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.

<b>Note</b>: 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.

<b>Note</b>: 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.

My Answer

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

Best Practice

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

반복문 안에서 조건문에 return처리를 해버리면 훨씬 깔끔했을텐데 멍청하게 그걸 이용 안하고 쓸모없는 result 변수에 할당하고, 마지막에 return하는 형식으로 구현해버렸다. 쓸데없이 변수에 할당하는 습관을 줄여야지..