Btw...another observation in case 1.2:
I wrote:
Now, let v=min(x[1],x[2],x[3],.x[k]), and 'h' be the index of the
minimum element(x[h] is minimum)
Then, dp[i][j]=v+sigma(f!=h)[min(x[f],y[f])]
Here, just setting dp[i][j]=v will do (athough the complexity is same
in both the cases)
because for
Sort first array preserving the initial position of the elements.
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We don't know how many elements in the stream. All decisions should be done
at online.
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Well All the subsets, not the number
On Dec 25, 8:48 pm, juver++ avpostni...@gmail.com wrote:
What you need to get for the answer - amount of such subsets or display
them?
First problem can be solved using DP.
Second - brute force.
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There are 3 paths exist in bst, post, pre and inorder.
store all these paths and find contiguous sum(and set of elements
which leads to this sum) and check if it equals to given sum, then
that is path.
On Dec 27, 6:48 pm, mohit ranjan shoonya.mo...@gmail.com wrote:
any hint for below question
The better option will be to shift and divide.
As you can store the arbitrary large number in a link list or int
array. Left shift the right array and subtract, this will give you a
better approach, rathar than just subtracting each time. Add them in
the end ..
On Dec 27, 1:47 am, Aniket
@ pacific:
Also consider the choice of flipping the node itself.
And if an internal node cannot be flipped, it is still possible to flip
its value, only the above choice is not taken into consideration..
On 2010-12-28 13:27, pacific pacific wrote:
here is my approach :
Starting from the root
Thanks , I need to learn sml now :).
I was looking for the conversation over algo development and hints to
do that.
On Dec 25, 10:04 pm, Puneet Ginoria punnu.gino...@gmail.com wrote:
This attachment contains the code for the above program in SML language and
it uses lambda calculus.
On
On what basis we will divide the array.
(Eg 1. Sort and divide them on positive negative )
On Dec 25, 10:52 pm, radha krishnan radhakrishnance...@gmail.com
wrote:
This s similar to the problemhttps://www.spoj.pl/problems/SUBSUMS/
we have to split the array into 2 arrays
say A,B
generate all
The description on internal nodes indicates this:
The value of an AND gate node is given by the logical AND of its TWO
children's values.
The value of an OR gate likewise is given by the logical OR of its TWO
children's values.
On 2010-12-28 13:35, suhash wrote:
Your approach is for a
Let cst[i][j] store the cost to flip node i to given gate j (0-'AND',
1-'OR').
Then: cst[i][j] = 0,if j==gate[i];
cst[i][j] = 1,if j!=gate[i] and ok[i];
cst[i][j] = INFINITY, if j!=gate[i] and !ok[i];
1. To get value 1:
1.1 flip current gate to AND, and change all
Is second array sorted?
On Tue, Dec 28, 2010 at 1:57 PM, juver++ avpostni...@gmail.com wrote:
Sort first array preserving the initial position of the elements.
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Incorrect. Path can be combined from the several traversal algorithm's
output.
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@Terence: I like your explanation. Very short and crisp! :)
On Dec 28, 12:10 pm, Terence technic@gmail.com wrote:
Let cst[i][j] store the cost to flip node i to given gate j (0-'AND',
1-'OR').
Then: cst[i][j] = 0, if j==gate[i];
cst[i][j] = 1, if j!=gate[i] and ok[i];
Sorry my mistake! But the general problem with more than 2 children
possible is more interesting! :)
On Dec 28, 10:58 am, Terence technic@gmail.com wrote:
The description on internal nodes indicates this:
The value of an AND gate node is given by the logical AND of its TWO
children's
Not clear what path you are referring to.
Question. Should the path include root value always? (What is problem
with only left or only right path (not containing root))
In your example for 16 one more path can be 0 1 5 10 as
well. Should algo return all the paths or just first one.
Consider your example:
One is
2 5 1 6 4 3
other is
1 2 3 4 5 6.
After sorted the first array and keep the position you will have:
(1,3)
(2,1)
(3,6)
(4,5)
(5,2)
(6,3)
O(n log n)
for each value of the second array do a binary search in the first array
and discover the position
O (log n)
@ Wladimir
How do you get
(1,3)
(2,1)
(3,6)
(4,5)
(5,2)
(6,3)
B'cos once you sort the array you lose its original index.
On Tue, Dec 28, 2010 at 5:50 AM, Wladimir Tavares wladimir...@gmail.comwrote:
Consider your example:
One is
2 5 1 6 4 3
other is
1 2 3 4 5 6.
After sorted the
Now the question boils down to how to sorted the array preserving its index.
On Tue, Dec 28, 2010 at 12:27 AM, juver++ avpostni...@gmail.com wrote:
Sort first array preserving the initial position of the elements.
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I think the original question says Path can go from left subtree tree ,
include root and go to right tree as well. This should mean the path must
include the root.
On Tue, Dec 28, 2010 at 4:52 AM, shanushaan er.srivastavaro...@gmail.comwrote:
Not clear what path you are referring to.
And of course boundary cases(leaf nodes) are to be handled. For a leaf
node 'i', ok[i][j]=1(if j==v[i]), 0 otherwise!!!
On Dec 28, 11:04 pm, suhash suhash.venkat...@gmail.com wrote:
I think this can be solved using dp. Consider the subtree rooted at
node 'i'. Let ok[i][j] be a boolean (0 or 1)
Hi Anad,
You can create a struct with this information
typedef struct TArray{
int value;
int index;
};
TArray a[100];
Wladimir Araujo Tavares
http://www.si.ufc.br/~wladimir http://www.si.ufc.br/%7Ewladimir/
Fiz uma faculdade! Só não fiz a segunda porque acabaram os tijolos.
On Tue, Dec
if I already have a structure indicating the position of the element in the
array. Then why do we need to sort. Question is to provide index of element
in O(nlogn).
On Tue, Dec 28, 2010 at 2:20 PM, Wladimir Tavares wladimir...@gmail.comwrote:
Hi Anad,
You can create a struct with this
for each element in the second array apply binary search in first array. ie,
For X[1] find 1 in first array with binary search.
Time Complexity=O(nlogn)
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Hello,
2010/12/29 Anand anandut2...@gmail.com:
if I already have a structure indicating the position of the element in the
array. Then why do we need to sort. Question is to provide index of element
in O(nlogn).
You do not have a structure before preprocessing the data, whose
complexity is
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