@Dave: I read the question correctly and for A = { 1 , 2} B = { 2 , 1}
the output is correct.
Maybe I didn't explain the steps correctly. This is the code:
for(int i = 0 ; i < arr1.Length ; i++)
{
arr1XOR ^= arr1[i];
arr1XOR ^= i;
arr1SUM += arr1[i];
arr1MUL *= arr1[i];
}
for (int i = 0; i < arr2.Length; i++)
{
arr2XOR ^= arr2[i];
arr2XOR ^= i;
arr2SUM += arr2[i];
arr2MUL *= arr2[i];
}
if(arr1XOR == arr2XOR && arr1SUM == arr2SUM && arr1MUL ==
arr2MUL)
{
//SAME VALUES - IDENTICAL ARRAYS
}
else
{
//NOT IDENTICAL
}
Please correct me if I'm wrong.
Marius.
On Aug 22, 3:45 am, Dave <dave_and_da...@juno.com> wrote:
> @UMarius: A = {1,2}, B = {2,1} fails your test. If you reread the
> original problem, you see that the question is not whether the arrays
> are identical (which is easily determined by simply comparing them
> element-by-element in O(n)), but whether they contain the same values,
> possibly in a different order.
>
> Dave
>
> On Aug 21, 11:01 am, UMarius <mar...@aseara.ro> wrote:
>
>
>
>
>
>
>
> > What about this?
>
> > 1. xor all elements of each array and their corresponding indexes &
> > sum all the elements of each array & mul all elements of each array
> > 2. if all results are the same then the arrays are identical
>
> > Nice to "meet" you all, I just joined and this is my first post :) ...
>
> > > 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?- 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 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.
