Re: [algogeeks] a google question
@Algoose Chase
point taken
Thanks
Mohit Ranjan
Samsung India Software Operations.
On Sat, May 1, 2010 at 8:43 PM, Algoose Chase harishp...@gmail.com wrote:
@mohit
The idea of DP is fine.
When you find the Max i dont think you need to include A[i+1]+B[j+1]
because it can never be greater than both A[i+1]+B[j] and A[i]+B[j+1] since
both the lists are sorted in decreasing order.
On Fri, Apr 30, 2010 at 8:47 PM, mohit ranjan shoonya.mo...@gmail.comwrote:
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] 400!
wat is tail recursion plz explan in detail On 2 May 2010 08:15, Rohit Saraf rohit.kumar.sa...@gmail.com wrote: @divya use tail recursion and rest should be fine.. -- -- 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 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.comalgogeeks%2bunsubscr...@googlegroups.com . For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] 400!
google it... u will gt it i am on mobile... cannot explain now.. On 5/2/10, divya jain sweetdivya@gmail.com wrote: wat is tail recursion plz explan in detail On 2 May 2010 08:15, Rohit Saraf rohit.kumar.sa...@gmail.com wrote: @divya use tail recursion and rest should be fine.. -- -- 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 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.comalgogeeks%2bunsubscr...@googlegroups.com . For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- -- Rohit Saraf Second Year Undergraduate, Dept. of Computer Science and Engineering IIT Bombay http://www.cse.iitb.ac.in/~rohitfeb14 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] a google question
@divya You're rite. Post a solution if you have one.
--
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Vignesh
On 2 May 2010 13:14, divya jain sweetdivya@gmail.com wrote:
@Mohit
according to ur algo if a[1], b[0] has sum greater than a[0],b[1]
then i is incremented i is now 2 so for next iteration u ll compare a[2]
b[0] and a[1] b1]. but what about a[0] b[1] this pair s lost forever. think
for ths case also.
On 2 May 2010 11:22, mohit ranjan shoonya.mo...@gmail.com wrote:
@Algoose Chase
point taken
Thanks
Mohit Ranjan
Samsung India Software Operations.
On Sat, May 1, 2010 at 8:43 PM, Algoose Chase harishp...@gmail.comwrote:
@mohit
The idea of DP is fine.
When you find the Max i dont think you need to include A[i+1]+B[j+1]
because it can never be greater than both A[i+1]+B[j] and A[i]+B[j+1] since
both the lists are sorted in decreasing order.
On Fri, Apr 30, 2010 at 8:47 PM, mohit ranjan
shoonya.mo...@gmail.comwrote:
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.comwrote:
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] 400!
I agree with abhijith. But given some very large x for which i would have to find factorial. I would either (i) use gmp in cpp or BigInteger or java if its not a lab exercise or an interview (ii) use simple brute multiplication algorithm. The second approach requires (a) The no. of digits in n! for making storage available (b) The calculation itself which I would brute force References: http://inder-gnu.blogspot.com/2009/08/find-number-of-digits-in-factorial-of.html http://stackoverflow.com/questions/1113167/can-one-know-how-large-a-factorial-would-be-before-calculating-it http://delphiforfun.org/programs/big_factorials.htm On 2 May 2010 13:59, Rohit Saraf rohit.kumar.sa...@gmail.com wrote: google it... u will gt it i am on mobile... cannot explain now.. On 5/2/10, divya jain sweetdivya@gmail.com wrote: wat is tail recursion plz explan in detail On 2 May 2010 08:15, Rohit Saraf rohit.kumar.sa...@gmail.com wrote: @divya use tail recursion and rest should be fine.. -- -- 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 http://www.cse.iitb.ac.in/%7Erohitfeb14 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.comalgogeeks%2bunsubscr...@googlegroups.com algogeeks%2bunsubscr...@googlegroups.comalgogeeks%252bunsubscr...@googlegroups.com . For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.comalgogeeks%2bunsubscr...@googlegroups.com . For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- -- 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 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.comalgogeeks%2bunsubscr...@googlegroups.com . For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- There are two kinds of people. Those who care for others and The others -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] a google question
This problem has been discussed before @
http://groups.google.co.in/group/algogeeks/browse_thread/thread/9c1e1aa8cf1ed437/d451cd9468d985f7
I believe, the problem can't be solved in O(n) but only in O(nlog n).
