@Nikhil Agarwal: You cannot take extra memory and neither the range of
numbers is specified.
Counting will not be a viable option.
On Fri, Aug 20, 2010 at 9:38 AM, Nikhil Agarwal
<nikhil.bhoja...@gmail.com>wrote:
> 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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>> > > > >> > > >>> Thanks & Regards
>> > > > >> > > >>> Nikhil Agarwal
>> > > > >> > > >>> Senior Undergraduate
>> > > > >> > > >>> Computer Science & Engineering,
>> > > > >> > > >>> National Institute Of Technology, Durgapur,India
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>> > > > > Thanks & Regards
>> > > > > Nikhil Agarwal
>> > > > > Senior Undergraduate
>> > > > > Computer Science & Engineering,
>> > > > > National Institute Of Technology, Durgapur,India
>> > > > >http://tech-nikk.blogspot.com
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>
>
> --
> 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
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