Re: [algogeeks] find triplets in an integer array A[] which satisfy condition: a[i]^2 + a[j]^2 = a[k]^2
Thanks everyone; that O(n^2) is an awesome solution.
Regards,
Akash Agrawal
http://tech-queries.blogspot.com/
On Thu, Dec 2, 2010 at 12:21 PM, fenghuang fenghaungyuyi...@gmail.comwrote:
@anoop
when you find some i and j(i j) meet the condition i.e. asq[i] + asq[j]
== asq[k], you can merge the same value without rollback.
in this sense, you are right.
On Thu, Dec 2, 2010 at 2:26 PM, anoop chaurasiya
anoopchaurasi...@gmail.com wrote:
@fenghuang try this array:
a[]={3,3,3,3,3,3,4,4,4,4,4,4,5}
so asq[]={9,9,9,9,9,9,16,16,16,16,16,16,25}
here as u can see the total number of requisite triple pairs are 6*6=36,
in general for above array total number of pairs is (n/2)*(n/2) i.e. n^2/4
where n is the size of the array.
by using O(n) algo and since you are choosing them one by one,u can't
include all of them as they are of order O(n^2).
so removing repetitions is the only option i think..
On Wed, Dec 1, 2010 at 11:37 PM, fenghuang fenghaungyuyi...@gmail.comwrote:
@anoop
in fact, it always work even if there are repeated elements, because they
don't change the decision.
in detail, assume ii, ,jj, kk is one of the answers, then
a[ii]+a[jj]=a[kk]. since the array is sorted, so a[ii-1]+a[jj] = a[kk] and
a[ii] + a[jj+1] = a[kk].
so when you try the pair of 'ii-1' and 'jj', the next step must be
calculate a[ii] + a[jj] as long as a[ii-1]+a[jj] is not equal to a[kk]. the
same to the pair 'ii' and 'jj+1'.
the algorithm is correct and in the whole procedure, repeated elements
don't affect the decision.
I'm sorry for my poor English.
Thank You!
On Thu, Dec 2, 2010 at 12:14 PM, anoop chaurasiya
anoopchaurasi...@gmail.com wrote:
sorry for the interruption,we can make it work even if the elements are
repeated, by removing the duplicacy in linear time(as the array is already
sorted) and taking a count of no. of duplicates in the seperate array.
On Wed, Dec 1, 2010 at 9:37 PM, Senthilnathan Maadasamy
senthilnathan.maadas...@gmail.com wrote:
A small correction to the algorithm above. In Step 3, instead of
finding *any* pair (a,b) such that a+b = x we need to find *all* such
pairs. However the complexity remains the same.
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Re: [algogeeks] find triplets in an integer array A[] which satisfy condition: a[i]^2 + a[j]^2 = a[k]^2
@anoop in fact, it always work even if there are repeated elements, because they don't change the decision. in detail, assume ii, ,jj, kk is one of the answers, then a[ii]+a[jj]=a[kk]. since the array is sorted, so a[ii-1]+a[jj] = a[kk] and a[ii] + a[jj+1] = a[kk]. so when you try the pair of 'ii-1' and 'jj', the next step must be calculate a[ii] + a[jj] as long as a[ii-1]+a[jj] is not equal to a[kk]. the same to the pair 'ii' and 'jj+1'. the algorithm is correct and in the whole procedure, repeated elements don't affect the decision. I'm sorry for my poor English. Thank You! On Thu, Dec 2, 2010 at 12:14 PM, anoop chaurasiya anoopchaurasi...@gmail.com wrote: sorry for the interruption,we can make it work even if the elements are repeated, by removing the duplicacy in linear time(as the array is already sorted) and taking a count of no. of duplicates in the seperate array. On Wed, Dec 1, 2010 at 9:37 PM, Senthilnathan Maadasamy senthilnathan.maadas...@gmail.com wrote: A small correction to the algorithm above. In Step 3, instead of finding *any* pair (a,b) such that a+b = x we need to find *all* such pairs. However the complexity remains the same. -- 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. -- Anoop Chaurasiya CSE (2008-2012) NIT DGP -- 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] find triplets in an integer array A[] which satisfy condition: a[i]^2 + a[j]^2 = a[k]^2
@fenghuang try this array:
a[]={3,3,3,3,3,3,4,4,4,4,4,4,5}
so asq[]={9,9,9,9,9,9,16,16,16,16,16,16,25}
here as u can see the total number of requisite triple pairs are 6*6=36,
in general for above array total number of pairs is (n/2)*(n/2) i.e. n^2/4
where n is the size of the array.
by using O(n) algo and since you are choosing them one by one,u can't
include all of them as they are of order O(n^2).
so removing repetitions is the only option i think..
