[LeetCode February Challange] Day 19 - Minimum Remove to Make Valid Parentheses
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
Time complexity : O(n)
Space complexity : O(1)
class Solution {
public:
string minRemoveToMakeValid(string s) {
int close = count(s.begin(), s.end(), ')');
int open = 0;
string ans;
for (char c: s) {
if (c == '(') {
if (close <= open)
continue;
++open;
} else if (c ==')') {
--close;
if (open <= 0)
continue;
--open;
}
ans += c;
}
return ans;
}
};