Check this new algo: plz provide counter eg.(if any)
step 1 : count the no. of elements,0s,1s,2s,3s in arr1 and arry2.if yes
proceed to step 2 else print fail
step2: if(sum of the elements and product of the non zero elements are same
in both arrays ) print arrays are same else print fail.
On Fri, Aug 20, 2010 at 4:26 AM, Dave <dave_and_da...@juno.com> wrote:
> @Saikat: Rather than challenging everyone to keep coming up with
> counterexamples, please provide a rationale as to why an algorithm
> such as this should work. It looks as if you have two equations in 2N
> unknowns and are trying to assert that the only solution is A_1 =
> B_1,
> A_2 = B_2, etc. (where I have assumed that each array is sorted).
> Usually, it takes 2N equations to determine 2N unknowns, although
> other information about the solutions can lessen that number in
> certain circumstances.
>
> At least if you are going to propose something, do so only after you
> have tested it on all of the combinations of up to four numbers
> between -5 and 5.
>
> Dave
>
> On Aug 19, 11:52 am, Saikat Debnath <saikat....@gmail.com> wrote:
> > 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 <dave_and_da...@juno.com> 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 <fundoon...@yahoo.co.in> 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
> > > > <nikhil.bhoja...@gmail.com>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 <dave_and_da...@juno.com>
> 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 <dave_and_da...@juno.com> 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 <srinivaseev...@gmail.com>
> 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 <
> raiskhan.i...@gmail.com>
> > > > >> 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 <
> raiskhan.i...@gmail.com
> > > > >> >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 <
> > > > >> > > >> nikhil.bhoja...@gmail.com> 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 <
> amitjaspal...@gmail.com>
> > > > >> 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?
> >
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> > > > > National Institute Of Technology, Durgapur,India
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Senior Undergraduate
Computer Science & Engineering,
National Institute Of Technology, Durgapur,India
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