[algogeeks] a google question

2010-04-30 Thread divya
Given two sorted postive integer arrays A(n) and B(n) (W.L.O.G, let's
say they are decreasingly sorted), we define a set S = {(a,b) | a \in
A
and b \in B}. Obviously there are n^2 elements in S. The value of such
a pair is defined as Val(a,b) = a + b. Now we want to get the n pairs
from S with largest values. The tricky part is that we need an O(n)
algorithm.

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Re: [algogeeks] MXN matrix

2010-04-30 Thread Ashish Mishra
@amit

nice sit didn't know about that  :)
thanks

On Thu, Apr 29, 2010 at 3:59 PM, amit kumar loveami...@gmail.com wrote:

 Answer for question 1:  http://geeksforgeeks.org/?p=6257

 And, I think the same solution can be extended/manipulated for rectangular
 matrix with largest number of 1's.

 Regards,
 Amit


 On Wed, Apr 28, 2010 at 10:23 PM, Vivek S s.vivek.ra...@gmail.com wrote:

 Let Memo[i][j] be the sum of elements in the (sub) rectangle from (0,0) to
 (i,j)

 Then use principle of inclusion and exclusion to find the sum of elements
 from (a, b) to (c, d) in O(1)

 for N*N matrix, Complexity is O(N^4)

 On 28 April 2010 13:36, Ashish Mishra amishra@gmail.com wrote:

 you are given a M x N matrix with 0's and 1's
 find the matrix with largest number of 1,
 1. find the largest square matrix with 1's
 2. Find the largest rectangular matrix with 1's

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 Regards,
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Re: [algogeeks] Build BST from binary tree without extra space

2010-04-30 Thread Rajesh Patidar
thing is that u r using recursion and we don't have to use it(
recussion  use memory indirectly) as per the question

On Thu, Apr 29, 2010 at 3:55 PM, Algoose Chase harishp...@gmail.com wrote:
 If you mean to convert the binary tree to binary search tree directly , then

 BinarytoBST(Node* root)
 {
    if( root == nulll) return;

    BinarytoBST(root-left);
    BinarytoBST(root-right);

    if( root-left )
    Node* NodeL = MAX(root-left);
    if ( root-right )
    Node* NodeR = MIN(root-right);

    if( NodeL-Data  root-data)
    {
    // swap values.
    temp = NodeL-data;
    NodeL-data = root-data;
    root-data = temp;

    BinarytoBST(NodeL);
    }
    if( NodeR-Data = root-data)
    {
    // swap values.
    temp = NodeR-data;
    NodeR-data = root-data;
    root-data = temp;

    BinarytoBST(NodeR);
    }

 



 On Thu, Apr 29, 2010 at 11:23 AM, Algoose Chase harishp...@gmail.com
 wrote:

 Hi,

 How do you define without extra space ?
 Doing any order traversal either using recursion or using iteration is
 going to take extra space .

 If you are given a binary tree represented by pointers that points to
 children nodes is it possible to do a heap sort without an array ?

 On Thu, Apr 29, 2010 at 6:59 AM, sharad kumar aryansmit3...@gmail.com
 wrote:

 my choice is build  a min heap .sort the array with heap sort.then find
 the median of the sorted array and build tree

 On Wed, Apr 28, 2010 at 10:16 PM, Vivek S s.vivek.ra...@gmail.com
 wrote:

 @Rajesh Patidar
 I think we should do in Post order traversal alone. If we go by
 Preorder/Inorder we might lose track of children node that is currently
 being inserted into the BST. - correct me if im wrong :)

 On 28 April 2010 15:30, Rajesh Patidar patidarc...@gmail.com wrote:

 pickup node in any order no matter(pre,post,inorder)  and just one by
 one. start adding the node into bst no need to use extra space u have
 to just ditach the node from binary tree and attach it in bst.

 On Wed, Apr 28, 2010 at 1:18 AM, Ashish Mishra amishra@gmail.com
 wrote:
  How to build BST from binary tree in place i.e without extra space ??
 
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 Regards,
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[algogeeks] infy color balls

2010-04-30 Thread Ashish Mishra
One of my friend ask me this n i am bad in P  C (will love if smone can
provide me a link to learn it though)

nyways prob is:
there are infy color balls of k different color
you are allowed to pick n balls out of those infy(infinite) balls

cond is : you must have all k color balls with u
obviously kn

Question is how many possibilities for selection you have ?

