may be we can add condition that sum of squares and cubes be same.
On Sun, Jul 3, 2011 at 1:09 PM, Deoki Nandan <deok...@gmail.com> wrote:
> there is no possible solution for this question in less than O(nlgn) time.
> As by theorem given in cormen and solution is possible using xor
>
>
> On Sun, Jul 3, 2011 at 2:27 PM, Sandeep Jain <sandeep6...@gmail.com>wrote:
>
>> For case1) yes XOR works,
>> for Well, for the other two cases hash-maps may come in handy. :)
>>
>>
>> Regards,
>> Sandeep Jain
>>
>>
>>
>>
>>
>> On Sun, Jul 3, 2011 at 1:48 PM, sunny agrawal <sunny816.i...@gmail.com>wrote:
>>
>>> But i don't think xor method will work at all for all of the cases above
>>> you mentioned.
>>> setA = {4,7}
>>> setB = {5,6}
>>>
>>> -> all numbers in both set are nonzero and distinct
>>> -> all numbers are in some range :D
>>> and for character parts it will similarly fail....by taking character set
>>> of ascii values 4,5,6,7
>>>
>>> and about general solution i dont know how to do it in O(n)
>>> one thing i was thinking of goes this way, taking arrays A and B instead
>>> of sets.
>>> if we can prove these polynomial to be same in O(n) time.
>>> (x-a[0])*(x-a[1])*.................*(x-a[n-1]) ==
>>> (x-b[0])*(x-b[1])*.........(x-b[n-1])
>>> dont know if it can be done efficienty
>>>
>>>
>>> On Sun, Jul 3, 2011 at 1:25 PM, Sandeep Jain <sandeep6...@gmail.com>wrote:
>>>
>>>> Agreed, BUT if you don't add a stipulation. You won't be able to reduce
>>>> the complexity.
>>>> For a 100% general solution, I don't think you can reduce the complexity
>>>> more than O(nLgn.)
>>>> There are variations of this question:
>>>> --> All numbers are non-zero and distinct.
>>>> --> All numbers belong to given range
>>>> --> You can also have character's in place of numbers
>>>> In all the above cases, you will have time complexity O(n)
>>>>
>>>> PS: I'm definitely looking forward to learn a solution, better than
>>>> O(nLgn)
>>>>
>>>>
>>>>
>>>> Regards,
>>>> Sandeep Jain
>>>>
>>>>
>>>>
>>>>
>>>> On Sun, Jul 3, 2011 at 1:09 PM, sunny agrawal
>>>> <sunny816.i...@gmail.com>wrote:
>>>>
>>>>> @sandeep
>>>>> SET A -> {0,3,4,7}
>>>>> SET B -> {1,2,5,6}
>>>>>
>>>>> xor of all elements is zero
>>>>> sum of both the sets is same
>>>>> no of elements in both are same
>>>>>
>>>>> overall result : all Algorithm posted above Fails
>>>>>
>>>>> On Sun, Jul 3, 2011 at 12:59 PM, Sandeep Jain
>>>>> <sandeep6...@gmail.com>wrote:
>>>>>
>>>>>> I was thinking the same, BUT here the question is that we have two
>>>>>> *SETS* and that's the catch.
>>>>>> So, XORing all elements of SET A with SET B should result in ZERO only
>>>>>> when both the set have same elements.
>>>>>>
>>>>>>
>>>>>> Regards,
>>>>>> Sandeep Jain
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>> On Sun, Jul 3, 2011 at 11:25 AM, Pranav Agarwal <
>>>>>> meetpranav...@gmail.com> wrote:
>>>>>>
>>>>>>> I think that the above algo will fail for the following two arrays:
>>>>>>> a={2,2,3,3}
>>>>>>> b={4,4,1,1}
>>>>>>>
>>>>>>> sum(a)=sum(b);
>>>>>>> a^b=0;
>>>>>>> len(a)=len(b);
>>>>>>>
>>>>>>> Correct me if i am wrong!
>>>>>>>
>>>>>>> Pranav
>>>>>>>
>>>>>>>
>>>>>>> On Sun, Jul 3, 2011 at 7:43 AM, varun pahwa <
>>>>>>> varunpahwa2...@gmail.com> wrote:
>>>>>>>
>>>>>>>> @aditya. xor all elements mean that. take xor of each element of 1st
>>>>>>>> array store in a variable that take xor of variable and each element
>>>>>>>> of the
>>>>>>>> second array if all elements are common then the variable will be 0
>>>>>>>> some
>>>>>>>> where.
