Leetcode1249-minimumRemoveToMakeValidParentheses

Description

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

Example 1:

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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:

1 2	Input: s = "a)b(c)d" Output: "ab(c)d"

Example 3:

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

Example 4:

1 2	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.

Solution

class Solution {
    class Pair{
        public char ch;
        public int idx;
        public Pair(char ch, int idx){
            this.ch = ch;
            this.idx = idx;
        }
    }
    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) == '(') st.add(i + 1);
            if (sb.charAt(i) == ')') {
            if (!st.empty() && st.peek() >= 0) st.pop();
            else st.add(-(i + 1));
            }
        }
        while (!st.empty())
            sb.deleteCharAt(Math.abs(st.pop()) - 1);
        return sb.toString();
    }
}