@Divya: Are you sure the interviewer explicitly asked for an O(n) time
algorithm?
On Sun, May 2, 2010 at 1:48 PM, vignesh radhakrishnan
rvignesh1...@gmail.com wrote:
@divya You're rite. Post a solution if you have one.
--
Regards,
Vignesh
On 2 May 2010 13:14, divya jain sweetdivya@gmail.com wrote:
@Mohit
according to ur algo if a[1], b[0] has sum greater than a[0],b[1]
then i is incremented i is now 2 so for next iteration u ll compare a[2]
b[0] and a[1] b1]. but what about a[0] b[1] this pair s lost forever. think
for ths case also.
On 2 May 2010 11:22, mohit ranjan shoonya.mo...@gmail.com wrote:
@Algoose Chase
point taken
Thanks
Mohit Ranjan
Samsung India Software Operations.
On Sat, May 1, 2010 at 8:43 PM, Algoose Chase harishp...@gmail.comwrote:
@mohit
The idea of DP is fine.
When you find the Max i dont think you need to include A[i+1]+B[j+1]
because it can never be greater than both A[i+1]+B[j] and A[i]+B[j+1] since
both the lists are sorted in decreasing order.
On Fri, Apr 30, 2010 at 8:47 PM, mohit ranjan
shoonya.mo...@gmail.comwrote:
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.com
wrote:
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.comwrote:
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
Hi
will this work ?
since we need Set S with n pairs of largest values , any such pair within
the set would (always) contain A[0] or B[0] because they maximize the value
of the pair.
so
// code snippet
typedef std::vectorint,int Pairs;
Pairs.push(A[0],B[0])
int i = 1; // index for ListA
int j = 1; // index for list B
count = 1;
while( count = n)
{
if( A[0] + B[j] B[0] + A[i] )
{
Pairs.push(A[0],B[j])
j++;
}
else
{
Pairs.Add(A[i],B[0])
i++;
}
count++;
}
since there are n items in both the lists, j and i wont go beyond n in the
while loop
On Sun, May 2, 2010 at 3:44 PM, Shishir Mittal 1987.shis...@gmail.comwrote:
This problem has been discussed before @
http://groups.google.co.in/group/algogeeks/browse_thread/thread/9c1e1aa8cf1ed437/d451cd9468d985f7
I believe, the problem can't be solved in O(n) but only in O(nlog n).
@Divya: Are you sure the interviewer explicitly asked for an O(n) time
algorithm?
On Sun, May 2, 2010 at 1:48 PM, vignesh radhakrishnan
rvignesh1...@gmail.com wrote:
@divya You're rite. Post a solution if you have one.
--
Regards,
Vignesh
On 2 May 2010 13:14, divya jain sweetdivya@gmail.com wrote:
@Mohit
according to ur algo if a[1], b[0] has sum greater than a[0],b[1]
then i is incremented i is now 2 so for next iteration u ll compare
a[2] b[0] and a[1] b1]. but what about a[0] b[1] this pair s lost forever.
think for ths case also.
On 2 May 2010 11:22, mohit ranjan shoonya.mo...@gmail.com wrote:
@Algoose Chase
point taken
Thanks
Mohit Ranjan
Samsung India Software Operations.
On Sat, May 1, 2010 at 8:43 PM, Algoose Chase harishp...@gmail.comwrote:
@mohit
The idea of DP is fine.
