breaking a string equation into two parts in python3 - regex

Question

I have a string such as
'(((a+b)+a)+c)' which I'd like to break into two parts, the result would be ('((a+b)+a)','c').
If I were to run it again on the first element of the result it would give me ('(a+b)', 'a')
and if I ran it again on '(a+b)' it would return ('a', 'b').
I was thinking I could do this via a regular expression but I couldn't figure this out and went down the path of having many if statements checking for opening and closing brackets but it gets a bit messy

Answer 1

Here is an example that works on examples such as yours:
def breakit(s):
count = 0
for i, c in enumerate(s):
if count == 1 and c in '+-':
return s[1:i].strip(), s[i+1:-1].strip()
if c == '(': count +=1
if c == ')': count -= 1
return s
breakit(s)
>> ('((a+b)+a)', 'c')
breakit(_[0])
('(a+b)', 'a')
breakit(_[0])
('a', 'b')

Answer 2

voila:
#!/usr/bin/python3.5
def f(s):
p=s.rsplit('+',1)
return [p[0][1:],p[1][:-1]]
s='(((a+b)+a)+c)'
for i in range(3):
k=f(s)
s=k[0]
print(k)
output:
['((a+b)+a)', 'c']
['(a+b)', 'a']
['a', 'b']

Answer 3

I thought I'd post my answer as well, not quite as elegant as the chosen solution but it works
def break_into_2(s):
if len(s) == 1:
# limiting case
return s
# s[0] can either be a digit or '('
if s[0].isdigit():
# digit could be 10,100,1000,...
idx = 0
while s[idx].isdigit():
idx += 1
a = s[:idx]
b = s[idx+1:]
return a, b
# otherwise, s[0] = '('
idx = 1
counter = 1
# counter tracks opening and closing parenthesis
# when counter = 0, the left side expression has
# been found, return the idx at which this happens
while counter:
if s[idx] == '(':
counter+=1
elif s[idx] == ')':
counter -=1
idx +=1
if s[:idx] == s:
# this case occurs when brackets enclosing entire expression, s
# runs the function again with the same expression from idxs 1:-1
return break_into_2(s[1:-1])
return s[:idx], s[idx+1:]