Equal-Sides-Of-An-Array.js

/*
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.
*/

// Answer:
function findEvenIndex(arr) {
  for (let i = 0; i < arr.length; i++) {
    let leftPart = calculateSum(arr.slice(0,i));
    let rightPart = calculateSum(arr.slice(i + 1, arr.length));
    if (leftPart === rightPart) return i;
  }
  return -1;
}

function calculateSum(arr) {
  return arr.length <= 0 ? 0 : arr.reduce((acc, el) => acc + el,0)
}

// BigO: O(n2)