1249

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.

 

Constraints:

• 1 <= s.length <= 105
• s[i] is either'(' , ')', or lowercase English letter.

Companies:
Facebook, Bloomberg, Amazon, tiktok, Microsoft, Snapchat

Related Topics:
String, Stack

Similar Questions:

• Minimum Number of Swaps to Make the String Balanced (Medium)
• Check if a Parentheses String Can Be Valid (Medium)

Solution 1. Scan Left to Right then Right to Left

// OJ: https://leetcode.com/problems/minimum-remove-to-make-valid-parentheses/
// Author: github.com/lzl124631x
// 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;
    }
};

Or

// OJ: https://leetcode.com/problems/minimum-remove-to-make-valid-parentheses/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1) if changing input string is allowed; otherwise O(N)
class Solution {
public:
    string minRemoveToMakeValid(string s) {
        int left = 0, N = s.size();
        for (int i = 0; i < N; ++i) {
            if (isalpha(s[i])) continue;
            if (s[i] == '(') ++left;
            else if (left) --left;
            else s[i] = ' ';
        }
        left = 0;
        for (int i = N - 1; i >= 0; --i) {
            if (isalpha(s[i]) || s[i] == ' ') continue;
            if (s[i] == ')') ++left;
            else if (left) --left;
            else s[i] = ' ';
        }
        s.erase(remove(begin(s), end(s), ' '), end(s));
        return s;
    }
};

Solution 2. Stack

// OJ: https://leetcode.com/problems/minimum-remove-to-make-valid-parentheses/
// Author: github.com/lzl124631x
// 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;
    }
};