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
>>>>
>>>> People who fail to plan are those who plan to fail.
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>
>
>
> --
> Sunny Aggrawal
> B-Tech IV year,CSI
> Indian Institute Of Technology,Roorkee
>
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