lot of ebook links and reference materials here. and other languages
too
http://online-books-reference.blogspot.com/
http://www.googlebooks.tk
All the ebook links listed in this site is available in one zip
file(Size 10KB)
http://www.megaupload.com/?d=FN54H6ZZ
guess this will be useful to
rainix wrote:
let level_head_node(i) is the first node of the level i of the
perfectly balanced BST
..
node node_a = level_head_node(i);// get the first node
of the levle i
node node_b = level_head_node(i+1);// get the first node of
the level i+1
node
rainix wrote:
let level_head_node(i) is the first node of the level i of the
perfectly balanced BST
..
node node_a = level_head_node(i);// get the first node
of the levle i
node node_b = level_head_node(i+1);// get the first node of
the level i+1
node
rainix wrote:
1
/\
23
/\ /\
4 5 6 7
/\
8 9
(5)
time cost: 2n, O(n)
space cost: 3, O(1)
Unfortunately this is not a BST.
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You
Add 4 auxiliary nodes s',a,b and t' and split any existing node v[i] to
v1[i], v2[i]. If there's an edge from v[i] to v[j], there would be an
edge from v2[i] to v1[j] with capacity 1. The edges s'-s, t-t',
a-any v2[i], any v1[i]-b, any v1[i]-v2[i] have infinity capacity.
s'-a and b-t' has a
Forget itThis is obviously wrong, I seems to came up with a
solution to find the maximum s-t flow after deleting k edges.
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To post to this
Hi..frns...
here algorithms for this problem...
1st assume each row contains only 1.
store the position of 1's in each row in one dimensional array.
and use hash technique ..[ this is for to know whether any row is repeated or not ] if two numbers repeated
this is not rearrangble matrix.
SPX2 wrote:
what complexity aj ?
O(ne) where e is the n is the number of rows and e is the number of 1's
in the matrix.
worst case O(n^3).
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the matrix
1 1 1
1 1 1
1 1 1
has rank 1, but is still rearrangable, so it seems linear dependence
has not much to do with the problem.
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To
[EMAIL PROTECTED] wrote:
See, the idea is to build a max balanced tree.
For which, we need to check the height of the tree
only the first time
There is an on line algorithm for this problem. An on line algorithm
does not need to see all the input in advance. In this case, the on
line
I know that on line was not part of the initial spec, but the on line
algorithm is simpler than some that have been proposed, so I suggest
looking for it.
I came across this problem (a college senior of mine was asked this
very
question in his Amazon interview).
I settled for the following
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