The busiest band

Problem

The antenna at Kepler Base can only listen to one band at a time, and the band has a fixed width. You want to point it where the most ships are. Write a function that takes frequencies, a list of positive integers (one per ship; it may be empty and may have repeats), and width, an integer from 0 up. Return an integer: the most ships that fit together in one band, that is, in a group where the highest frequency minus the lowest is width or less. Repeats each count, and the edge is in: a difference equal to width does fit.

With 101, 98, 105, 99, 120, 104 and 103 and width 5, the band from 99 to 104 catches 99, 101, 103 and 104: that is 4, and no band catches 5.

With a single ship the answer is 1; with no ships, 0.

Examples

• The example

  [101, 98, 105, 99, 120, 104, 103], 5 → 4

• The edge is in

  [10, 13, 16], 3 → 2

• Repeats with width 0

  [50, 50, 50, 52], 0 → 3

• The best band is at the end

  [88, 90, 91, 95, 96, 97, 98], 4 → 4

• A single ship

  [7], 0 → 1

Besides these, the challenge has hidden tests that are revealed when you submit your solution.

You start with this

def band(frequencies, width):
    pass

function band(frequencies, width) {
}