sort array c only.
now select every pair from array a nd b ( o(n2))
for every pair find the element from c which gives sum of the three element
0. ( search using binary search as c is sorted)
so the algo is O (n^2 logn))

On 16 June 2010 13:47, Amir hossein Shahriari <
[email protected]> wrote:

> @algoose chase: you're right thx for the test case
>
> On Wed, Jun 16, 2010 at 11:45 AM, Algoose Chase <[email protected]>wrote:
>
>> @ Amir:
>> I think you cannot find two numbers in B and C such that their sum is
>> -A[k] in O(n) in all cases with the logic that you mentioned.
>>
>> for instance you can try with :
>> A : 2 7 8 10
>> B :-2, 0, 1, 9
>> C: -7 -2 1 3
>>
>> On Wed, Jun 16, 2010 at 5:56 AM, Amir hossein Shahriari <
>> [email protected]> wrote:
>>
>>> sort all arrays
>>> for each number in A , A[k]
>>> find two numbers in B and C such that their sum is -A[k]
>>> this can be done in O(n) time:
>>>
>>> set i at the beginning of B and j at the end of C
>>> while i<n or j>=0
>>>   if ( B[i] + C[j] == -A[k] ) return true
>>>   if ( B[i] + C[j] < -A[k] ) increase i
>>>   else decrease j
>>>
>>> we have to do this for each element of A so the overall complexity is:
>>> O(n2) time O(1) space
>>>
>>> correct me if I'm wrong
>>>
>>> On 6/15/10, sharad kumar <[email protected]> wrote:
>>> > plzzz comment on this question
>>> >
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