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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>>>>>>
>>>>>>
>>>>>> --
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>>>>>> B.Tech (IT)
>>>>>> 7th Sem.
>>>>>> Indian Institute of Information Technology Allahabad.
>>>>>> Ph : 09793899112 ,08011820777
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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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