Find Even Index Array.py

'''
PROBLEM STATEMENT:

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.

'''


def find_even_index(arr):
    for i in range(0,len(arr)):
        sum_left = sum(list(arr[:i]))
        sum_right = sum(list(arr[i+1:]))
        if sum_left == sum_right:
            return i
        elif sum_left == [] and sum_right == 0:
            return 0
        elif i == len(arr)-1:
            return -1