Check if two ranges overlap in ruby

Question

I know that I can do:

(1..30).cover?(2)
=> true

But when I try to do the same with another range it always returns false:

(1..30).cover?(2..3)
=> false

So my question is - is there any elegant way to compare two ranges in ruby? In my case I want to check if two DateTime-ranges overlap. Thanks in advance.

Answer 1

Two ranges overlap for a given range A when:

1. range B starts within range A,
2. range B ends within range A or
3. range B starts before range A and ends after range A

Examples:

Range A    |-----|
             |-----|  Case 1
         |-----|      Case 2
             |-|      Case 1 + 2
         |---------|  Case 3

Looking closer the rule is: Two ranges overlap when Range B starts before the range A ends and range B ends after the range A starts.

def ranges_overlap?(range_a, range_b)
  range_b.begin <= range_a.end && range_a.begin <= range_b.end 
end 

Answer 2

def overlap?(r1,r2)
  !(r1.first > r2.last || r1.last < r2.first)
end

overlap? 1..5, 4..10 #=> true
overlap? 1..5, 6..10 #=> false
overlap? 1..10, 4..8 #=> true
overlap? 1..4, 4..8  #=> true

The operative line is equivalent to:

r1.first <= r2.last && r1.last >= r2.first

I normally try to avoid negation, but in this case I think it reads better with it.

Another way:

def overlap?(r1,r2)
  !(([r1.first, r2.first].min..[r1.last, r2.last].max).size >= r1.size + r2.size)
end

overlap? 1..5, 4..10 #=> true
overlap? 1..5, 6..10 #=> false
overlap? 1..10, 4..8 #=> true
overlap? 1..4, 4..8  #=> true

The operative line is equivalent to:

([r1.first, r2.first].min..[r1.last, r2.last].max).size < r1.size + r2.size

Again, I prefer the one with negation.