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.
Solution
We can use a stack to trace the balanced parentheses also use a Hashset to record the index of characters need to be removed.
/*
time complexity O(N)
space complexity O(N)
*/
import java.util.*;
class Solution {
public String minRemoveToMakeValid(String s) {
if(s == null || s.length() == 0)
return s;
Stack<Integer> paren = new Stack<>();
Set<Integer> remove = new HashSet<>();
StringBuilder sb = new StringBuilder();
for(int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if(c == '(') {
paren.push(i);
} else if(c == ')') {
if(paren.isEmpty()) {
remove.add(i);
} else {
paren.pop();
}
}
}
if(!paren.isEmpty()) {
while(!paren.isEmpty()) {
remove.add(paren.pop());
}
}
for(int i = 0; i < s.length(); i++) {
if(!remove.contains(i))
sb.append(s.charAt(i));
}
return sb.toString();
}
}
The time complexity is O(N) we need to process each character.
The space complexity is O(N) we store at most N positions in the stack and set.