When you find the Max i dont think you need to include A[i+1]+B[j+1]
because it can never be greater than both A[i+1]+B[j] and A[i]+B[j+1]
since
both the lists are sorted in decreasing order.
On Fri, Apr 30, 2010 at 8:47 PM, mohit ranjan shoonya.mo...@gmail.com
wrote:
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.com wrote:
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.comwrote:
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] Where does OS scheduling run??
@ Pradeep *CPU stop its current processing and goes to the interrupt subroutine* you have mentioned that the CPU stops its current processing and goes to the interrupt subroutine.. My Question is how does the CPU stops its execution(any special hardware involved) because it is busy in executing the current instruction. With Regards, Prabagaran. On Sun, May 2, 2010 at 4:02 AM, pradeep verma ppradeep...@gmail.com wrote: lets suppose Processor executing a instruction(process1) and another process2 tries to take the control of CPU so inorder to inform CPU it has to interrupt the CPU right now we know that if interrupt comes CPU stop its current processing and goes to the interrupt subroutine...now CPU knows that its a pre-emption interrupt so CPU first run its short term scheduler(this will inform CPU that the interruting process priority is less or greater ..n if greater than CPU goes to previous process1 preempt it and start executing higher priority process2 ) I think its clear Regards Pradeep On Sun, May 2, 2010 at 3:06 AM, praba garan prabagara...@gmail.comwrote: @ Guillermo Garcia The link gives the overall abstract idea. I am talking in register level. When a user process executes 1. PC program counter will contain the address of the next instruction in user code. 2. Processor registers(accumulator ...) contain the current instruction data. Then where does the interrupt actually arrives?? And by that time the user process the control, then who does the preempting and how?? With Regards, Prabagaran. On Sun, May 2, 2010 at 2:35 AM, Guillermo Garcia gegarci...@gmail.comwrote: read here - http://en.wikipedia.org/wiki/Preemption_%28computing%29 Time slice The period of time for which a process is allowed to run in a preemptive multitasking system is generally called the *time slice*, or *quantum*. The scheduler is run once every time slice to choose the next process to run. If the time slice is too short then the scheduler will consume too much processing time. An interrupt http://en.wikipedia.org/wiki/Interrupt is scheduled to allow the operating systemhttp://en.wikipedia.org/wiki/Operating_system kernel http://en.wikipedia.org/wiki/Kernel_%28computer_science%29 to switch between processes when their time slices expire, effectively allowing the processor’s time to be shared between a number of tasks, giving the illusion that it is dealing with these tasks simultaneously, or concurrently. The operating system which controls such a design is called a multi-tasking system. On Sat, May 1, 2010 at 5:26 PM, praba garan prabagara...@gmail.comwrote: @ Guillermo Garcia Suppose a user program is executing and and clock interrupt arrives.. Then who receives the interrupt?? Can you xplain me the clock interrupt(like any hardwares involved) bit detailed?? With Regards, Prabagaran. On Sun, May 2, 2010 at 1:38 AM, Guillermo Garcia gegarci...@gmail.comwrote: The scheduler takes control with a clock interruption. Then it analyzes if it has to preempt or not the running task. On Sat, May 1, 2010 at 5:00 PM, praba garan prabagara...@gmail.comwrote: Hi all, I have a doubt in OS. The scheduler does the process of preemption. And one processor can run atmost 1 instruction at a time. Then how where does the scheduler run?? With Regards, Prabagaran. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.comalgogeeks%2bunsubscr...@googlegroups.com . For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.comalgogeeks%2bunsubscr...@googlegroups.com . For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.comalgogeeks%2bunsubscr...@googlegroups.com . For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.comalgogeeks%2bunsubscr...@googlegroups.com . For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message
[algogeeks] Re: a google question
This problem can be simplified to a variation of merge part of merge
sort algorithm.
- Break the set S having n^2 values into n individual arrays of length
n, say S1, S2,..Sn, where S1 = [a1+b1, a1+b2,...a1+bn].