Thanks
Ashish

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Re: [algogeeks] a google question

2010-04-30 Thread Amir hossein Shahriari
On Fri, Apr 30, 2010 at 4:35 PM, divya sweetdivya@gmail.com wrote:

 Given two sorted postive integer arrays A(n) and B(n) (W.L.O.G, let's
 say they are decreasingly sorted), we define a set S = {(a,b) | a \in
 A
 and b \in B}. Obviously there are n^2 elements in S. The value of such
 a pair is defined as Val(a,b) = a + b. Now we want to get the n pairs
 from S with largest values. The tricky part is that we need an O(n)
 algorithm.

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Re: [algogeeks] infy color balls

2010-04-30 Thread Kishen Das
Answer is k! * k^(n-k).
You can select k number of balls in K! ways and then rest of (n-k) balls in
k ways.
This way you are ensuring that you have all k color balls and n number of
balls in total.

- Kishen Das

On Fri, Apr 30, 2010 at 6:04 AM, Ashish Mishra amishra@gmail.comwrote:

 One of my friend ask me this n i am bad in P  C (will love if smone can
 provide me a link to learn it though)

 nyways prob is:
 there are infy color balls of k different color
 you are allowed to pick n balls out of those infy(infinite) balls

 cond is : you must have all k color balls with u
 obviously kn

 Question is how many possibilities for selection you have ?

 Thanks
 Ashish

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Re: [algogeeks] infy color balls

2010-04-30 Thread saurabh gupta
C(n-1,k-1) if balls are similar

On Fri, Apr 30, 2010 at 4:34 PM, Ashish Mishra amishra@gmail.comwrote:

 One of my friend ask me this n i am bad in P  C (will love if smone can
 provide me a link to learn it though)

 nyways prob is:
 there are infy color balls of k different color
 you are allowed to pick n balls out of those infy(infinite) balls

 cond is : you must have all k color balls with u
 obviously kn

 Question is how many possibilities for selection you have ?

 Thanks
 Ashish

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Re: [algogeeks] a google question

2010-04-30 Thread Kishen Das
Basically the index of ( a + b) in the final array will be ceil(( index of a
+ index of b )/2).
Also if there is a clash of indices you just have to compare the values at
those indices and adjust them.
I have run my concept with below two arrays and it works !!!

Arary A: 1 2 3 4 5 6
Array B: 2 3 5 6 8 9



addition of indices   84511611

addition of values (2+9) ( 1+5)   (4+2)   (6+8)   ( 3+3) ( 5 + 9)


values: 11 6 6 14 9 14


Added indices:   4 5  6 8 11 11 ( These are not sorted indices, you will
know the indices of values in the final array right away after looking at
the indices of a and b )
indices/2:  2  3  3 4  6  6

corresponding final values6 6 6 11 14 14

- Kishen Das


On Fri, Apr 30, 2010 at 7:05 AM, divya sweetdivya@gmail.com wrote:

 Given two sorted postive integer arrays A(n) and B(n) (W.L.O.G, let's
 say they are decreasingly sorted), we define a set S = {(a,b) | a \in
 A
 and b \in B}. Obviously there are n^2 elements in S. The value of such
 a pair is defined as Val(a,b) = a + b. Now we want to get the n pairs
 from S with largest values. The tricky part is that we need an O(n)
 algorithm.

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Re: [algogeeks] Finding the mode in a set of integers

2010-04-30 Thread Dheeraj Jain
Here are 3-4 methods to solve

http://geeksforgeeks.org/?p=503

On Sat, Apr 17, 2010 at 5:45 AM, Ashim Kapoor ashimkap...@gmail.com wrote:

 are you referring to the lectures by Dr Naveen Garg ? Or are these lectures
 different? Please clarify.

 Thank you,
 Ashim.


 On Sat, Apr 17, 2010 at 5:43 AM, rahul rai raikra...@gmail.com wrote:

 On 4/16/10, Rohit Saraf rohit.kumar.sa...@gmail.com wrote:
  Just got another O(n) solution.
 
  Find the n/2 th largest element in the array using the Median of Medians
  Selection algorithm.   =? takes O(n)
  That's It !
 
