algo.js

const algo = arr => {
  let sum_half = 0
  let i = 0
  let p = 0
  while(i < arr.length)
    sum_half += arr[i++]
  sum_half /= 2
  let left_side_sum = 0
  let last_diff = 0
  for(i in arr) {
    left_side_sum += arr[i]
    let diff = sum_half - left_side_sum
    if (last_diff > 0 && diff < 0) {
      p = i
      break;
    }
    last_diff = diff 
  }
  return p
}

const arr = [5,3,7,10,1,4,6]
let p = algo(arr)

I found this challenge exciting and decided to solve.
A non-empty zero-indexed array A consisting of N integers is given. Array A represents
numbers on a tape.
Any integer P, such that 0 < P < N, splits this tape into two non-empty parts: A[0], A[1], ..., A[P −
1] and A[P], A[P + 1], ..., A[N − 1].
The dif ference between the two parts is the value of: |(A[0] + A[1] + ... + A[P − 1]) − (A[P] + A[P+ 1] + ... + A[N − 1])|
In other words, it is the absolute difference between the sum of the first part and the sum of the
second part.
For example, consider array A such that:
A[0] = 3
A[1] = 1
A[2] = 2
A[3] = 4
A[4] = 3
We can split this tape in four places:
● P = 1, difference = |3 − 10| = 7
●
● P = 2, difference = |4 − 9| = 5
●
● P = 3, difference = |6 − 7| = 1
●
● P = 4, difference = |10 − 3| = 7
●
Write a function:
class Solution { public int solution(int[] A); }
that, given a non-empty zero-indexed array A of N integers, returns the minimal difference that
can be achieved.
For example, given:
A[0] = 3
A[1] = 1
A[2] = 2
A[3] = 4
A[4] = 3
the function should return 1, as explained above.
Assume that:
● N is an integer within the range [2..100,000];
● each element of array A is an integer within the range [−1,000..1,000].
Complexity:
● expected worst-case time complexity is O(N);
● expected worst-case space complexity is O(N), beyond input storage (not counting
the storage required for input arguments).
Elements of input arrays can be modified.