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