Codility303-Time Complexity-TapeEquilibrium.js

//https://app.codility.com/demo/results/trainingK46UY6-UHA/
// Task description
// A non-empty 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 difference 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:

// function solution(A);

// that, given a non-empty 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.

// Write an efficient algorithm for the following assumptions:

// 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].
function solution(A) {
  // write your code in JavaScript (Node.js 8.9.4)
  let sum1 = 0
  let sum2 = A.reduce(( a , b ) => ( a + b))
  let mindiff = Infinity
  for (i = 0; i < (A.length-1); i++){
    sum1 += A[i]
    sum2 -= A[i]
    let diff = sum2 - sum1
    mindiff = Math.min(mindiff, Math.abs(diff))
  }
  return mindiff
}