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
    

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