6kyu-equalSidesOfAnArray.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.

ALGORITHMSARRAYSFUNDAMENTALS
*/



//P: one input, an array of integers with length between 0 and 1000, integers can be positive or negative
//R: return the lowest index N where the left side of the array to N is equal to the right side of the array. If none exists, return -1
//E: [20, 10, -80, 10, 10, 15, 35] => 0 (at index 0, left side is empty, which in this challenge equals 0 and the right side totals out to 0 as well)
  // [1, 2, 3, 4, 3, 2, 1] => 3 (index 3 has  1, 2,3 to the left and 3,2,1 to the right)
//P: iterate through a for loop, splitting the array at i each time into two
  // calculate the sum for each side of the split array
  // if the sides equal, return i, if not continue on through arr.length
  // if no index found where sides equal, return -1

  function findEvenIndex(arr){
    let result = -1;
    
    if(arr.reduce((a,b) => a + b, 0) === 0){return 0}
    if(arr.slice(0, arr.length-1).reduce((a,b) => a+b, 0) === 0){return arr.length-1}
    
    for(let i = 0; i < arr.length; i++){
      let left = arr.slice(0, i).reduce((a,b) => a+b, 0)
      let right = arr.slice(i+1).reduce((a,b) => a+b, 0)
      if(left === right){
        return i
      }
    };
    
    return result; 
  };