sry for last post:
the code bye sravan reddy would not work for negative nos.
but slight modification in if conditions make it work for negative nos too.
by the way nice algo suggested
#include<stdio.h>
int main(int argc, char **argv){
int a[] = {-2,-3,-4,-19,10};
//int a[]={2,8,-6};
//int a[]={2,2,13,4,7,3,8,12,9,1,5};
int len = sizeof(a)/sizeof(a[0]);
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("Possible for \tLeft : %d, Right: %d, Center: %d
\n",p1,p2,i);
break;
//return 0;
}
else if(((left > right && p2 < len-1 &&
a[p2+1]>0))|| ((a[p2+1]<0))){
p2++;
right = right+ a[p2];
}
else if(((left < right && p1 > 0)&& a[p1-1]>0)||
((a[p1-1]<0))){
p1--;
left = left + a[p1];
}
else{
printf("Not Possible for \t Left : %d, Right: %d, Center:
%d \n",p1,p2,i);
break;
//return 0;
}
}
}
On Tue, Jan 10, 2012 at 10:42 AM, priyanka jaggi
<priyankajag...@gmail.com>wrote:
> @ankur : in this question, the elements of the array are continuous
>
> i think the solution of shravan reddy is correct and works for negative
> nos too.
>
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