minimum_add_to_make_parentheses_valid

View Source

 1# @leet start
 2class Solution:
 3    def minAddToMakeValid(self, s: str) -> int:
 4        """
 5        To find the minimum amount of operations to make a set of parens valid,
 6        we need to count the number of valid parens as we would normally, but
 7        also handle when there are unbalanced closing parens.
 8        We then add up the length of the stack (unbalanced opening parens) and
 9        the length of the unbalanced closing parens.
10        """
11        stack = []
12        unbalanced = 0
13
14        for c in s:
15            if c == "(":
16                stack.append(c)
17            else:
18                if stack:
19                    stack.pop()
20                else:
21                    unbalanced += 1
22        return len(stack) + unbalanced
23
24
25# @leet end
26
27
28def test():
29    assert 2 + 2 == 4

class Solution: View Source

 3class Solution:
 4    def minAddToMakeValid(self, s: str) -> int:
 5        """
 6        To find the minimum amount of operations to make a set of parens valid,
 7        we need to count the number of valid parens as we would normally, but
 8        also handle when there are unbalanced closing parens.
 9        We then add up the length of the stack (unbalanced opening parens) and
10        the length of the unbalanced closing parens.
11        """
12        stack = []
13        unbalanced = 0
14
15        for c in s:
16            if c == "(":
17                stack.append(c)
18            else:
19                if stack:
20                    stack.pop()
21                else:
22                    unbalanced += 1
23        return len(stack) + unbalanced

def minAddToMakeValid(self, s: str) -> int: View Source

 4    def minAddToMakeValid(self, s: str) -> int:
 5        """
 6        To find the minimum amount of operations to make a set of parens valid,
 7        we need to count the number of valid parens as we would normally, but
 8        also handle when there are unbalanced closing parens.
 9        We then add up the length of the stack (unbalanced opening parens) and
10        the length of the unbalanced closing parens.
11        """
12        stack = []
13        unbalanced = 0
14
15        for c in s:
16            if c == "(":
17                stack.append(c)
18            else:
19                if stack:
20                    stack.pop()
21                else:
22                    unbalanced += 1
23        return len(stack) + unbalanced

To find the minimum amount of operations to make a set of parens valid, we need to count the number of valid parens as we would normally, but also handle when there are unbalanced closing parens. We then add up the length of the stack (unbalanced opening parens) and the length of the unbalanced closing parens.

def test(): View Source

29def test():
30    assert 2 + 2 == 4