gthOfString +1 ( +1 for '\0')
strlen(anyString) = simple length
sizeof(anyArray) = numberofElementPresentInArray*sizeof(dataType)
--
*Thanks,*
*Vikas Kumar*
*
*
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The question was from startup company verego (may be misspelled),
you understood it in correct way.
We are given 'k' number of ranges in the fashion L, R
they are all integers in big ranges , (assume long long for the case)
then we are given move operation defined as
move( i, +1/-1) : so range
There are a number of integer ranges say [ L, R]i denoting left and
right of segment i, lets says we are given K such segments and we
define one operation as move which makes our chosen segment to move
either +1 or -1 so after
move(i, +1) segment i will be [L+1, R+1]. There are some particular
num
x is a pointer which can be changed from out side and stores the
pointers for int*
On Jun 8, 10:58 am, Vishal Thanki wrote:
> Following declaration makes the "x" as a volatile pointer to an integer.
>
> int *volatile x;
>
> But what does following means?
>
> int **volatile x;
>
> ~Vishal
--
You
@priyaranjan
No,Your triangle will be right angle triangle.
In fact any triangle chosen will be right angle triangle.
Proof:
suppose there are 2 planes kept at parallel with each other.(top bottom)
join corresponding vertices of up to down and form cube.
Now 3 points chosen can be on same plan
f(n)=n-1.
On Wed, Feb 16, 2011 at 7:39 PM, Akshata Sharma
wrote:
> please help..
>
> if f(n+1) = max{ f(k) + f(n-k+1) + 1} for 1 <= k <= n; f(1) = 0.
> Find f(n+1) in terms of n.
> Eg: f(4) = ? n = 3; 1<= k <= 3; f(4) = max{f(1) + f(3) + 1, f(2) +
> f(2)+1, f(3) + f(1) +1}
>
> --
> You received
@Snehal
just a hint: there is no need of that channel 1 & channel 2.
just treat each program as independent program.
On Mon, Jan 31, 2011 at 10:44 PM, snehal jain wrote:
> or provide some link
>
>
> On Mon, Jan 31, 2011 at 10:44 PM, snehal jain wrote:
>
>> @ juver++
>>
>> can you please share
use it like this
char* strrev(char *)
strrev(str)
void strrev(char *str)
{
xstrrev(str , 0, strlen(str)); // code posted by Nishant
}
:-)
On Sep 23, 11:35 pm, albert theboss wrote:
> Ur function prototype is not similar to one i posted before
>
> check it out
--
You rece
we can approach it like
*lets start with index 0
if (found) return index
else
index += INCR_;
*we will try to increment with INCR_ until one of the following
conditions met
index goes out of bounds
found the words which must come after the value eg "meet" should
occur prior to "must"
found
int getMaxOneRow()
{
...int maxOne = 0, maxRow = 0;
...for (curRow=0; curRow maxOne)
..{
.maxOne = n;
.maxRow = curRow;
..}
...}
return maxRow;
}
int getOneCount(int row)
{
...int count =0;
...for(int i =0; i
wrote:
> There is a 2 dimensional array with each cell containin
you can search for box stacking problem in google. There is a DP
method.
On Sep 22, 12:11 am, Dave wrote:
> Certainly having a smaller volume is necessary for a box to fit in
> another box, but it is not sufficient. E.g., a box of size 1 x 1 x 1
> will not fit in a box of size 2 x 2 x 1/2.
>
> Da
use a bst and insert in it reverse of array. use a count in bst node
and incr count at each right child insertion
On Sep 14, 11:00 am, "Shiv ..." wrote:
> A pseudo code-
>
> int n; //= number of inputs.
> cin>>*a; // the inputs.
>
> int ** invArr;
> *inVArr[n-1] = NULL;
> //ini
take this approach
fill array of snakes starting position in snake[num_snake]
for each snake[i] , take the end of snake and fill in some other array
take random number gen and fill these arrays-->
e.g. end_snake[i] = ran(start_snake[i]-10) // so that snake does not
end up in same row
same logic
use opengl library to draw the graph or whatever. take RED BOOK for
reference.
On Sep 14, 5:54 pm, Mithun wrote:
> Can anyone help me with the code for drawing a graph in C or CPP
> (using graphics)
>
> Input is an Adjacency matrix
--
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struct list
{
median --> median from the start to num
number --->number
list *next
};
during insertion : insert in sorted LL
after that all subseq node's median has to be modified
O(n)
during median_retriev
check the median value and return that
O(1)
I assumed deletion is not happening.
@Bittu
I am confused about one point you need to process atleast n*k
elements so how will you do it in nlogk time. It seems really
tricky if possible.it's min time should be O(nk). correct me if I
am wrong.
On Sep 17, 6:34 am, tkcn wrote:
> Use k-way merging +1.
>
> 1. Before the merg
nice recurrence
On Sep 14, 9:29 pm, Gene wrote:
> You can approach this the same way you'd do it by hand. Build up the
> string of brackets left to right. For each position, you have a
> decision of either ( or ) bracket except for two constraints:
> (1) if you've already decided to use
I think Gene is right. they are asking more than search here. Thanks
to Gene for suggesting this idea :)
now the how part,
we can have a level of abstraction:
no of products matching to name ---> use trie, suffix tree, some
genetic algo, string matching, ms tree..
based on products ge
can you define what here subsequence means?
On Wed, Aug 25, 2010 at 9:32 PM, Rahul wrote:
> @Jaswanth
> It will be really kind if you will state the algorithm rather than
> providing codes, as it is tedious.