 
  --
  Rohit Saraf
  Second Year Undergraduate,
  Dept. of Computer Science and Engineering
  IIT Bombay
  http://www.cse.iitb.ac.in/~rohitfeb14http://www.cse.iitb.ac.in/%7Erohitfeb14
 
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 Work out the CDEEP Algorithms course lectures . It will give all basics

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[algogeeks] Re: a google question

2010-04-30 Thread banu
Nice question:

1. |A| = |B| i.e I assume their cardinality is equal

2. A(n) = { a1, a2  a3, ...aN} such that a1=a2=a3...=aN
3. B(n) = { b1, b2  b3, ...bN} such that b1=b2=b3...=bN

4. S(n^2) = A x B = {(a1,b1), (a1,b2)(a1,bN),
  (a2,b1), (a2,b2)(a2,bN),
   
  (aN,b1), (aN,b2)(aN,bN)}
  I assume each pair is stored as a sum i.e ai + bi and also that I
have paired them in a way such that we find a pair  ai  bi  for some
i in 1..N and and a(i-1) = b(i-1)

A first observation is that in the worst case, the first 2N numbers in
S will contain the final result of N numbers.
i.e in  (a1,b1), (a1,b2)(a1,bN), (a2,b1), (a2,b2)(a2,bN)

In the best case first N numbers in S will contain the final N numbers
(already sorted in decreasing order)
i.e in  (a1,b1), (a1,b2)(a1,bN)

Now, if we consider again the worst case scenario, then we can first
divide 2N numbers in two groups of size N each and each of this group
is already sorted in decreasing order.
i.e  (a1,b1), (a1,b2)(a1,bN) and (a2,b1), (a2,b2)(a2,bN)

Now we can simply apply Merge Algorithm on these 2 already sorted
arrays of size N each in O(N) time, which solves the problem

I can be wrong only if the the results lie outside first 2N numbers,
which I hope is not the case(then I think its not possible to solve in
O(n) but in O(nlogn)).


On Apr 30, 2:05 pm, divya sweetdivya@gmail.com wrote:
 Given two sorted postive integer arrays A(n) and B(n) (W.L.O.G, let's
 say they are decreasingly sorted), we define a set S = {(a,b) | a \in
 A
 and b \in B}. Obviously there are n^2 elements in S. The value of such
 a pair is defined as Val(a,b) = a + b. Now we want to get the n pairs
 from S with largest values. The tricky part is that we need an O(n)
 algorithm.

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[algogeeks] Re: a google question

2010-04-30 Thread banu
Nice question:

1. |A| = |B| i.e I assume their cardinality is equal

2. A(n) = { a1, a2  a3, ...aN} such that a1=a2=a3...=aN
3. B(n) = { b1, b2  b3, ...bN} such that b1=b2=b3...=bN

4. S(n^2) = A x B = {(a1,b1), (a1,b2)(a1,bN),
  (a2,b1), (a2,b2)(a2,bN),
   
  (aN,b1), (aN,b2)(aN,bN)}

  assuming we have added in a way such that we find a pair  ai  bi,
for some i in 1..N such that a(i-1) = b(i-1)

A first observation is that in the worst case, the first 2N numbers in
S will contain the final result of N numbers.
i.e in  (a1,b1), (a1,b2)(a1,bN), (a2,b1), (a2,b2)(a2,bN)

In the best case first N numbers in S will contain the final N numbers
(already sorted in decreasing order)
i.e in  (a1,b1), (a1,b2)(a1,bN)

Now, if we consider again the worst case scenario, then we can first
divide 2N numbers in two groups of size N each and each of this group
is already sorted in decreasing order.
i.e  (a1,b1), (a1,b2)(a1,bN) and (a2,b1), (a2,b2)(a2,bN)

Now we can simply apply Merge Algorithm on these 2 already sorted
arrays of size N each in O(N) time, which solves the problem

I can be wrong only if the the results lie outside first 2N
numbers(which I hope is not the case).


On Apr 30, 2:05 pm, divya sweetdivya@gmail.com wrote:
 Given two sorted postive integer arrays A(n) and B(n) (W.L.O.G, let's
 say they are decreasingly sorted), we define a set S = {(a,b) | a \in
 A
 and b \in B}. Obviously there are n^2 elements in S. The value of such
 a pair is defined as Val(a,b) = a + b. Now we want to get the n pairs
 from S with largest values. The tricky part is that we need an O(n)
 algorithm.

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[algogeeks] value of n

2010-04-30 Thread Amit Agarwal
how do I compute n from this equation.
n  8lg(n)

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Re: [algogeeks] infy color balls

2010-04-30 Thread mohit ranjan
@Ashish

Check this

http://www.codechef.com/problems/MARBLES/

Mohit Ranjan
Samsung India Software Operations.


On Fri, Apr 30, 2010 at 4:34 PM, Ashish Mishra amishra@gmail.comwrote:

 One of my friend ask me this n i am bad in P  C (will love if smone can
 provide me a link to learn it though)

 nyways prob is:
 there are infy color balls of k different color
 you are allowed to pick n balls out of those infy(infinite) balls

 cond is : you must have all k color balls with u
 obviously kn

 Question is how many possibilities for selection you have ?

