Java

import java.util.Stack;

/**

 Given a string S of '(' and ')' parentheses, we add the minimum number of parentheses ( '(' or ')', and in any positions ) so that the resulting parentheses string is valid.

 Formally, a parentheses string is valid if and only if:

 It is the empty string, or
 It can be written as AB (A concatenated with B), where A and B are valid strings, or
 It can be written as (A), where A is a valid string.
 Given a parentheses string, return the minimum number of parentheses we must add to make the resulting string valid.


 Example 1:

 Input: "())"
 Output: 1

 Example 2:

 Input: "((("
 Output: 3

 Example 3:

 Input: "()"
 Output: 0

 Example 4:

 Input: "()))(("
 Output: 4


 Note:

 S.length <= 1000
 S only consists of '(' and ')' characters.

 */

public class Minimum_Add_to_Make_Parentheses_Valid {
    class Solution {
        public int minAddToMakeValid(String S) {
            int ans = 0, bal = 0;
            for (int i = 0; i < S.length(); ++i) {
                bal += S.charAt(i) == '(' ? 1 : -1;
                // It is guaranteed bal >= -1
                if (bal == -1) {
                    ans++;
                    bal++;
                }
            }

            return ans + bal;
        }
    }

    class Solution2 {

        // I'm thinking using stack
        public int minAddToMakeValid(String S) {

            Stack<Character> sk = new Stack<>();

            int count = 0;

            for (char each: S.toCharArray()) {
                if (each == '(') {
                    sk.push(each);
                } else if (each == ')') {

                    if (sk.isEmpty()) {
                        // key is this case
                        sk.push(each);
                    }
                    else if (sk.peek() == '(') {
                        sk.pop();
                    } else {
                        count += 1;
                    }
                }
            }

            return count + sk.size();
        }
    }
}

// OJ: https://leetcode.com/problems/minimum-add-to-make-parentheses-valid/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int minAddToMakeValid(string s) {
        int left = 0, invalidRight = 0;
        for (char c : s) {
            if (c == '(') ++left;
            else if (left) --left;
            else ++invalidRight;
        }
        return invalidRight + left;
    }
};

class Solution(object):
    def minAddToMakeValid(self, S):
        """
        :type S: str
        :rtype: int
        """
        if not S: return 0
        stack = []
        res = 0
        for i, s in enumerate(S):
            if '(' == s:
                if stack and (stack[-1] == ')'):
                    cnt = 0
                    while stack:
                        if stack.pop() == ')':
                            cnt -= 1
                        else:
                            cnt += 1
                        if cnt == 0:
                            break
                    res += abs(cnt)
                stack.append('(')
            else:
                stack.append(')')
        cnt = 0
        while stack:
            if stack.pop() == ')':
                cnt -= 1
            else:
                cnt += 1
        res += abs(cnt)
        return res

Java

class Solution {
    public int minAddToMakeValid(String S) {
        int addCount = 0;
        int curCount = 0;
        int length = S.length();
        for (int i = 0; i < length; i++) {
            char c = S.charAt(i);
            if (c == '(')
                curCount++;
            else {
                if (curCount > 0)
                    curCount--;
                else
                    addCount++;
            }
        }
        addCount += curCount;
        return addCount;
    }
}

// OJ: https://leetcode.com/problems/minimum-add-to-make-parentheses-valid/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int minAddToMakeValid(string s) {
        int left = 0, invalidRight = 0;
        for (char c : s) {
            if (c == '(') ++left;
            else if (left) --left;
            else ++invalidRight;
        }
        return invalidRight + left;
    }
};

class Solution(object):
    def minAddToMakeValid(self, S):
        """
        :type S: str
        :rtype: int
        """
        if not S: return 0
        stack = []
        res = 0
        for i, s in enumerate(S):
            if '(' == s:
                if stack and (stack[-1] == ')'):
                    cnt = 0
                    while stack:
                        if stack.pop() == ')':
                            cnt -= 1
                        else:
                            cnt += 1
                        if cnt == 0:
                            break
                    res += abs(cnt)
                stack.append('(')
            else:
                stack.append(')')
        cnt = 0
        while stack:
            if stack.pop() == ')':
                cnt -= 1
            else:
                cnt += 1
        res += abs(cnt)
        return res

921 - Minimum Add to Make Parentheses Valid

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