I think this wont work for cases with negetive nos

Ex

-2,8,-6

On Tue, Jan 10, 2012 at 6:51 AM, sravanreddy001 <sravanreddy...@gmail.com>wrote:

>
> solution at this link:
> http://ideone.com/ifVIv
>
> for every position, (iteration)
> maitain left, right for the sums,
> keep adding elements towards the begenning to left, and towards the end to
> right, (based on the conditions in the code)
>
> Complexity: outer forloop : O(n)
> inner while loop O(n)
> total O(n^2) and O(1) space.
>
> its currently printing all the position, & center is included to the left
> side,
>
> Left : 3, Right: 9, Center: 6
> this reads as.. sum(elements at 3,4,5,6) == sum(elements at 7,8,9)
>
> let me know if it needs more explanantion.
>
>
>
>
> for(int i=0;i<len;i++){
>                 int left=0,right=0;
>                 int p1 = i;
>                 int p2 = i+1;
>                 left = left + a[p1];
>                 right = right + a[p2];
>                 while(p1>=0 && p2< len){
>                         if( left == right){
>                                 printf("Left : %d, Right: %d, Center: %d 
> \n",p1,p2,i);
>                                 break;
>                                 //return 0;
>                         }
>                         else if(left > right && p2 < len-1){
>                                 p2++;
>                                 right = right+ a[p2];
>                         }
>                         else if(left < right && p1 > 0){
>                                 p1--;
>                                 left = left + a[p1];
>                         }
>                         else{
>                                 //printf("Left : %d, Right: %d, Center: %d 
> \n",p1,p2,i);
>                                 //printf("Not Possible\n");
>                                 break;
>                                 //return 0;
>                         }
>                 }
>
>         }
>
>
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