Formatted question description: https://leetcode.ca/all/1249.html

1249. Minimum Remove to Make Valid Parentheses (Medium)

Given a string s of '(' , ')' and lowercase English characters. 

Your task is to remove the minimum number of parentheses ( '(' or ')', in any positions ) so that the resulting parentheses string is valid and return any valid string.

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

  • It is the empty string, contains only lowercase characters, 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.

 

Example 1:

Input: s = "lee(t(c)o)de)"
Output: "lee(t(c)o)de"
Explanation: "lee(t(co)de)" , "lee(t(c)ode)" would also be accepted.

Example 2:

Input: s = "a)b(c)d"
Output: "ab(c)d"

Example 3:

Input: s = "))(("
Output: ""
Explanation: An empty string is also valid.

Example 4:

Input: s = "(a(b(c)d)"
Output: "a(b(c)d)"

 

Constraints:

  • 1 <= s.length <= 10^5
  • s[i] is one of  '(' , ')' and lowercase English letters.

Related Topics:
String, Stack

Solution 1. Scan Left to Right then Right to Left

// OJ: https://leetcode.com/problems/minimum-remove-to-make-valid-parentheses/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    string minRemoveToMakeValid(string s) {
        int left = 0, i, j;
        for (i = 0, j = 0; i < s.size(); ++i) {
            if (isalpha(s[i]) || (s[i] == '(' && ++left) || (s[i] == ')' && --left >= 0)) s[j++] = s[i];
            left = max(left, 0);
        }
        s.resize(j);
        reverse(begin(s), end(s));
        left = 0;
        for (i = 0, j = 0; i < s.size(); ++i) {
            if (isalpha(s[i]) || (s[i] == ')' && ++left) || (s[i] == '(' && --left >= 0)) s[j++] = s[i];
            left = max(left, 0);
        }
        s.resize(j);
        reverse(begin(s), end(s));
        return s;
    }
};

Solution 2. Stack

// OJ: https://leetcode.com/problems/minimum-remove-to-make-valid-parentheses/
// Time: O(N)
// Space: O(N)
class Solution {
public:
    string minRemoveToMakeValid(string s) {
        stack<int> st;
        for (int i = 0; i < s.size(); ++i) {
            if (s[i] == '(') st.push(i);
            else if (s[i] == ')') {
                if (st.empty()) s[i] = '*';
                else st.pop();
            }
        }
        while (st.size()) {
            s[st.top()] = '*';
            st.pop();
        }
        s.erase(remove(begin(s), end(s), '*'), end(s));
        return s;
    }
};

Java

  • class Solution {
        public String minRemoveToMakeValid(String s) {
            Stack<Integer> stack = new Stack<Integer>();
            StringBuffer sb = new StringBuffer();
            int length = s.length();
            int deleteCount = 0;
            for (int i = 0; i < length; i++) {
                char c = s.charAt(i);
                if (Character.isLetter(c))
                    sb.append(c);
                else if (c == '(') {
                    sb.append(c);
                    stack.push(i - deleteCount);
                } else if (c == ')') {
                    if (!stack.isEmpty()) {
                        sb.append(c);
                        stack.pop();
                    } else
                        deleteCount++;
                }
            }
            while (!stack.isEmpty()) {
                int index = stack.pop();
                sb.deleteCharAt(index);
            }
            return sb.toString();
        }
    }
    
  • // OJ: https://leetcode.com/problems/minimum-remove-to-make-valid-parentheses/
    // Time: O(N)
    // Space: O(1) if changing input string is allowed; otherwise O(N)
    class Solution {
    public:
        string minRemoveToMakeValid(string s) {
            int j = 0, left = 0;
            for (int i = 0; i < s.size(); ++i) {
                if (s[i] == '(') ++left;
                else if (s[i] == ')') {
                    if (left == 0) continue;
                    --left;
                }
                s[j++] = s[i];
            }
            s.resize(j);
            reverse(begin(s), end(s));
            j = 0, left = 0;
            for (int i = 0; i < s.size(); ++i) {
                if (s[i] == ')') ++left;
                else if (s[i] == '(') {
                    if (left == 0) continue;
                    --left;
                }
                s[j++] = s[i];
            }
            s.resize(j);
            reverse(begin(s), end(s));
            return s;
        }
    };
    
  • # 1249. Minimum Remove to Make Valid Parentheses
    # https://leetcode.com/problems/minimum-remove-to-make-valid-parentheses/
    
    class Solution:
        def minRemoveToMakeValid(self, s: str):
            stack = []
            s = list(s)
            
            for i,c in enumerate(s):
                if c == ")":
                    if stack: 
                        stack.pop()
                    else:
                        s[i] = ""
    
                elif c == "(":
                    stack.append(i)
                
    
            while stack:
                s[stack.pop()] = ""
    
            return "".join(s)
            
        def minRemoveToMakeValid(self, s: str):
            stack = []
            res = [""] * len(s)
            forbidden = set()
            
            for i,c in enumerate(s):
                if c == ")":
                    if stack and stack[-1][1] == "(": 
                        stack.pop()
                        res[i] = c
                    else:
                        forbidden.add(i)
    
                elif c == "(":
                    res[i] = c
                    stack.append((i,c))
                
                else:
                    res[i] = c
    
            if len(stack) > 0:
                forbidden |= set([i for i,_ in stack])
    
                for i in forbidden:
                    res[i] = ""
    
            return "".join(res)
            
        def minRemoveToMakeValid(self, s: str):
            stack = []
            res = []
            forbidden = set()
            
            for i,c in enumerate(s):
                if c != "(" and c != ")": 
                    res.append(c)
                else:
                    if c == ")":
                        if stack and stack[-1][1] == "(": 
                            stack.pop()
                            res.append(c)
                        else:
                            forbidden.add(i)
                        
                    else:
                        res.append(c)
                        stack.append((i,c))
    
            if len(stack) > 0:
                forbidden |= set([i for i,_ in stack])
                ans = []
                for i,x in enumerate(s):
                    if i not in forbidden:
                        ans.append(x)
                
                return "".join(ans)
    
            
            return "".join(res)
    

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