I need to add a few lines at the end
of my code, so that it becomes:
for( i = 1 ; i < n ; ++i )
for( j = 1 ; j < m ; ++j )
if( a[i][j] != 0 )
a[i][0] = a[0][j] = 1;
for( i = 1 ; i < n ; ++i )
for( j = 1 ; j < m ; ++j )
if( a[i][0
You can use the first row and column as a scratchpad to keep track of
which rows and columns need to be 1.
Make one pass over the matrix to put 1's in the scratchpad locations.
Then make a second pass checking the scratchpad to see of the square
needs to be flipped to a 1.
To make this scratchpad
Dave has a nice idea but I cant get it to work =/
[[1, 0, 0, 1], [0, 0, 1, 0], [0, 0, 0, 0]] original matrix
[[1, 0, 1, 1], [1, 1, 1, 1], [0, 0, 1, 1]] dave's
[[1, 1, 1, 1], [1, 1, 1, 1], [1, 0, 1, 1]] expected
Maybe I converted it wrong. My method was basically the same as
Anup's --
1st
@dave...additionally u ve to do this...checking the 1st row nd 1st
column...
if(a[0][0])
set both first row and first column;
else
for(i=1;i wrote:
> dave s algo is nice :)
>
> On Aug 31, 10:09 pm, Dave wrote:
>
>
>
>
>
>
>
> > @Ashima: Scan all but the first row and the first column. If
dave s algo is nice :)
On Aug 31, 10:09 pm, Dave wrote:
> @Ashima: Scan all but the first row and the first column. If there is
> a 1 in a row, set the first element of that row to 1. If there is a 1
> in a column, set the first element of that column to zero. Now, set
> any element in all but th
@Ashima: Scan all but the first row and the first column. If there is
a 1 in a row, set the first element of that row to 1. If there is a 1
in a column, set the first element of that column to zero. Now, set
any element in all but the first row and the first column of the
matrix that has a 1 it the
@dave wats d logic behind ur code
Ashima
M.Sc.(Tech)Information Systems
4th year
BITS Pilani
Rajasthan
On Wed, Aug 31, 2011 at 9:05 AM, Dave wrote:
> @Manish:
>
> for( i = 1 ; i < n ; ++i )
>for( j = 1 ; j < m ; ++j )
>if( a[i][j] != 0 )
>a[i][0] = a[0][j] = 1;
> for(
@Manish:
for( i = 1 ; i < n ; ++i )
for( j = 1 ; j < m ; ++j )
if( a[i][j] != 0 )
a[i][0] = a[0][j] = 1;
for( i = 1 ; i < n ; ++i )
for( j = 1 ; j < m ; ++j )
if( a[i][0] + a[0][j] != 0 )
a[i][j] = 1;
Dave
On Aug 31, 8:40 am, manish kapur wrote:
>
@Ashish: if i got ur algo correct, contrary to all the above examples,
u r forming a linked list of level order traversal of the tree. m i
right?
On Jul 17, 8:49 pm, Ashish Goel wrote:
> 1. PUSH ROOT IN Q
> 2. PUSH DUMMY NODE IN Q, DEFINE PREVIOUS NODE AS NULL
> 3. WHILE Q IS NOT EMPTY
> 3A. POP
im a bit confused with child-sibling term, this expects output for
input:
A
/\
B C
/ \ / \
DE F G
output:
A
/
B C
/ /
DE FG
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mine solution will give sample o/p 2.
On Sun, Jul 17, 2011 at 9:08 PM, naveen ms wrote:
> im a bit confused with child-sibling term, this expects output for
>
> input:
> A
> /\
> B C
> / \ / \
> DE F G
>
>
> output:
> A
> /
> B--
plz give some example..sry but what do you mean by child sibling
version??
On Jul 16, 3:46 pm, Reynald wrote:
> Given a Parent -Child binary tree ,build the child -sibling version of
> it?
> Minimize the space requirements wherever possible.
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This can be done using single array too... :) :)
Do anybody wants the code?
On Sun, Jul 17, 2011 at 3:04 PM, sagar pareek wrote:
> This can be done like this
>
> 1. find out the height of the tree
> 2. make the number of arrays(node* pointers)=height of tree
> 3. traverse the tree from root as
This can be done like this
1. find out the height of the tree
2. make the number of arrays(node* pointers)=height of tree
3. traverse the tree from root as
arr0[0]=root;
arr1[0]=root->left;
arr1[1]=root->right
arr2[0]=arr1[0]->left
arr2[1]=arr1[1]->right
.
.
. and so on
not
in this recursive code...the right link node will point to its sibling
to the right (if it has) or else it will be null.
the left link of the node will point to its child(if it has) or else
it will be null.
regards,
Naveen
CSE
R.V.C.E, Bangalore.
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im a bit confused with child-sibling term, this expects output for
A
/\
B C
/ \ / \
DE F G
1
A
/
B C
/ /
DE FG
2 A
/
B-- C
/
DEFG
is output expected 1 or 2
sure
void convert(Node * root, Node* sibling) {
if(root == NULL) return;
convert(root->left, root->right);
convert(root->right, NULL);
root->right = sibling;
}
convert(root, NULL);
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DK
http://twitter.com/divyekapoor
http://www.divye.in
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I would assume that in addition to functional testing through
automation(combination of various fonts,sizes etc) the tenets like
reliability, accessibility, interoperability, security(fuzz testing),
different architectures(amd, intel 32 bit, 64 bit etc), stress(extremely
long file reaching space co
Well, this is a white box test. In a wbt, you look for edge and corner
cases. Edge cases in this scenario are the largest and smallest point sizes.
