It is indeed an elegant interview question with respect to hiring for
designing the systems.It was asked when I was interviewed for the final
rounds of on-site interview at Google.
On Thu, Jul 15, 2010 at 11:47 PM, Tech Id wrote:
> Hi Ashish,
>
> Your algo does not have a chance for jump_back.
>
Hello .
How to decide insertion sort takes less cost of time
for small input size.
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that would be the Catlan number...
Anil
On Thu, Jul 15, 2010 at 11:41 PM, mohit ranjan wrote:
> @Ajeet
> *
> Google "Dyck words*"
>
>
> Mohit
>
>
>
> On Mon, Jul 12, 2010 at 1:43 PM, aejeet wrote:
>
>> A string can contain only the characters A and B.
>>
>> The string formation must follow the
Hello:
This is a problem in theoretical computer science and algorithms.
When solving the 0-1 knapsack problem using dynamic programming, how
can you use the resulting table to determine whether the optimal
solution is unique?
The problem is the following: A thief is going to steal some items
an
http://groups.google.com/group/comp.lang.misc/browse_thread/thread/35967349de8446c1#
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@Ajeet
*
Google "Dyck words*"
Mohit
On Mon, Jul 12, 2010 at 1:43 PM, aejeet wrote:
> A string can contain only the characters A and B.
>
> The string formation must follow the following rules
> 1. The number of occurrences of A and that of B must be equal in the
> string
> 2. For any prefix
Construct the transitive closure of the graph. You can use Warshall's
algorithm, which is O(v^3) in run time. If any row of the adjacency
matrix is all 1's, the corresponding vertex can reach all others. You
can ignore the diagonal element if you don't care whether vertices are
reachable from th
Think about it more carefully. If you are 1m south of the north pole,
then walk pi meters east or west, your longitude changes by 180
degrees. At the equator it changes hardly at all. This is hard to do
accurately and correctly over the whole earth. Google geographic
coordinate transformations.
Hi Ashish,
Your algo does not have a chance for jump_back.
This may be required for certain cases, especially when multiple cars
are running on one lane.
My solution was for one car on one lane, hence car_times[i] gives the
car in i-th lane will take to hit the path frog is trying to cross.
It sh
http://code.google.com/p/frogger-game/source/browse/trunk/
can it be interview question??
the frong will wait and this would vary at run time, the car_times is
storing just one value, multiple cars can come in a lane so it should be
hash map
one thought: for each lane we know when the car would
An inefficient but working method is the brute force one.
Do a recursive test for all the cases possible and if you
find one which helps the frog cross the road, then that is it.
Let int car_times[numLanes] store the time each car will take to reach
the path where the frog is crossing the road.
F
good one
shifting array is the solution but i want to do it without shifting, is
there a solution!!!
Best Regards
Ashish Goel
"Think positive and find fuel in failure"
+919985813081
+919966006652
On Thu, Jul 15, 2010 at 4:26 AM, jalaj jaiswal wrote:
> try merging this array
> [{5,9,10},{3,6,7
Hi
On 15 July 2010 07:37, Tech Id wrote:
> This seems to be asked in context of google maps.
>
I asked in terms of OSM.
>
> While drawing map at a particular zoom level, ratio of map-area-on-
> screen and shown-latitudes/longitudes-area is known.
> So, x meters can be converted to latitiude/lo
technically i can sell first and buy later also
so this solution doesnot consider shorts and long..
my algo would be
return max(MaxStockGain(a,size), MaxStockGain(reverse(a,size),size));
can be done better..without reverse too :)
lol
Best Regards
Ashish Goel
"Think positive and find fuel in fai
This seems to be asked in context of google maps.
While drawing map at a particular zoom level, ratio of map-area-on-
screen and shown-latitudes/longitudes-area is known.
So, x meters can be converted to latitiude/longitude difference easily
by this ratio.
On Jul 15, 12:59 pm, siddharth srivasta
sorry i didnt see it ..
On Thu, Jul 15, 2010 at 4:18 AM, jalaj jaiswal wrote:
> its b tree not binary tree
>
>
> On Wed, Jul 14, 2010 at 10:45 AM, naveen wrote:
>
>> if I do two traversals of a tree inorder wise i can get it in O(n).
>> Questions should be one traversal i think . Then we should
Hey moderator..
try 2 avoid these members...
On Thu, Jul 15, 2010 at 1:01 AM, X +18 wrote:
> http://bit.ly/aRDr4E
>
> Streaming entertainment network with lots of video clips. ... Use a
> software called qvod player, you can watch the DVD movies online for free.
> ...
>
> http://bit.ly/aRDr4E
>
@varun
c should also be in the array
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F
Hi,
suppose we have given n distinct numbers are a[n].
1. sort them using quick or any other sort in O(nlogn)
2. use then
int find(int *arr, int n) /* return largest number if fing it otheriwse
retirn 0*/
{
int i,j,k, temp;
for(k=n-1; k>2;k--)
{
j = k-1
i=0;
while(i!=j)
{
te
Hi
If I have a (latitude, langitude) as well as pixel positions of those
locations, then how can (longitude, latitude) or pixel coordinates of a
point x metres away can be found ?
If this information is not enough, what else information might be required ?
--
Siddharth Srivastava
When you have
int MaxStockGain(int arr[], int size)
{
if(size <= 0)
return 0;
int curMin = arr[0];
int MaxGain = 0;
for(int i = 1; i < size; ++i)
{
if (arr[i] < curMin)
{
curMin = arr[i];
}
int currGain = arr[i] - curMin;
if(cur
try merging this array
[{5,9,10},{3,6,7,9}]...
On Wed, Jul 14, 2010 at 9:30 AM, Ashish Goel wrote:
>
>
> int dstart = -1;
> int dend = -1;
> int istart=-1; int iend = -1;
> bool decreasing = false;
> for (int i=1;i {
> if (a[i] >=a[i-1])
> { if (decreasing)
> {
> dend =i-1; istart=i
Another way in which this can be thought is in terms of Tower of Hanoi
problem.Just introduce two more stacks of same size as input stack and get
the sorted output as result.
On Mon, Jun 14, 2010 at 9:18 AM, Anurag Sharma wrote:
> Why not just pop all elements from stack ( O(n) ) and insert it i
its b tree not binary tree
On Wed, Jul 14, 2010 at 10:45 AM, naveen wrote:
> if I do two traversals of a tree inorder wise i can get it in O(n).
> Questions should be one traversal i think . Then we should use a tree where
> nodes contain links to parents.
>
>
> On Tue, Jul 13, 2010 at 6:58 PM,
>
> I have one more question here , suppose we have some dynamic array
>> ( of const size ) where insertions and removal is happening
>> dynamically --->
>> you want the 2 elements from the array having least difference. Design
>> a data structure for this. Less than O(n) solution appreciated.
>>
>
@harit..
logic plz.. not the code..
On Wed, Jul 14, 2010 at 9:50 PM, harit agarwal wrote:
> this is O(n) solution and using O(n) space...
>
> #include
> #include
> #include
> using namespace std;
> void leader_count(vector v,int *ar)
> {
> stack s;
> int n=v.size();
> int i=n-1;
> while(i>=0
I think this will help:
http://en.wikipedia.org/wiki/Frogger
Consider the roads to be n-laned and of constant width with constant time
traffic coming on each lane.For example,say after 1 minute,a car comes on
each lane but in a arithmetic sequence and not all at same time.To make it
more clear,
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