wouldn't your algo take n^3 time as well given the fact that the lists are
not sorted? searching for the matching value fr z should be taking O(n) time
and to do so for each pair of x and y shall take O(n^2).
Please correct me if i got something wrong here

On Wed, Dec 15, 2010 at 11:44 AM, Prashant Kulkarni <
prashant.r.k...@gmail.com> wrote:

> Hi,
> Very good question.
>
> i think very knows brute-force soln of n^3 time complexity.
>
> here is a alternative method/soln of O(n^2 * log(n)) time complexity Algo
>
> 1) pick any number from list A (One by one)...... let say 'x'
>
> 2) pick any number from list B (one by one )...... let say 'y'
>
> 3) z=x+y
>
> 4)look for 'z' in the list by considering all numbers as unsigned if u
> found then check sign of that number it be opposite of 'z' if so return 1
> else 0
>
> [above steps is repeated for all elements]
>
> *correct me if I am wrong*
>
> --
> Prashant Kulkarni
> IISc Bangalore
>
>
>
>
> On Wed, Dec 15, 2010 at 10:36 AM, Soumya Prasad Ukil <
> ukil.sou...@gmail.com> wrote:
>
>> Are the linked list sorted?
>>
>>
>> On 14 December 2010 05:33, divya <sweetdivya....@gmail.com> wrote:
>>
>>> Given three lists: A, B and C of length n each. Write a function which
>>> returns true if any 3 three numbers (1 from each list), sum up to
>>> zero. Else return false, if there is no such combination.
>>>
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>>
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>> soumya prasad ukil
>>
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