You guy check this solution. It is expected to be run in
O(n lg n) for random permutation in average time. For
worst case, I think we can improve it for that.
Let's do an example firstly.
7 3 4 1 2 6 5 8
For *1*, it self is a block. Let's count the blocks containing *1* firstly.
If a block
These are the steps for the O(n^2) solution
n=length of A
for each subarray A[i,j] where ji
min=min(A[i,j])
max=max(A[i,j])
if(max - min==size (A[i,j]) print A[i,j]
min[A[i,j]]=min( A[j], min(A[i,j-1])
similar one for max
Note:
A[i,j] = A[i],A[i+1]A[j]
I was
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Hardyhttp://eaziweb.blogspot.com/2009/12/how-to-install-dreamweaver-cs3-in.html
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Shahzeb Farooq Chohan
Software Engineer
Cogilent Solutions
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Shahzeb Farooq Chohan
Software Engineer
Cogilent Solutions
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Hi, Well
Recaman's sequence is defined by: a(0) = 0
a(k) = a(k-1) -
k, k0, a(k) new term in series or else a(k-1) + k
So sequence is 0, 1, 3,6, 2, 7, 13, 20, 12, 21, 11, 22, 10,
23,
My approach used memoization using map in c++
On Dec 2, 6:45 am, Vinoth Kumar vinoth.ratna.ku...@gmail.com wrote:
These are the steps for the O(n^2) solution
n=length of A
for each subarray A[i,j] where ji
min=min(A[i,j])
max=max(A[i,j])
if(max - min==size (A[i,j]) print A[i,j]
min[A[i,j]]=min( A[j],
On Dec 2, 10:42 am, Geoffrey Summerhayes sumr...@gmail.com wrote:
It's a binary tree, [ 7 3 4 1 2 6 5 8] has children
[ 7 3 4 1 2 6 5] and [ 3 4 1 2 6 5 8], all the way
down to [ 7 3] [3 4] [4 1] ...
If you start at the bottom keeping track of min and max
for each node, if max-min == node
