Probes before the deadline

Problem

Ilse runs the workshop on Kepler Station, where one probe is tuned per day: one on day 1, another on day 2, and so on, with no days off. Each probe launches the same day it is tuned, and it has a deadline: the last day it can still launch. Ilse chooses the order to tune them in. Write a function that takes deadlines, a list of integers from 1 up (it may be empty), and returns true if there is some order in which every probe launches on or before its deadline, or false if there is none.

With [3, 1, 2], the order 1, 2, 3 launches them on days 1, 2 and 3, each one just in time: true. With [2, 1, 1], two probes are due on day 1 and only one can launch that day: false. With no probes nothing can be late, so the empty list gives true.

Examples

• The example

  [3, 1, 2] → true

• Two due on day 1

  [2, 1, 1] → false

• Plenty of time

  [5, 5, 5] → true

• Three for two days

  [2, 2, 2] → false

• A single probe, deadline 1

  [1] → true

• One far deadline cannot save the near ones

  [1, 1, 5] → false

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

You start with this

def on_time(deadlines):
    pass

function onTime(deadlines) {
}