The fonts with largest kerning values. Largest ascenders and descenders.
Largest number of printable characters. You'd probably concentrate on
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 KUM
@ligerdave
agree with u :)
>
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Regards
Aditya Kumar
B-tech 3rd year
Computer Science & Engg.
MNNIT, Allahabad.
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@aditya
there is always trade-off. for what it's asking, TRIE is probably the
fastest. the problem already stated, using "data structure", which to
me, means, you index a document. indexing is expensive, but it's
overhead process and it has nothing to do w/ finding an existing word
in a doc.
On De
It sets the rightmost bit to 0. Could also be done with
q = len & ~1;
Dave
On Dec 7, 12:50 am, ritesh wrote:
> q = (len >> 1) << 1;
>
> what this line is accomplishing?
>
> On Dec 4, 12:38 pm, Abioy Sun wrote:
>
>
>
> > Hello,
>
> > 2010/12/4 siva viknesh :
>
> > > Modified 2 color sort probl
2010/12/7 ritesh :
> q = (len >> 1) << 1;
> what this line is accomplishing?
q = len & 0xFFFE;
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q = (len >> 1) << 1;
what this line is accomplishing?
On Dec 4, 12:38 pm, Abioy Sun wrote:
> Hello,
>
> 2010/12/4 siva viknesh :
>
> > Modified 2 color sort problem i.e. you are given an array of integers
> > containing only 0s and 1s.You have to place all the 0s in even
> > position and 1s i
@ ashita and vrinda
can u please write ur ms written and interview questions..
i ll be really thankful to both of u as ms is going to visit my campus
soon.. so plz help...
On Sep 26, 1:58 pm, ashita dadlani wrote:
> @Mohit:
> I dont think it really matters here.We just have to validate the snaps
@Mohit:
I dont think it really matters here.We just have to validate the snapshot of
the game board.Number of players should not have any relevance here.
On Sat, Sep 25, 2010 at 2:46 PM, mohit ranjan wrote:
> @Ashita,
>
> Your logic is fine for one vs one game, but as per question it's "one vs
>
@Ashita,
Your logic is fine for one vs one game, but as per question it's "one vs
many game"
Any idea what is that ?
Mohit
On Sat, Sep 25, 2010 at 1:18 PM, ashita dadlani wrote:
> 1.The soldiers are initially placed at row 2 or row 7th(each-one of white
> and either of black).Also let white o
1.The soldiers are initially placed at row 2 or row 7th(each-one of white
and either of black).Also let white ones be at row 2.So they can never be at
row 1st.Incase it is so in the game,its not a valid game.
2.There are Bishops.Each color has one of its Bishop which moves diagonally
on all white s
Valid must mean that you can get from an initial board to the the
current game state by a series of legal moves.
This is a classic branch and bound game tree search problem. You
could search either from a starting configuration and try to "find"
the current game state. Or start from the current
its an infinite loop. Beware.
On Mon, Aug 23, 2010 at 5:32 AM, Gene wrote:
> This doesn't work on "abb" for example.
>
> On Aug 22, 9:28 am, Ashish Goel wrote:
> > use a array
> > arr[char]=count char represent say a-z
> > count is # of occurances
> >
> > while (*s!='\0')
> > {
> > arr[*s-'a']+
This doesn't work on "abb" for example.
On Aug 22, 9:28 am, Ashish Goel wrote:
> use a array
> arr[char]=count char represent say a-z
> count is # of occurances
>
> while (*s!='\0')
> {
> arr[*s-'a']++;
> if (arr[*s-'a']==1) lastchar=*s;
>
> }
>
> lastchar is the last non repeating char
>
> Best
optimization, use bitmap instead of array...
char can be unicode, char may take 1 or 2 bytes, that can be written, big
deal..
Best Regards
Ashish Goel
"Think positive and find fuel in failure"
+919985813081
+919966006652
On Sun, Aug 22, 2010 at 7:59 PM, Saikat Debnath wrote:
> This is the answe
Maybe to reduce the memory usage and to include all types of characters we
could create a one to one mapping between the character and the number
of occurrences.And while retrieving start from reverse checking the mapping
value,print if it's one.
On Sun, Aug 22, 2010 at 7:59 PM, Saikat Debnath wro
This is the answer i have given and interviewer said, "Optimise it
further in terms of memory and character need not be ASCII"
On Aug 22, 6:28 pm, Ashish Goel wrote:
> use a array
> arr[char]=count char represent say a-z
> count is # of occurances
>
> while (*s!='\0')
> {
> arr[*s-'a']++;
> if (a
@Luciano
Your method seems very vague. Can you elaborate ? The
input can have elements larger than 9. So will you keep count of those
elements with a numeric array of size 10. Also, your second step is
not clear.
Pls elaborate
On Jul 8, 8:17 pm, jalaj jaiswal wrote:
> @ above
> the
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