- One can observe that each Si has entries which are themselves sorted
within Si.
- Perform merge of S1, S2,..Sn where take the largest values of each
Si, and create a heap of these individual max values.
- In each step, return the max value from heap and then add the next
max value from the Si that had the current max value and add it in the
max-heap.
- Repeat this step n times and then exit.
Time: O(n).
Satish
On May 2, 5:41 pm, Algoose Chase harishp...@gmail.com wrote:
Hi
will this work ?
since we need Set S with n pairs of largest values , any such pair within
the set would (always) contain A[0] or B[0] because they maximize the value
of the pair.
so
// code snippet
typedef std::vectorint,int Pairs;
Pairs.push(A[0],B[0])
int i = 1; // index for ListA
int j = 1; // index for list B
count = 1;
while( count = n)
{
if( A[0] + B[j] B[0] + A[i] )
{
Pairs.push(A[0],B[j])
j++;
}
else
{
Pairs.Add(A[i],B[0])
i++;
}
count++;
}
since there are n items in both the lists, j and i wont go beyond n in the
while loop
On Sun, May 2, 2010 at 3:44 PM, Shishir Mittal 1987.shis...@gmail.comwrote:
This problem has been discussed before @
http://groups.google.co.in/group/algogeeks/browse_thread/thread/9c1e1...
I believe, the problem can't be solved in O(n) but only in O(nlog n).
@Divya: Are you sure the interviewer explicitly asked for an O(n) time
algorithm?
On Sun, May 2, 2010 at 1:48 PM, vignesh radhakrishnan
rvignesh1...@gmail.com wrote:
@divya You're rite. Post a solution if you have one.
--
Regards,
Vignesh
On 2 May 2010 13:14, divya jain sweetdivya@gmail.com wrote:
@Mohit
according to ur algo if a[1], b[0] has sum greater than a[0],b[1]
then i is incremented i is now 2 so for next iteration u ll compare
a[2] b[0] and a[1] b1]. but what about a[0] b[1] this pair s lost forever.
think for ths case also.
On 2 May 2010 11:22, mohit ranjan shoonya.mo...@gmail.com wrote:
@Algoose Chase
point taken
Thanks
Mohit Ranjan
Samsung India Software Operations.
On Sat, May 1, 2010 at 8:43 PM, Algoose Chase
harishp...@gmail.comwrote:
@mohit
The idea of DP is fine.
When you find the Max i dont think you need to include A[i+1]+B[j+1]
because it can never be greater than both A[i+1]+B[j] and A[i]+B[j+1]
since
both the lists are sorted in decreasing order.
On Fri, Apr 30, 2010 at 8:47 PM, mohit ranjan shoonya.mo...@gmail.com
wrote:
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.com wrote:
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.comwrote:
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] tree from linked list
u are given a sorted lnked list construct a balanced binary search tree from it. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] a google question
i found this question on some site and it was mentioned there todo in o(n).
i dont have the solution
@ above
ur solution doesnt include the case of ncluding a[i] b[j] it takes only a[i]
b[0] or b[j] a[0]
On 2 May 2010 18:11, Algoose Chase harishp...@gmail.com wrote:
Hi
will this work ?
since we need Set S with n pairs of largest values , any such pair within
the set would (always) contain A[0] or B[0] because they maximize the value
of the pair.
so
// code snippet
typedef std::vectorint,int Pairs;
Pairs.push(A[0],B[0])
int i = 1; // index for ListA
int j = 1; // index for list B
count = 1;
while( count = n)
{
if( A[0] + B[j] B[0] + A[i] )
{
Pairs.push(A[0],B[j])
j++;
}
else
{
Pairs.Add(A[i],B[0])
i++;
}
count++;
}
since there are n items in both the lists, j and i wont go beyond n in the
while loop
On Sun, May 2, 2010 at 3:44 PM, Shishir Mittal 1987.shis...@gmail.comwrote:
This problem has been discussed before @
http://groups.google.co.in/group/algogeeks/browse_thread/thread/9c1e1aa8cf1ed437/d451cd9468d985f7
I believe, the problem can't be solved in O(n) but only in O(nlog n).
