This is the trivial case. There is only one tree with 0 nodes.
But you can prove by contradiction if you like.
On Wed, Jun 11, 2008 at 3:12 AM, zee <[EMAIL PROTECTED]> wrote:
>
> referring to problem 12-4 of CLR
>
> how does we show that b0 ( the number of binary trees formed by 0 nodes is
> 1)
>
This is the idea my code used, though I kept track of the start of the
array with a pointer and the end with an integer length.
You seem to imply that the median of an array is always a single
element (at n1 or n2). Care is necessary when the lengths of the
arrays are even. In this case the med
Hello all, thanks for providing some solutions, this is the solution to
the problem that I had given earlier although a informal one
let the two arrays be A1 and A2 with p1, q1 he start and end of the
first array and p2, q2 be the start and end of the second array.
let n1 be the median of the fi
Just noticed again that I made a silly algebra error and this code
doesn't work right when the operands are of even length. Sorry. If
anyone is interested I'll work out the correction.
I just noticed that the problem requires the arrays to be the same
length. Obviously I solved it for different lengths.
Here's one (untested). Suppose the two arrays are a and b and WLOG |a|
<= |b|.
First consider the "divide and conquer" step where |a| > 1 (and
therefore |b|>1 also). Let A be the middle element of a and B the
middle element of B.
Then there are two cases that let you throw away |a|-1 elements
why not post your answer here by the way?
I already found the solution on this group only. Thanks a lot.
And no its not homework
Thanks Rohit
This is a slick problem. Is it homework?