 Thanks
 Ashish

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Re: [algogeeks] a google question

2010-04-30 Thread mohit ranjan
oops

Sorry didn't read properly
last algo was for array sorted in ascending order

for this case, just reverse the process

A[n] and B[n] are two array

loop=n, i=0, j=0;

while(loop0)  // for n largest pairs
{
  print A[i]+B[j];  // sum of first index from both array will be
max

  foo = MAX ( A[i+1]+B[j], A[i+1]+B[j+1], A[i]+B[j+1] ) // using DP,
moving forward

  if foo==A[i+1]+B[j]; i++   // only increment A
  if foo==A[i+1]+B[j+1]; i++; j++   // increment both A and B
  if foo==A[i]+B[j+1]; j++  // increment only B
}



Mohit Ranjan
Samsung India Software Operations.


On Fri, Apr 30, 2010 at 8:40 PM, mohit ranjan shoonya.mo...@gmail.comwrote:

 Hi Divya,


 A[n] and B[n] are two array

 loop=n, i=n-1, j=n-1;

 while(loop0)  // for n largest pairs
 {
   print A[i]+B[j];  // sum of last index from both array will be
 max

   foo = MAX ( A[i-1]+B[j], A[i-1]+B[j-1], A[i]+B[j-1] ) // using DP
 moving backward

   if foo=A[i-1]+B[j]; i--   // only reduce A
   if foo=A[i-1]+B[j-1]; i--; j--   // reduce both A and B
   if foo=A[i]+B[j-1]; j--  // reduce only B
 }

 Time: O(n)


 Mohit Ranjan



 On Fri, Apr 30, 2010 at 5:35 PM, divya sweetdivya@gmail.com wrote:

 Given two sorted postive integer arrays A(n) and B(n) (W.L.O.G, let's
 say they are decreasingly sorted), we define a set S = {(a,b) | a \in
 A
 and b \in B}. Obviously there are n^2 elements in S. The value of such
 a pair is defined as Val(a,b) = a + b. Now we want to get the n pairs
 from S with largest values. The tricky part is that we need an O(n)
 algorithm.

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Re: [algogeeks] a google question

2010-04-30 Thread Rajesh Patidar
ignore the previous mail it wrongly send.

On Fri, Apr 30, 2010 at 11:12 PM, Rajesh Patidar patidarc...@gmail.com wrote:
 let consider the list in two different part one traversing list B with
 respect to A and A with B
 (a.len,b.len) is always solution
 a1=a2=a.len

 On Fri, Apr 30, 2010 at 5:35 PM, divya sweetdivya@gmail.com wrote:
 Given two sorted postive integer arrays A(n) and B(n) (W.L.O.G, let's
 say they are decreasingly sorted), we define a set S = {(a,b) | a \in
 A
 and b \in B}. Obviously there are n^2 elements in S. The value of such
 a pair is defined as Val(a,b) = a + b. Now we want to get the n pairs
 from S with largest values. The tricky part is that we need an O(n)
 algorithm.

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Re: [algogeeks] a google question

2010-04-30 Thread mohit ranjan
Hi Divya,


A[n] and B[n] are two array

loop=n, i=n-1, j=n-1;

while(loop0)  // for n largest pairs
{
  print A[i]+B[j];  // sum of last index from both array will be max

  foo = MAX ( A[i-1]+B[j], A[i-1]+B[j-1], A[i]+B[j-1] ) // using DP
moving backward

  if foo=A[i-1]+B[j]; i--   // only reduce A
  if foo=A[i-1]+B[j-1]; i--; j--   // reduce both A and B
  if foo=A[i]+B[j-1]; j--  // reduce only B
}

Time: O(n)


Mohit Ranjan


On Fri, Apr 30, 2010 at 5:35 PM, divya sweetdivya@gmail.com wrote:

 Given two sorted postive integer arrays A(n) and B(n) (W.L.O.G, let's
 say they are decreasingly sorted), we define a set S = {(a,b) | a \in
 A
 and b \in B}. Obviously there are n^2 elements in S. The value of such
 a pair is defined as Val(a,b) = a + b. Now we want to get the n pairs
 from S with largest values. The tricky part is that we need an O(n)
 algorithm.

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 You received this message because you are subscribed to the Google Groups
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 To post to this group, send email to algoge...@googlegroups.com.
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Re: [algogeeks] value of n

2010-04-30 Thread abhijith reddy
binary search on n

On Fri, Apr 30, 2010 at 10:15 PM, Amit Agarwal lifea...@gmail.com wrote:

 how do I compute n from this equation.
 n  8lg(n)

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