you can arrange them with equal distances !
if n=1 then, it does not matter where you put the point !
if n1 then, put them with distances = (r2i-r1i) / (n-1) !
it means ou put the first point on r1i and the last point on r2i, the
remaining point are distributed with equal distances !
On Oct 5,
@harit
random number means its output cant be predicted,it doesnt mean that the
probability will be equal for any number of test runs.
e.g consider random number between 3-5 for 3 test runs
if the first output is 4
second is 3
then to ensure same probability third output must be 5
which makes
I'm trying to wrap my head around this problem. I compute a visibility
graph in order for a robot to move around, and an obvious application
in this graph is to find a shortest path. Now finding the start and
end of this path is not difficult, but to the best of my knowledge the
shortest-path
I found this question in a interview blog.
http://placementsindia.blogspot.com/2007/09/google-top-interview-puzzles.html
Question No. 19
My understanding for retrieve minimum in constant time is, we should
be able to find the minimum element in stack and to remove (pop) it in
constant time.
Thanks a Ton everybody!
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brother try this
push(x)
{
if stack.empty()
{
stack.push(x);
stack.push(x);
}
else
{
if x stack.top()
{
min = stack.top()
stack.push(x)
stack.push(min)
}
else
{
The basic idea would be to do push / pop the minimum in stack along with the
push / pop elements. It is like maintaining a parallel stack of minimums. A
naive implementation would involve two stacks. All operations should be O(1)
time complexity.
On Tue, Oct 6, 2009 at 1:31 PM, Manisha
