package LeetCode_1249
/**
* 1249. Minimum Remove to Make Valid Parentheses
* https://leetcode.com/problems/minimum-remove-to-make-valid-parentheses/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:
1. It is the empty string, contains only lowercase characters, or
2. It can be written as AB (A concatenated with B), where A and B are valid strings, or
3. 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.
* */
class Solution {
/*
* solution 1: BFS, TLE, Time complexity:O(2^n), Space complexity:O(2^n);
*
* solution 2: two list, keep tracking the index of open and close parentheses,
* Time complexity:O(n), Space complexity:O(n);
* */
fun minRemoveToMakeValid(s: String): String {
//solution 2:
val open = ArrayList<Int>()
val close = ArrayList<Int>()
for (i in s.indices) {
val c = s[i]
if (c == '(') {
open.add(i)
} else if (c == ')') {
if (open.isNotEmpty()) {
//for keep both balance
open.removeAt(open.lastIndex)
} else {
close.add(i)
}
}
}
val charArray = s.toCharArray()
//replace the invalid parentheses
for (i in open) {
charArray[i] = '#'
}
for (i in close) {
charArray[i] = '#'
}
val result = StringBuilder()
for (ch in charArray) {
if (ch != '#') {
result.append(ch)
}
}
return result.toString()
}
}