@Divya: Are you sure the interviewer explicitly asked for an O(n) time
algorithm?
On Sun, May 2, 2010 at 1:48 PM, vignesh radhakrishnan
rvignesh1...@gmail.com wrote:
@divya You're rite. Post a solution if you have one.
--
Regards,
Vignesh
On 2 May 2010 13:14, divya jain sweetdivya@gmail.com wrote:
@Mohit
according to ur algo if a[1], b[0] has sum greater than a[0],b[1]
then i is incremented i is now 2 so for next iteration u ll compare
a[2] b[0] and a[1] b1]. but what about a[0] b[1] this pair s lost forever.
think for ths case also.
On 2 May 2010 11:22, mohit ranjan shoonya.mo...@gmail.com wrote:
@Algoose Chase
point taken
Thanks
Mohit Ranjan
Samsung India Software Operations.
On Sat, May 1, 2010 at 8:43 PM, Algoose Chase harishp...@gmail.comwrote:
@mohit
The idea of DP is fine.
When you find the Max i dont think you need to include A[i+1]+B[j+1]
because it can never be greater than both A[i+1]+B[j] and A[i]+B[j+1]
since
both the lists are sorted in decreasing order.
On Fri, Apr 30, 2010 at 8:47 PM, mohit ranjan
shoonya.mo...@gmail.com wrote:
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.com wrote:
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.comwrote:
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] 400!
I think challenge here is not the Execution time, but the storage. 300 ! or 400! should generally go beyond the storage capabilities of long long ints in cpp. @ Rohit Saraf: Hence, I don't know if even tail recursion will ultimately be able to store the output. I think Rajesh Patidar's answer fits the bill well, in terms of storage. On Sun, May 2, 2010 at 2:23 PM, vignesh radhakrishnan rvignesh1...@gmail.com wrote: I agree with abhijith. But given some very large x for which i would have to find factorial. I would either (i) use gmp in cpp or BigInteger or java if its not a lab exercise or an interview (ii) use simple brute multiplication algorithm. The second approach requires (a) The no. of digits in n! for making storage available (b) The calculation itself which I would brute force References: http://inder-gnu.blogspot.com/2009/08/find-number-of-digits-in-factorial-of.html http://stackoverflow.com/questions/1113167/can-one-know-how-large-a-factorial-would-be-before-calculating-it http://delphiforfun.org/programs/big_factorials.htm On 2 May 2010 13:59, Rohit Saraf rohit.kumar.sa...@gmail.com wrote: google it... u will gt it i am on mobile... cannot explain now.. On 5/2/10, divya jain sweetdivya@gmail.com wrote: wat is tail recursion plz explan in detail On 2 May 2010 08:15, Rohit Saraf rohit.kumar.sa...@gmail.com wrote: @divya use tail recursion and rest should be fine.. -- -- 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 http://www.cse.iitb.ac.in/%7Erohitfeb14 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.comalgogeeks%2bunsubscr...@googlegroups.com algogeeks%2bunsubscr...@googlegroups.comalgogeeks%252bunsubscr...@googlegroups.com . For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.comalgogeeks%2bunsubscr...@googlegroups.com . For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- -- 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 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.comalgogeeks%2bunsubscr...@googlegroups.com . For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- There are two kinds of people. Those who care for others and The others -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.comalgogeeks%2bunsubscr...@googlegroups.com . For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] a google question
OOPs.. sorry
this doesnt work !
