()[]{}<> : 0
([{<>}]) : 0
<>{[]}() : 0
{<>([])} : 0
<(>)[{}] : 1
<[({)}]> : 1
[{<}]>() : 2
{<>([}]) : 2
<{(>})[] : 3
[(]<){>} : 3
<([>{)}] : 4
(<{[>})] : 4
(<[{)>}] : 5
<{[(>})] : 5
[{<(]}>) : 6
(<{[)>}] : 6

l7~f&_f{/~;&}s,2/

l         read a line of input
7~f&      clear the lowest 3 bits of each character
           the goal is to convert brackets of the same type to the same char
_         duplicate the resulting string, let's call it S
f{…}      for each character in S, and S (the char and S are pushed every time)
         swap the character with S
  /       split S around that character, resulting in 3 pieces:
           before, between, after
  ~       dump the pieces on the stack
  ;       pop the last piece
  &       intersect the first 2 pieces
          after the loop, we have an array of strings
          containing the chars interlocking to the left with each char of S
s         join all the string into one string
,         get the string length
2/        divide by 2, because S has duplicated characters

y/([{</)]}>/
s/.*/t& & & & /
:b
y/)]}>/]}>)/
s/S*>(S*)>S* /1t/
t
s/S* //
:
s/(tS*)(S)(S*)2(S*t)/134/
t
s/S *$/&/
tb
s/s//g
s/../1/g

perl -pe 'y/)]}>/([{</;for$x(/./g){$h{$x="\$x"}++&&s!$x(.*)$x!$z+=length$1,$1!e}$_=$z'

JmC/CdTzlsm@FPcsJd{J

(T`)]>}`([<{
(D)(.*)1(.*)
n$2n$3
(?=(D).*n.*1)
1
n
<empty>

(T`)]>}`([<{
(D)(.*)1(.*)
#$2#$3
(?=(D)[^#]*#[^#]*1)
1
#
<empty>

T`)]>}`([<{

(D)(.*)1(.*)
n$2n$3

(?=(D).*n.*1)
1

n
<empty>

x=>(a=b=0,[for(c of x)for(d of'1234')(e=c.charCodeAt()/26|0)==d?a^=1<<d:b^=(a>>d&1)<<d*4+e],f=y=>y&&y%2+f(y>>1))(b)/2

() => 40,41
<> => 60,62
[] => 91,93
{} => 123,125

b^=(a>>d&1)<<d*4+e

f=y=>y&&y%2+f(y>>1)

f(y)         => y && y%2 + f(y>>1)
f(0b1001101) =>       1  + f(0b100110) = 4
f(0b100110)  =>       0  + f(0b10011)  = 3
f(0b10011)   =>       1  + f(0b1001)   = 3
f(0b1001)    =>       1  + f(0b100)    = 2
f(0b100)     =>       0  + f(0b10)     = 1
f(0b10)      =>       0  + f(0b1)      = 1
f(0b1)       =>       1  + f(0b0)      = 1
f(0b0)       => 0                      = 0

WITH v AS(SELECT b,MIN(p)i,MAX(p)a FROM(SELECT SUBSTR(TRANSLATE(:1,'])>}','[(<{'),LEVEL,1)b,LEVEL p FROM DUAL CONNECT BY LEVEL<9)GROUP BY b)SELECT COUNT(*)FROM v x,v y WHERE x.i<y.i AND x.a<y.a AND y.i<x.a;

WITH v AS( -- Compute min and max pos for each bracket type
           SELECT b,MIN(p)i,MAX(p)a
           FROM   ( -- replace ending brackets by opening brakets and split the string
                    SELECT SUBSTR(TRANSLATE(:1,'])>}','[(<{'),LEVEL,1)b,LEVEL p
                    FROM DUAL
                    CONNECT BY LEVEL<9
                  )
           GROUP BY b
         )
SELECT COUNT(*)
FROM   v x,v y
WHERE  x.i<y.i AND x.a<y.a AND y.i<x.a -- Apply restrictions for interlocking brackets