Recently, I was solving a problem that involved disjoint ranges. And while stress testing my solution I found a simple trick to fairly and efficiently generate such ranges.

Let us first examine a set of $$$N = 3$$$ disjoint ranges: [{2,3}, {6, 7}, {4, 5}]. If we list their endpoints in order, we get a sequence of $$$2 \times N$$$ integers, where adjacent integers are paired together to form a range: [2,3, 4,5, 6,7]

It follows from this observation that you can generate N disjoint ranges by first generating $$$2 \times N$$$ unique integers (duplicates will affect the disjoint condition), and then "matching" adjacent integers together.

Here is some Python code:

from random import randint
def gen_ranges(n):
    x = set(())
    while len(x) != 2*n:
        x.add(randint(1, 100))
    y = sorted(list(x))
    z = []
    for i in range(0, 2*n, 2):
        z.append([y[i], y[i+1]])
    return z

print(gen_ranges(5))
# [[11, 19], [45, 62], [65, 67], [72, 76], [93, 98]]

As an additional advantage, the output is sorted.