50 lines
1.3 KiB
Python
50 lines
1.3 KiB
Python
def valid_parens(n):
|
|
if n == 0:
|
|
return []
|
|
if n == 1:
|
|
return ["()"]
|
|
|
|
result = set()
|
|
for x in valid_parens(n - 1):
|
|
result.add("({})".format(x))
|
|
result.add("(){}".format(x))
|
|
result.add("{}()".format(x))
|
|
return result
|
|
|
|
def valid_parens_efficient(n):
|
|
result = []
|
|
curr = [''] * n**2
|
|
do_valid_parens_efficient(result, curr, 0, n, n)
|
|
return result
|
|
|
|
def do_valid_parens_efficient(result, curr, i, lhs, rhs):
|
|
if lhs == 0 and rhs == 0:
|
|
result.append(''.join(curr))
|
|
else:
|
|
if lhs > 0:
|
|
curr[i] = '('
|
|
do_valid_parens_efficient(result, curr, i + 1, lhs - 1, rhs)
|
|
if rhs > lhs:
|
|
curr[i] = ')'
|
|
do_valid_parens_efficient(result, curr, i + 1, lhs, rhs - 1)
|
|
|
|
# Avoids recursion by using either a stack or a queue. I think this version is
|
|
# easier to understand.
|
|
def valid_parens_efficient_2(n):
|
|
result = []
|
|
xs = []
|
|
xs.append(('', n, n))
|
|
while xs:
|
|
curr, lhs, rhs = xs.pop()
|
|
print(curr)
|
|
if lhs == 0 and rhs == 0:
|
|
result.append(''.join(curr))
|
|
if lhs > 0:
|
|
xs.append((curr + '(', lhs - 1, rhs))
|
|
if rhs > lhs:
|
|
xs.append((curr + ')', lhs, rhs - 1))
|
|
return result
|
|
|
|
# print(valid_parens(4))
|
|
print(valid_parens_efficient(3))
|
|
print(valid_parens_efficient_2(3))
|