a+b=-c . Therefore the solution is reducible to finding two numbers a,b
such that a+b equals to -c.
1) sort the array : nlogn
2) Given a number k , you can find two numbers a,b such that a+b =k in
O(n);
3) take a number x , repeat step 2 for k = -x such that a and b should not
be x.
4) repeat step 3 for all the numbers in the array
The complexity is n log n + n*n
On Sun, Jul 22, 2012 at 9:44 AM, Victor Manuel Grijalva Altamirano <
kavic1.mar...@gmail.com> wrote:
> on solution it's with brute force, with three for's,
> for i=0; i<lengthS (to find a)
> for j=0; j<lengthS (to find b)
> for k=0; k<lengthS (to find c)
> if(i!=j and j!=k and i!=k)
> if(s[i]+s[j]+s[k]==0)printf("find it")
>
>
> other solution it's with binary search.
>
> 2012/7/22 lomash goyal <lomesh.go...@gmail.com>
>
>> Given an array *S* of *n* integers, are there elements *a*, *b*, *c* in *
>> S* such that *a* + *b* + *c* = 0? Find all unique triplets in the array
>> which gives the sum of zero.
>>
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>
>
> --
> Victor Manuel Grijalva Altamirano
> Universidad Tecnologica de La Mixteca
>
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