My solution to codility test MinAvgTwoSlice: https://app.codility.com/programmers/lessons/5-prefix_sums/min_avg_two_slice/

min-avg-two-slice.js

/**
 * Given:
 *
 * A non-empty array A consisting of N integers is given. A pair of integers (P, Q),
 * such that 0 ≤ P < Q < N, is called a slice of array A (notice that the slice contains
 * at least two elements). The average of a slice (P, Q) is the sum of
 * A[P] + A[P + 1] + ... + A[Q] divided by the length of the slice. To be precise,
 * the average equals (A[P] + A[P + 1] + ... + A[Q]) / (Q − P + 1).
 *
 * Implied assumption:
 * - a slice will never be longer than 3 items, as any slice longer will contain
 *   a slice of 2 or 3 items which has a lower average.
 *
 * Chosen approach:
 * - iterate over index of items
 * - keep index of lowest average double
 * - keep index -1 of lowest average triple if lower than double
 *
 */
const summer = (val, sum) => (sum += val)
const sumSlice = slice => slice.reduce(summer, 0)

function solution(A) {
  let minAvg
  let minIndex
  for (let i = 0; i <= A.length - 2; i++) {
    const slice2 = A.slice(i, i + 2)
    const sum2 = sumSlice(slice2)
    const avg2 = sum2 === 0 ? sum2 : sum2 / slice2.length
    if (minAvg === undefined || avg2 < minAvg) {
      minIndex = i
      minAvg = avg2
    }
    if (i > 0) {
      const slice3 = A.slice(i - 1, i + 2)
      const sum3 = sumSlice(slice3)
      const avg3 = sum3 === 0 ? sum3 : sum3 / slice3.length
      if (avg3 < minAvg) {
        minIndex = i - 1
        minAvg = avg3
      }
    }
  }
  return minIndex
}