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 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 > 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 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 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 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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