On Sun, May 2, 2010 at 6:11 PM, Algoose Chase harishp...@gmail.com wrote:
Hi
will this work ?
since we need Set S with n pairs of largest values , any such pair within
the set would (always) contain A[0] or B[0] because they maximize the value
of the pair.
so
// code snippet
typedef std::vectorint,int Pairs;
Pairs.push(A[0],B[0])
int i = 1; // index for ListA
int j = 1; // index for list B
count = 1;
while( count = n)
{
if( A[0] + B[j] B[0] + A[i] )
{
Pairs.push(A[0],B[j])
j++;
}
else
{
Pairs.Add(A[i],B[0])
i++;
}
count++;
}
since there are n items in both the lists, j and i wont go beyond n in the
while loop
On Sun, May 2, 2010 at 3:44 PM, Shishir Mittal 1987.shis...@gmail.comwrote:
This problem has been discussed before @
http://groups.google.co.in/group/algogeeks/browse_thread/thread/9c1e1aa8cf1ed437/d451cd9468d985f7
I believe, the problem can't be solved in O(n) but only in O(nlog n).
@Divya: Are you sure the interviewer explicitly asked for an O(n) time
algorithm?
On Sun, May 2, 2010 at 1:48 PM, vignesh radhakrishnan
rvignesh1...@gmail.com wrote:
@divya You're rite. Post a solution if you have one.
--
Regards,
Vignesh
On 2 May 2010 13:14, divya jain sweetdivya@gmail.com wrote:
@Mohit
according to ur algo if a[1], b[0] has sum greater than a[0],b[1]
then i is incremented i is now 2 so for next iteration u ll compare
a[2] b[0] and a[1] b1]. but what about a[0] b[1] this pair s lost forever.
think for ths case also.
On 2 May 2010 11:22, mohit ranjan shoonya.mo...@gmail.com wrote:
@Algoose Chase
point taken
Thanks
Mohit Ranjan
Samsung India Software Operations.
On Sat, May 1, 2010 at 8:43 PM, Algoose Chase harishp...@gmail.comwrote:
@mohit
The idea of DP is fine.
When you find the Max i dont think you need to include A[i+1]+B[j+1]
because it can never be greater than both A[i+1]+B[j] and A[i]+B[j+1]
since
both the lists are sorted in decreasing order.
On Fri, Apr 30, 2010 at 8:47 PM, mohit ranjan
shoonya.mo...@gmail.com wrote:
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.com wrote:
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.comwrote:
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
Here is a solution of O(n) , taking 4 pointers 2 for each array
#include cstdio
#includeiostream
using namespace std;
#define N 10
int main(void)
{
int arr1[N] = {8,7,4,3,2,1,1,1,1,1};
int arr2[N] = {34,23,21,19,15,13,11,8,4,2};
int *p11,*p12,*p21,*p22;
p11 = p12 = arr1;
p21 = p22 = arr2;
int f1;
f1 = 0;
for(int i=0;iN;i++) {
int ans=0;
int a,b,c,d;
a = *p11 + *p21;
b = *p11 + *p22;
c = *p21 + *p12;
d = *(p11+1) + *(p21+1);
//printf(a=%d b=%d c=%d d=%d\n,a,b,c,d); //debug
if(f1==0)ans = a ,p12++ , p22++ , f1=1;
else if(b = c b = d )ans = b , p22++ ;
else if(c = b c = d )ans = c , p12++ ;
elseans = d , p11++ , p21++ ,printf(4 );
printf(%d\n,ans);
}
}
Regards
Jitendra Kushwaha
Undergradute Student
Computer Science Eng.
MNNIT, Allahabad
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Re: [algogeeks] Where does OS scheduling run??