>
> --
> You received this message because you are subscribed to the Google Groups
> "Al
you are right in the way in cpp code it is not required but in C , you
can not write like node * start. it only recognise new type when
typedef is applied. second one is already explained by vijay
On Aug 18, 6:16 pm, Vijay wrote:
> 1. typedef is used to rename the data type. Here struct node is a
the method of farword seek is inefficient. consider case of 10
lines and you want to display only 3-4 lines. better seek from end.
use a buffer[buf_size].
let size =filesize.
lc = 0;
while(lc <= given line input)
{
fseek(fp, size);
if(size < buf_size)
fread(fp, size, buffer);
else
fread(f
what about printing these strings onto terminal. can printing of the
string is also similar. I tried to develop the algo but got messed
up.can some one help?
On Aug 8, 2:53 am, Amir hossein Shahriari
wrote:
> @ankur: dp[i][j] number of (prefix) strings that have i As and j Bs is:
> dp[i-1][j] num
1) Yes ofcourse
2) yes you can
3) yes ofcourse
4) yes
5) i dont think . May we our default constructor def not matching ;)
On Aug 3, 8:42 pm, Raj N wrote:
> 1) Can a constructor call another constructor to initialize the same
> object?
> 2) Can a struct variable be assigned to another if the stru
@ Ashish:
make a balanced binary tree of depth >3 and try the approach, u
need to save the pointers somewhere if they are in different subtrees
and at same level. Stack will be most appropriate one to use for this
purpose.
On Aug 2, 7:34 pm, Ashish Goel wrote:
> why stack? that is DFS
>
> it w
driver(Tree *root)
{
node *array = malloc(sizeof(node *) *H ); /* take the height node
pointer array. this will act as source pointer of each list */
makeList(root, 0, array)
}
makeList(Tree *root, int level, node *array)
{
if(! root) return ;
/*do a Depth traversal and store the list
int countTree(int num)
{
if(num <= 1) return 1;
int sum =0;
for(i =1; i wrote:
> Follow up on Catalon Nubmer...
>
> On Fri, Jul 30, 2010 at 10:44 AM, Amit Jaspal wrote:
>
>
>
> > n is clearly a number lets say 3 then BST's with 1,2,3 values as key are to
> > be calculated
>
> > On Fri, Jul
if it is a simple BT then you can simply attach the root to either
child ( which is null ) of other tree . just simply go leftmost and
then assign root of other tree as left child, as suggested by Gene.
On Jul 27, 8:23 am, Gene wrote:
> You actually only need a singly linked list. See and old di
are you asking for BST or simple BT. what are the conditions you want
to follow if simple BT
On Jul 26, 12:59 am, AlgoBoy wrote:
> Does anyone know how to merge two binary trees in O(n logn logn)
> complexity..
> intiutive solution is to flatten both the trees (by inorder
> traversal ) and then c
Thanks Amit and srikanth for pointing that out. I should have done
some analysis before posting this solution.
we can do this problem in 2 steps:
find-min-diff(arr a)
{
sort(a)
find(a, 0, min=INF, i=0, j=0)
return {min, i, j}
}
find(a, index, min, i, j)
{
if(a[index+1] - a[index] < mi
I did not get your point.
for 2 6 3 7
min 2
sec min 3
difference is 1
answer is 2 and 3
what more is asked??
On Jul 12, 2:21 pm, srikanth sg wrote:
> 2 6 3 7
> check for this
>
> On Mon, Jul 12, 2010 at 12:46 AM, vikas kumar wrote:
>
> > traverse array with 2 elements k
traverse array with 2 elements keeping track of 2 min elements. time
O(n) space O(1)
On Jul 11, 9:34 pm, Amit Jaspal wrote:
> Constraint - O(n)
>
> On Sun, Jul 11, 2010 at 9:24 AM, amit wrote:
> > Given an array of size n.find 2 numbers from array whose difference is
> > least.
>
> > --
> > You
is it something different than this
foo(char *p, int num)
{
if(!p) return "error"
if(num > 0 && p[num-1] == 0) print " 2 byte value"
else
print "1 byte value"
}
On Jul 11, 11:18 pm, Tech Id wrote:
> Question is not clear to me :(
> Can you explain a li
traverse array with 2 elements keeping track of 2 min elements. time
O(n) space O(1)
On Jul 11, 9:34 pm, Amit Jaspal wrote:
> Constraint - O(n)
>
> On Sun, Jul 11, 2010 at 9:24 AM, amit wrote:
> > Given an array of size n.find 2 numbers from array whose difference is
> > least.
>
> > --
> > You
adder with add value 1
On Jun 24, 8:30 pm, jaladhi dave wrote:
> Another approach :http://codepad.org/CqKIZJeO
>
> On Thu, Jun 24, 2010 at 7:31 PM, mohit ranjan wrote:
>
> > @Dave
>
> > Can u plz explain the logic behind this..
>
> > Mohit
>
> > On Thu, Jun 24, 2010 at 12:44 AM, Dave wrote:
>
>
adder with add value 1
On Jun 24, 8:30 pm, jaladhi dave wrote:
> Another approach :http://codepad.org/CqKIZJeO
>
> On Thu, Jun 24, 2010 at 7:31 PM, mohit ranjan wrote:
>
> > @Dave
>
> > Can u plz explain the logic behind this..
>
> > Mohit
>
> > On Thu, Jun 24, 2010 at 12:44 AM, Dave wrote:
>
>
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