>>>>>>>> var = a[0];
>>>>>>>> for(i = 1; i < sizeof(a)/sizeof(a[0]); i++)
>>>>>>>> var = var ^ a[i];
>>>>>>>> for(i = 0; i < sizeof(b)/sizeof(b[0]); i++)
>>>>>>>> var = var ^ b[i];
>>>>>>>>
>>>>>>>>
>>>>>>>>
>>>>>>>> On Sat, Jul 2, 2011 at 2:19 PM, aditya kumar <
>>>>>>>> aditya.kumar130...@gmail.com> wrote:
>>>>>>>>
>>>>>>>>> @mohit..:i dint get the logic behind XOR plz explain ..nd ya i dont
>>>>>>>>> think dat you can find second largest in less than O(n).
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> On Sun, Jul 3, 2011 at 2:43 AM, mohit mittal <
>>>>>>>>> mohitm.1...@gmail.com> wrote:
>>>>>>>>>
>>>>>>>>>> Dont think that the corresponding elements should be same.
>>>>>>>>>> XOR Should do it anyway.
>>>>>>>>>>
>>>>>>>>>> Btw other question "How would you find the second largest element
>>>>>>>>>> in an array using minimum no of comparisons?Any thing better than
>>>>>>>>>> O(n)."?
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> On Sun, Jul 3, 2011 at 2:41 AM, aditya kumar <
>>>>>>>>>> aditya.kumar130...@gmail.com> wrote:
>>>>>>>>>>
>>>>>>>>>>> xor will only result if corresponding elements are same . what if
>>>>>>>>>>> in both the array set of integers are same but they arnt
>>>>>>>>>>> corresponding to
>>>>>>>>>>> each other ??
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> On Sun, Jul 3, 2011 at 2:37 AM, Dumanshu <duman...@gmail.com>wrote:
>>>>>>>>>>>
>>>>>>>>>>>> xor all the elements of both arrays ==0
>>>>>>>>>>>> sum of 1st array == sum of 2nd array
>>>>>>>>>>>> no. of elements in 1st == no. of elements in 2nd
>>>>>>>>>>>> if the above conditions are met, they have the same set.
>>>>>>>>>>>> m i missin sth?
>>>>>>>>>>>> On Jul 3, 1:23 am, mittal <mohitm.1...@gmail.com> wrote:
>>>>>>>>>>>> > Given two arrays of numbers, find if each of the two arrays
>>>>>>>>>>>> have the same
>>>>>>>>>>>> > set of ntegers ? Suggest an algo which can run faster than
>>>>>>>>>>>> NlogN without
>>>>>>>>>>>> > extra space?
>>>>>>>>>>>>
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>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> --
>>>>>>>>>> Mohit Mittal
>>>>>>>>>> 4th year , Computer Engineering
>>>>>>>>>> Student-Coordinator , DTU WebTeam
>>>>>>>>>> Delhi Technological University
>>>>>>>>>>
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>>>>>>>>
>>>>>>>>
>>>>>>>> --
>>>>>>>> Varun Pahwa
>>>>>>>> B.Tech (IT)
>>>>>>>> 7th Sem.
>>>>>>>> Indian Institute of Information Technology Allahabad.
>>>>>>>> Ph : 09793899112 ,08011820777
>>>>>>>> Official Email :: rit2008...@iiita.ac.in
>>>>>>>> Another Email :: varunpahwa.ii...@gmail.com
>>>>>>>>
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>>>>>>>>
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>>>>>
>>>>>
>>>>>
>>>>> --
>>>>> Sunny Aggrawal
>>>>> B-Tech IV year,CSI
>>>>> Indian Institute Of Technology,Roorkee
>>>>>
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>>>
>>>
>>>
>>> --
>>> Sunny Aggrawal
>>> B-Tech IV year,CSI
>>> Indian Institute Of Technology,Roorkee
>>>
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>
>
>
> --
> **With Regards
> Deoki Nandan Vishwakarma
>
> *
> *
>
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