@Prabagaran Execution of an instruction, by the CPU, is an ATOMIC operation. Interrupts, if any, will be processed after the execution of the current instruction. CPU has interrupt pins attached to it. Whenever an interrupt occurs, the CPU will be informed about the interrupt through these hardwired pins. That's how the CPU will get to know the occurrence of an interrupt. For now, please assume that somehow, these pins will be set to the corresponding values, if any interrupt occurs. Please refer to the Intel's Manual for further details on Interrupt Handling. Now, when a timer interrupt occurs, the kernel will get to know that the time quantum assigned to the current process has expired and then it loads the next waiting process to execute on the CPU. Hope this helps. -- CM Saikanth Varma On Sun, May 2, 2010 at 10:25 PM, praba garan prabagara...@gmail.com wrote: @ Pradeep *CPU stop its current processing and goes to the interrupt subroutine* you have mentioned that the CPU stops its current processing and goes to the interrupt subroutine.. My Question is how does the CPU stops its execution(any special hardware involved) because it is busy in executing the current instruction. With Regards, Prabagaran. On Sun, May 2, 2010 at 4:02 AM, pradeep verma ppradeep...@gmail.comwrote: lets suppose Processor executing a instruction(process1) and another process2 tries to take the control of CPU so inorder to inform CPU it has to interrupt the CPU right now we know that if interrupt comes CPU stop its current processing and goes to the interrupt subroutine...now CPU knows that its a pre-emption interrupt so CPU first run its short term scheduler(this will inform CPU that the interruting process priority is less or greater ..n if greater than CPU goes to previous process1 preempt it and start executing higher priority process2 ) I think its clear Regards Pradeep On Sun, May 2, 2010 at 3:06 AM, praba garan prabagara...@gmail.comwrote: @ Guillermo Garcia The link gives the overall abstract idea. I am talking in register level. When a user process executes 1. PC program counter will contain the address of the next instruction in user code. 2. Processor registers(accumulator ...) contain the current instruction data. Then where does the interrupt actually arrives?? And by that time the user process the control, then who does the preempting and how?? With Regards, Prabagaran. On Sun, May 2, 2010 at 2:35 AM, Guillermo Garcia gegarci...@gmail.comwrote: read here - http://en.wikipedia.org/wiki/Preemption_%28computing%29 Time slice The period of time for which a process is allowed to run in a preemptive multitasking system is generally called the *time slice*, or *quantum*. The scheduler is run once every time slice to choose the next process to run. If the time slice is too short then the scheduler will consume too much processing time. An interrupt http://en.wikipedia.org/wiki/Interrupt is scheduled to allow the operating systemhttp://en.wikipedia.org/wiki/Operating_system kernel http://en.wikipedia.org/wiki/Kernel_%28computer_science%29 to switch between processes when their time slices expire, effectively allowing the processor’s time to be shared between a number of tasks, giving the illusion that it is dealing with these tasks simultaneously, or concurrently. The operating system which controls such a design is called a multi-tasking system. On Sat, May 1, 2010 at 5:26 PM, praba garan prabagara...@gmail.comwrote: @ Guillermo Garcia Suppose a user program is executing and and clock interrupt arrives.. Then who receives the interrupt?? Can you xplain me the clock interrupt(like any hardwares involved) bit detailed?? With Regards, Prabagaran. On Sun, May 2, 2010 at 1:38 AM, Guillermo Garcia gegarci...@gmail.com wrote: The scheduler takes control with a clock interruption. Then it analyzes if it has to preempt or not the running task. On Sat, May 1, 2010 at 5:00 PM, praba garan prabagara...@gmail.comwrote: Hi all, I have a doubt in OS. The scheduler does the process of preemption. And one processor can run atmost 1 instruction at a time. Then how where does the scheduler run?? With Regards, Prabagaran. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.comalgogeeks%2bunsubscr...@googlegroups.com . For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.comalgogeeks%2bunsubscr...@googlegroups.com . For
Re: [algogeeks] Where does OS scheduling run??
