*[locale]*);
}
return toReturn;
}
- similarly for other combinations ---
--
Abhay Prakash
IV Year (B.Tech + M.Tech Dual Degree)
Computer Science And Engineering
Indian Institute of Technology, Roorkee.
On Fri, Apr 11, 2014 at 10:34 AM, Narsimha Raju
cnarsimhar
need to show the result which satisfy all the 3
conditions.
2. Politics, Hyderabad
In this case we need to show all the results of politics
Hyderabad irrespective of language.
--
Regards,
C. Narsimha Raju
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The answer is single queue...
If we fill the queue fully and then dequeue and try again to enqueue... We
will get the error.. This is because when we dequeue,we move the front
pointer of the queue but not the rear pointer.. But enqueue is at rear
end.. The condition for enqueue also checks the
@icy
It's still there except that you'll get a different question.
That page promises you a telephone interview if you solve the challenge
but I don't know how true that is for non-US guys ..
i solved one question two weeks back .. and no one contacted me till now ..
~raju
On Tue, Oct 25, 2011
hi all ,
can anyone pls share citrix interview questions ...
thanks
raju
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@nitin ..
Output array is not a new array ... you can do anything to input array ..
~raju
On Fri, Sep 30, 2011 at 1:24 PM, Nitin Garg nitin.garg.i...@gmail.comwrote:
Can we assume the output array is a new array and we can distort the
originial array???
On Fri, Sep 30, 2011 at 9:14 AM
Had this question been already discussed .. someone pls give me the
algo/logic.
Given a Binary Tree, Convert it into Doubly Linked List where the nodes are
represented Spirally.
For Example :-
A
B C ABCGED || ACBDEG
D E G
~raju
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using two stacks or using a queue and a stack ... these are obvious
solutions ..
Just want to know if there exists an iterative way without extra space !!!
I should've mentioned these details earlier .. sorry for that !!
~raju
On Tue, Sep 27, 2011 at 10:09 PM, geeks ankurshukla.h...@gmail.com
Rotated array = [ 0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0 ]
find 1 using binary search ...
turns out we cant use binary search if we've repeated elements in rotated
array ...
~raju
On Tue, Sep 27, 2011 at 8:18 PM, akanksha akanksha.271...@gmail.com wrote:
/*
A program in which an array have
hi all,
can anyone pls share the questions amazon has been asking in online written
tests.
thanks
raju
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*
*
*#include fstream
#include iostream
using namespace std;
bool flag=false;
void check(int i)
{
int sum=0;
char n[5];
itoa(i,n,10);
for(int j=0;j5 n[j] != '\0';j++)
{
char p = n[j];
sum += atoi(p);
}
if(sum==3 || sum==6 || sum==9)
flag = true;
else if(sum9)
check(sum);
}
void main()
{
int i;
yes..you are right ..sukaran...that is depends on compiler which one
we are using...6/5 or 6/6 only two possibilities come but answer
will come 2 only for this qn on every compiler.. ..
On Sep 4, 8:22 pm, sukran dhawan sukrandha...@gmail.com wrote:
order of evaluation for such expressions is
int a,b,max,min;
max=(a+b+abs(a-b))/2;
min=(a+b-abs(a-b))/2;
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cheers
priyanka
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It will work...
a=3
b=-2
max=(a+b+abs(a-b))/2
so
max=(3-2+abs(3-(-2)))/2
=(1+abs(5))/2
=(1+5)/2
=3
Same in case of min
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cheers
priyanka
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void mirror(struct node* node)
{
if (node==NULL)
{
return;
}
else
{
struct node* temp;
mirror(node-left);
mirror(node-right);
temp = node-left;
node-left = node-right;
node-right = temp;
}
}
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wat about a double ended queue?
cheers
priyanka
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tats bcos...
if it is a^21...
it will be (a^21/2)*(a^21/2)*(a^21mod2)
= (a^10)*(a^10)*(a^1)
correct me if i'm wrong
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priyanka
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write a c program to print the following
*
* *
* * *
* *
*
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cheers
priyanka
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HI
A BST is given you need to print all the negative numbers.
Sol: Inorder traversal will work. but i want to ignore the right sub trees
based on the value at the root. (i.e efficient sol)
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Regards,
C. Narsimha Raju
MS, IIIT Hyderabad.
http://research.iiit.ac.in/~narsimha_raju
/algogeeks?hl=en.
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Regards,
C. Narsimha Raju
MS, IIIT Hyderabad.
http://research.iiit.ac.in/~narsimha_raju/
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Handling ' 0 's
Case1: array containing single zero.
Caae 2: array containing multiple zeros.
int numberofZeros =0, index =0 ;
before[0]=1
after[length-1]=1;
for (int i=1; ilength;i++)
{
if( !a[i] ) //encountered zero
{
numberofZeros++;
if(numberofZeros == 1) index =i; //Case1
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