1. Add sum of squares of all numbers in respective groups, if equal goto step 2. 2. XOR all elements of both groups, (if==0) elements are same.
On Aug 19, 9:21 pm, Dave <[email protected]> wrote: > @Nikhil Jindal: What you say is true for 2 numbers, but not for more > than 2. > 1 + 6 + 6 = 2 + 2 + 9 = 13, and 1 * 6 * 6 = 2 * 2 * 9 = 36. > > Dave > > On Aug 19, 6:00 am, Nikhil Jindal <[email protected]> wrote: > > > > > Nikhil's algo is correct if the following is always true: > > > Given: x + y = S, x * y = M > > and x' + y' = S', x' * y' = M' > > > If: S' = S and M' = M, then x = x' and y = y' > > i.e for given sum and product, the elements are unique. > > > On Thu, Aug 19, 2010 at 12:54 PM, Nikhil Agarwal > > <[email protected]>wrote: > > > > @Dave : my algo won't fail for {0,1,-1} and {0,2,-2} . > > > > S1=0;S2=0; > > > M1=-1 and M2 =-4 (excluding 0 multiplication which i had corrected) > > > M1!=M2 there fore it is correct. > > > > Code: > > > > bool check_arrays(vector<int> v1,vector<int> v2){ > > > if(v1.size()!=v2.size()) > > > return 0; > > > if(v1.size()==0&&v2.size()==0) > > > return 1; > > > int sum,product1,product2; > > > sum=0;product1=1;product2=1; > > > for(int i=0;i<v1.size();i++){ > > > sum+=v1[i]; > > > sum-=v2[i]; > > > if(v1[i]!=0) > > > product1*=v1[i]; > > > if(v2[i]!=0) > > > product2*=v2[i]; > > > } > > > if(sum==0&&(product1==product2)) > > > return 1; > > > return 0; > > > } > > > On Thu, Aug 19, 2010 at 11:26 AM, Dave <[email protected]> wrote: > > > >> @Srinivas, Make that: Your algorithm seems to fail on A = {0,1,-2), B > > >> = > > >> (0,2,-3). I was thinking ones-complement arithmetic instead of twos- > > >> complement. > > > >> Dave > > > >> On Aug 18, 11:59 pm, Dave <[email protected]> wrote: > > >> > @Srinivas, Your algorithm seems to fail on A = {0,1,-1), B = > > >> > (0,2,-2). > > > >> > Dave > > > >> > On Aug 18, 10:53 pm, srinivas reddy <[email protected]> wrote: > > > >> > > add one more thing to the solution suggested by nikhil i.e;count the > > >> number > > >> > > of elements in array 1 and number of elements in array2 if these two > > >> values > > >> > > are equal then after follow the algo proposed by nikhil agarwal.. > > > >> > > On Wed, Aug 18, 2010 at 8:50 PM, Rais Khan <[email protected]> > > >> wrote: > > >> > > > @Chonku: Your algo seems to fail with following input. > > >> > > > Arr1[]= {1,6} > > >> > > > Arr2[]={7} > > > >> > > > On Wed, Aug 18, 2010 at 8:42 PM, Rais Khan <[email protected] > > >> >wrote: > > > >> > > >> @Nikhil: Your algo seems to fail with following input. What do you > > >> say? > > >> > > >> Arr1[]= {1,2,3} > > >> > > >> Arr2[]={6} > > > >> > > >> On Wed, Aug 18, 2010 at 7:17 AM, Nikhil Agarwal < > > >> > > >> [email protected]> wrote: > > > >> > > >>> Sum all the elements of both the arrays..let it be s1 and s2 > > >> > > >>> Multiply the elements and call as m1 and m2 > > >> > > >>> if(s1==s2) &&(m1==m2) > > >> > > >>> return 1;else > > >> > > >>> return 0; > > > >> > > >>> O(n) > > > >> > > >>> On Tue, Aug 17, 2010 at 11:33 PM, amit <[email protected]> > > >> wrote: > > > >> > > >>>> Given two arrays of numbers, find if each of the two arrays have > > >> the > > >> > > >>>> same set of integers ? Suggest an algo which can run faster than > > >> NlogN > > >> > > >>>> without extra space? > > > >> > > >>>> -- > > >> > > >>>> You received this message because you are subscribed to the > > >> Google > > >> > > >>>> Groups "Algorithm Geeks" group. > > >> > > >>>> To post to this group, send email to [email protected]. > > >> > > >>>> To unsubscribe from this group, send email to > > >> > > >>>> [email protected]<algogeeks%2bunsubscr...@googlegroups > > >> > > >>>> .com> > > >> <algogeeks%2bunsubscr...@googlegroups.com> > > >> > > >>>> . > > >> > > >>>> For more options, visit this group at > > >> > > >>>>http://groups.google.com/group/algogeeks?hl=en. > > > >> > > >>> -- > > >> > > >>> Thanks & Regards > > >> > > >>> Nikhil Agarwal > > >> > > >>> Senior Undergraduate > > >> > > >>> Computer Science & Engineering, > > >> > > >>> National Institute Of Technology, Durgapur,India > > >> > > >>>http://tech-nikk.blogspot.com > > >> > > >>>http://beta.freshersworld.com/communities/nitd > > > >> > > >>> -- > > >> > > >>> You received this message because you are subscribed to the > > >> > > >>> Google > > >> Groups > > >> > > >>> "Algorithm Geeks" group. > > >> > > >>> To post to this group, send email to [email protected]. > > >> > > >>> To unsubscribe from this group, send email to > > >> > > >>> [email protected]<algogeeks%2bunsubscr...@googlegroups > > >> > > >>> .com> > > >> <algogeeks%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 [email protected]. > > >> > > > To unsubscribe from this group, send email to > > >> > > > [email protected]<algogeeks%2bunsubscr...@googlegroups > > >> > > > .com> > > >> <algogeeks%2bunsubscr...@googlegroups.com> > > >> > > > . > > >> > > > For more options, visit this group at > > >> > > >http://groups.google.com/group/algogeeks?hl=en.-Hidequotedtext - > > > >> > > - Show quoted text -- Hide quoted text - > > > >> > - Show quoted text - > > > >> -- > > >> You received this message because you are subscribed to the Google Groups > > >> "Algorithm Geeks" group. > > >> To post to this group, send email to [email protected]. > > >> To unsubscribe from this group, send email to > > >> [email protected]<algogeeks%2bunsubscr...@googlegroups > > >> .com> > > >> . > > >> For more options, visit this group at > > >>http://groups.google.com/group/algogeeks?hl=en. > > > > -- > > > Thanks & Regards > > > Nikhil Agarwal > > > Senior Undergraduate > > > Computer Science & Engineering, > > > National Institute Of Technology, Durgapur,India > > >http://tech-nikk.blogspot.com > > >http://beta.freshersworld.com/communities/nitd > > > > -- > > > You received this message because you are subscribed to the Google Groups > > > "Algorithm Geeks" group. > > > To post to this group, send email to [email protected]. > > > To unsubscribe from this group, send email to > > > [email protected]<algogeeks%2bunsubscr...@googlegroups > > > .com> > > > . > > > > For more options, visit this group at > > >http://groups.google.com/group/algogeeks?hl=en. > > > Please access the attached hyperlink for an important electronic > > communications > > disclaimer:http://dce.edu/web/Sections/Standalone/Email_Disclaimer.php-Hide > > quoted text - > > > - Show quoted text - -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. 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