On Wed, Dec 1, 2010 at 11:37 PM, fenghuang fenghaungyuyi...@gmail.comwrote:
@anoop
in fact, it always work even if there are repeated elements, because they
don't change the decision.
in detail, assume ii, ,jj, kk is one of the answers, then
a[ii]+a[jj]=a[kk]. since the array is sorted, so a[ii-1]+a[jj] = a[kk] and
a[ii] + a[jj+1] = a[kk].
so when you try the pair of 'ii-1' and 'jj', the next step must be
calculate a[ii] + a[jj] as long as a[ii-1]+a[jj] is not equal to a[kk]. the
same to the pair 'ii' and 'jj+1'.
the algorithm is correct and in the whole procedure, repeated elements
don't affect the decision.
I'm sorry for my poor English.
Thank You!
On Thu, Dec 2, 2010 at 12:14 PM, anoop chaurasiya
anoopchaurasi...@gmail.com wrote:
sorry for the interruption,we can make it work even if the elements are
repeated, by removing the duplicacy in linear time(as the array is already
sorted) and taking a count of no. of duplicates in the seperate array.
On Wed, Dec 1, 2010 at 9:37 PM, Senthilnathan Maadasamy
senthilnathan.maadas...@gmail.com wrote:
A small correction to the algorithm above. In Step 3, instead of finding
*any* pair (a,b) such that a+b = x we need to find *all* such pairs.
However the complexity remains the same.
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Anoop Chaurasiya
CSE (2008-2012)
NIT DGP
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Re: [algogeeks] find triplets in an integer array A[] which satisfy condition: a[i]^2 + a[j]^2 = a[k]^2
@anoop
when you find some i and j(i j) meet the condition i.e. asq[i] + asq[j] ==
asq[k], you can merge the same value without rollback.
in this sense, you are right.
On Thu, Dec 2, 2010 at 2:26 PM, anoop chaurasiya anoopchaurasi...@gmail.com
wrote:
@fenghuang try this array:
a[]={3,3,3,3,3,3,4,4,4,4,4,4,5}
so asq[]={9,9,9,9,9,9,16,16,16,16,16,16,25}
here as u can see the total number of requisite triple pairs are 6*6=36,
in general for above array total number of pairs is (n/2)*(n/2) i.e. n^2/4
where n is the size of the array.
by using O(n) algo and since you are choosing them one by one,u can't
include all of them as they are of order O(n^2).
so removing repetitions is the only option i think..
On Wed, Dec 1, 2010 at 11:37 PM, fenghuang fenghaungyuyi...@gmail.comwrote:
@anoop
in fact, it always work even if there are repeated elements, because they
don't change the decision.
in detail, assume ii, ,jj, kk is one of the answers, then
a[ii]+a[jj]=a[kk]. since the array is sorted, so a[ii-1]+a[jj] = a[kk] and
a[ii] + a[jj+1] = a[kk].
so when you try the pair of 'ii-1' and 'jj', the next step must be
calculate a[ii] + a[jj] as long as a[ii-1]+a[jj] is not equal to a[kk]. the
same to the pair 'ii' and 'jj+1'.
the algorithm is correct and in the whole procedure, repeated elements
don't affect the decision.
I'm sorry for my poor English.
Thank You!
On Thu, Dec 2, 2010 at 12:14 PM, anoop chaurasiya
anoopchaurasi...@gmail.com wrote:
sorry for the interruption,we can make it work even if the elements are
repeated, by removing the duplicacy in linear time(as the array is already
sorted) and taking a count of no. of duplicates in the seperate array.
On Wed, Dec 1, 2010 at 9:37 PM, Senthilnathan Maadasamy
senthilnathan.maadas...@gmail.com wrote:
A small correction to the algorithm above. In Step 3, instead of
finding *any* pair (a,b) such that a+b = x we need to find *all* such
pairs. However the complexity remains the same.
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NIT DGP
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