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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