1249. Minimum Remove to Make Valid Parentheses

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LeetCode

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Problem

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.

Code

class Solution {
    public String minRemoveToMakeValid(String s) {
        StringBuilder sb = new StringBuilder(s);
        Stack<Integer> st = new Stack<>();

        for (int i = 0; i < sb.length(); ++i) {
            if (sb.charAt(i) == '(') {
                // add index to stack
                st.add(i);
            }
            if (sb.charAt(i) == ')') {
                if (!st.empty()) {
                    st.pop();
                } else {
                    // need to replace
                    sb.setCharAt(i, '*');
                }
            }
        }

        while (!st.empty()) {
            sb.setCharAt(st.pop(), '*');
        }

        return sb.toString().replaceAll("\\*", "");
    }
}