Oprea

Equal_Sides_Of_An_Array

---

/*You are going to be given an array of integers. Your job is to take that array and find an index N where the sum
of the integers to the left of N is equal to the sum of the integers to the right of N. If there is no index that
would make this happen, return -1.

For example:

Let's say you are given the array {1,2,3,4,3,2,1}:
Your function will return the index 3, because at the 3rd position of the array, the sum of left side of the index
({1,2,3}) and the sum of the right side of the index ({3,2,1}) both equal 6.

Let's look at another one.
You are given the array {1,100,50,-51,1,1}:
Your function will return the index 1, because at the 1st position of the array, the sum of left side of the index
({1}) and the sum of the right side of the index ({50,-51,1,1}) both equal 1.

Last one:
You are given the array {20,10,-80,10,10,15,35}
At index 0 the left side is {}
The right side is {10,-80,10,10,15,35}
They both are equal to 0 when added. (Empty arrays are equal to 0 in this problem)
Index 0 is the place where the left side and right side are equal.
Note: Please remember that in most programming/scripting languages the index of an array starts at 0.
Input:
An integer array of length 0 < arr < 1000. The numbers in the array can be any integer positive or negative.
Output:
The lowest index N where the side to the left of N is equal to the side to the right of N. If you do not find an
index that fits these rules, then you will return -1.

Note:
If you are given an array with multiple answers, return the lowest correct index.
An empty array should be treated like a 0 in this problem.
*/

#include <vector>
using namespace std;

int find_even_index (const vector <int> numbers) {

int ls=0, rs=0, x=0, tmp=0;
for(vector<int>::const_iterator j = numbers.begin(); j<= numbers.end(); j++,x++)
{
ls+=tmp;
tmp = *j;
rs =0;

if(j == numbers.end()) rs =0;
else
for( vector<int>::const_iterator w=numbers.end()-1; w>j;w--)
{
rs+=*w;
}
if(ls==rs) return x;
}
return -1;

}