ASCII-Art Venn Diagram

Question

Given two lists that contain no duplicate elements a and b, find the crossover between the two lists and output an ASCII-Art Venn Diagram. The Venn Diagram will use a squarified version of the traditional circles for simplicity.

Example

Given:

a = [1, 11, 'Fox', 'Bear', 333, 'Bee']
b = ['1', 333, 'Bee', 'SchwiftyFive', 4]

Output (Order is 100% arbitrary, as long as the Venn Diagram is correct):

+-----+----+-------------+
|11   |333 |SchwiftyFive |
|Fox  |Bee |4            |
|Bear |1   |             |
+-----+----+-------------+

The program may either consider '1' == 1 or '1' != 1, up to your implementation. You may also choose to just handle everything as strings, and only accept string input.

---

Given:

a=[]
b=[1,2,3]

Output (Notice how the two empty parts still have the right-pad space):

+-+-+--+
| | |1 |
| | |2 |
| | |3 |
+-+-+--+

---

Given:

a=[1]
b=[1]

Output:

+-+--+-+
| |1 | |
+-+--+-+

Rules

• Elements of the Venn Diagram are left-aligned and padded to the max length entry plus 1.
• The ordering of the elements within sub-sections of the Venn-Diagram are arbitrary.
• Corners of the Venn Diagram (where | meets -) must be represented by a +.
• You are garuanteed that a.join(b).length() > 0, if both are empty, you may do whatever.
  • You may even print a picture of Abe Lincoln, don't care.
• This is code-golf, ascii-art and set-theory.

Bonus

Charcoal renders boxes like this naturally, but the whole set theory part... Don't know how well it does that. +100 bounty for the shortest charcoal submission before I am able to add a bounty to the question (2 days from being asked).

Answer 1

Python 2, 221 210 212 bytes

m=map
A,B=m(set,input())
d=A-B,B&A,B-A
e=[max(m(len,s))+1for s in d]
p,i,n='+|\n'
o=b=p+p.join(m('-'.__mul__,e))+p+n
while sum(m(len,d)):o+=i+i.join(m(str.ljust,[len(s)and s.pop()or''for s in d],e))+i+n
print o+b

Try it online!

Answer 2

PHP>=7.1, 287 Bytes

<?for([$a,$b]=$_GET,$x=max(($m=array_map)(count,$r=[($d=array_diff)($a,$b),array_intersect($a,$b),$d($b,$a)]));$n<3;$n++)for(sort($r[+$n]),$i=-1;$i<=$x;$i++){$o[$i].="|+"[$b=$i<0||$i==$x].str_pad($b?"":$r[+$n][$i],max($m(strlen,$r[+$n]))+1," -"[$b]).("|+"[$b][$n<2]);}echo join("
",$o);

Online Version

Expanded

for([$a,$b]=$_GET, # store input arrays in shorter variables
$x=max(($m=array_map)(count,   # get maximum of 
$r=[($d=array_diff)($a,$b),array_intersect($a,$b),$d($b,$a)])); #the set array
$n<3;$n++)
  for(sort($r[+$n]),$i=-1;$i<=$x;$i++){ # sort array to remove keys
    $o[$i].="|+"[$b=$i<0||$i==$x].   # concat line $b boolean for first and last line beginning char 
    str_pad($b?"":$r[+$n][$i]   # string of item in array if not first or last line
    ,max($m(strlen,$r[+$n]))+1  # fill till maximum length of items in array
    ," -"[$b]) # with char depends on first/last line or item line
    .("|+"[$b][$n<2]); # make end of string if last array is reach
}
echo join("   
",$o); #Output

Answer 3

Charcoal, 106 89 87 bytes:

Ａ⟦⟧ςＡ⟦⟧λＡ⟦⟧ρＡ⟦⟧τＷＳ⊞ςιＷＳ⊞⎇№ςιτριＦς⊞⎇№τι⟦⟧λιＦ⟦λτρ⟧«Ｆι«↓Ｐκ»ＭＬι↑←Ａ⁺⌈ＥιＬκ³ζＵＲζ⁺⌈⟦ＬλＬτＬρ⟧²Ｍζ→

Try it online! Note that link is to verbose code for explanatory purposes, with the -sl option that shows the equivalent native Charcoal code. Takes input as newline-separated strings with a blank line after each set.

Edit: Saved 11 bytes thanks to @ASCII-only. Previous version actually had a bug when the last word in the first set was not in the second set and also the first column was the tallest, which manifested as an apparently inability to optimise away a temporary. Saved 2 bytes by optimising two Move commands (the deverbosifier now does this automatically but the resulting code was always valid so the answer is still competing).

Edit: I don't think Multiprint used to work with multiline output but it does currently and making use of that would save 6 bytes, plus a further 4 bytes because current Charcoal preinitialises the u variable to the empty list: Try it online!