1249.Minimum-Remove-to-Make-Valid-Parentheses

1249. Minimum Remove to Make Valid Parentheses

题目地址

https://leetcode.com/problems/minimum-remove-to-make-valid-parentheses/

题目描述

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

代码

Approach #1 Stack

Time: O(n) Space: O(n)

class Solution {
  public String minRemoveToMakeValid(String s) {
        Set<Integer> indexesToRemove = new HashSet<>();
    Stack<Integer> stack = new Stack<>();
    for (int i = 0; i < s.length(); i++) {
      if (s.charAt(i) == '(') {
        stac,push(i);
      } else if (s.charAt(i) == ')') {
        if (stack.isEmpty()) {
          indexesToRemove.add(i);
        } else {
          stack.pop();
        }
      }
    }

    while (!stack.isEmpty()) {
      indexesToRemove.add(stack.pop());
    }

    StringBuilder sb = new StringBuilder();
    for (int i = 0; i < s.length(); i ++) {
      if (!indexesToRemove.contains(i)) {
        sb.append(s.charAt(i));
      }
    }

    return sb.toString();
  }
}