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package Leetcode;
import java.util.ArrayDeque;
import java.util.Deque;
/**
* @author kalpak
*
* 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.
*/
public class MinimumRemoveValidParenthesis {
public static String minRemoveToMakeValid(String s) {
StringBuilder result = new StringBuilder(s);
Deque stack = new ArrayDeque<>();
for(int i = 0; i < result.length(); i++) {
if(result.charAt(i) == '(')
stack.push(i);
else if (result.charAt(i) == ')') {
if(!stack.isEmpty())
stack.pop(); // A matching '(' has been found. Pop the stack.
else
result.setCharAt(i, '*'); // Unnecessary ')'. Mark for removal
}
}
// Now for all '(' in the stack, replace the character with '*' because corresponding ')' doesn't exist
while(!stack.isEmpty())
result.setCharAt(stack.pop(), '*');
return result.toString().replaceAll("\\*", "");
}
public static void main(String[] args) {
String str = "lee(t(c)o)de)";
System.out.println(minRemoveToMakeValid(str));
}
}