we know that there are many pins available in microprocessor chips one of them is INTR(interrupt Req) When a CPU receives an Interrupt Request (IRQ), it first checks if it must react to the interrupt. So-called Maskable Interrupts allow a programmer to specify that the CPU does ignore it, while Non-Maskeable Interrupt requests must be serviced. now if a process wants a control of CPU it sends a positive signal on INTR pin this will interrupt CPU n this is how CPU is being interrupted. after that CPU stops its current processing I think now its clear Regards Pradeep On Sun, May 2, 2010 at 10:25 PM, praba garan prabagara...@gmail.com wrote: @ Pradeep *CPU stop its current processing and goes to the interrupt subroutine* you have mentioned that the CPU stops its current processing and goes to the interrupt subroutine.. My Question is how does the CPU stops its execution(any special hardware involved) because it is busy in executing the current instruction. With Regards, Prabagaran. On Sun, May 2, 2010 at 4:02 AM, pradeep verma ppradeep...@gmail.comwrote: lets suppose Processor executing a instruction(process1) and another process2 tries to take the control of CPU so inorder to inform CPU it has to interrupt the CPU right now we know that if interrupt comes CPU stop its current processing and goes to the interrupt subroutine...now CPU knows that its a pre-emption interrupt so CPU first run its short term scheduler(this will inform CPU that the interruting process priority is less or greater ..n if greater than CPU goes to previous process1 preempt it and start executing higher priority process2 ) I think its clear Regards Pradeep On Sun, May 2, 2010 at 3:06 AM, praba garan prabagara...@gmail.comwrote: @ Guillermo Garcia The link gives the overall abstract idea. I am talking in register level. When a user process executes 1. PC program counter will contain the address of the next instruction in user code. 2. Processor registers(accumulator ...) contain the current instruction data. Then where does the interrupt actually arrives?? And by that time the user process the control, then who does the preempting and how?? With Regards, Prabagaran. On Sun, May 2, 2010 at 2:35 AM, Guillermo Garcia gegarci...@gmail.comwrote: read here - http://en.wikipedia.org/wiki/Preemption_%28computing%29 Time slice The period of time for which a process is allowed to run in a preemptive multitasking system is generally called the *time slice*, or *quantum*. The scheduler is run once every time slice to choose the next process to run. If the time slice is too short then the scheduler will consume too much processing time. An interrupt http://en.wikipedia.org/wiki/Interrupt is scheduled to allow the operating systemhttp://en.wikipedia.org/wiki/Operating_system kernel http://en.wikipedia.org/wiki/Kernel_%28computer_science%29 to switch between processes when their time slices expire, effectively allowing the processor’s time to be shared between a number of tasks, giving the illusion that it is dealing with these tasks simultaneously, or concurrently. The operating system which controls such a design is called a multi-tasking system. On Sat, May 1, 2010 at 5:26 PM, praba garan prabagara...@gmail.comwrote: @ Guillermo Garcia Suppose a user program is executing and and clock interrupt arrives.. Then who receives the interrupt?? Can you xplain me the clock interrupt(like any hardwares involved) bit detailed?? With Regards, Prabagaran. On Sun, May 2, 2010 at 1:38 AM, Guillermo Garcia gegarci...@gmail.com wrote: The scheduler takes control with a clock interruption. Then it analyzes if it has to preempt or not the running task. On Sat, May 1, 2010 at 5:00 PM, praba garan prabagara...@gmail.comwrote: Hi all, I have a doubt in OS. The scheduler does the process of preemption. And one processor can run atmost 1 instruction at a time. Then how where does the scheduler run?? With Regards, Prabagaran. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.comalgogeeks%2bunsubscr...@googlegroups.com . For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.comalgogeeks%2bunsubscr...@googlegroups.com . For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send
Re: [algogeeks] Where does OS scheduling run??
although CPU is busy in exexcution...it check's its registers values for the pending interrupts .. if any interrupt is pending at the end of the current CPU cycle...it shedules the interrupt handler to further execute the interrupt subroutine... -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] What happens when you double click a file??
What happens when you double click a file?? (changes in Kernel DATA STRUCTURES and interrupts) With Regards, Prabagaran. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
