Formatted question description: https://leetcode.ca/all/1249.html
1249. Minimum Remove to Make Valid Parentheses (Medium)
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 withB
), whereA
andB
are valid strings, or - It can be written as
(A)
, whereA
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 1. Scan Left to Right then Right to Left
// OJ: https://leetcode.com/problems/minimum-remove-to-make-valid-parentheses/
// Time: O(N)
// Space: O(1)
class Solution {
public:
string minRemoveToMakeValid(string s) {
int left = 0, i, j;
for (i = 0, j = 0; i < s.size(); ++i) {
if (isalpha(s[i]) || (s[i] == '(' && ++left) || (s[i] == ')' && --left >= 0)) s[j++] = s[i];
left = max(left, 0);
}
s.resize(j);
reverse(begin(s), end(s));
left = 0;
for (i = 0, j = 0; i < s.size(); ++i) {
if (isalpha(s[i]) || (s[i] == ')' && ++left) || (s[i] == '(' && --left >= 0)) s[j++] = s[i];
left = max(left, 0);
}
s.resize(j);
reverse(begin(s), end(s));
return s;
}
};
Solution 2. Stack
// OJ: https://leetcode.com/problems/minimum-remove-to-make-valid-parentheses/
// Time: O(N)
// Space: O(N)
class Solution {
public:
string minRemoveToMakeValid(string s) {
stack<int> st;
for (int i = 0; i < s.size(); ++i) {
if (s[i] == '(') st.push(i);
else if (s[i] == ')') {
if (st.empty()) s[i] = '*';
else st.pop();
}
}
while (st.size()) {
s[st.top()] = '*';
st.pop();
}
s.erase(remove(begin(s), end(s), '*'), end(s));
return s;
}
};
Java
-
class Solution { public String minRemoveToMakeValid(String s) { Stack<Integer> stack = new Stack<Integer>(); StringBuffer sb = new StringBuffer(); int length = s.length(); int deleteCount = 0; for (int i = 0; i < length; i++) { char c = s.charAt(i); if (Character.isLetter(c)) sb.append(c); else if (c == '(') { sb.append(c); stack.push(i - deleteCount); } else if (c == ')') { if (!stack.isEmpty()) { sb.append(c); stack.pop(); } else deleteCount++; } } while (!stack.isEmpty()) { int index = stack.pop(); sb.deleteCharAt(index); } return sb.toString(); } }
-
// OJ: https://leetcode.com/problems/minimum-remove-to-make-valid-parentheses/ // Time: O(N) // Space: O(1) if changing input string is allowed; otherwise O(N) class Solution { public: string minRemoveToMakeValid(string s) { int j = 0, left = 0; for (int i = 0; i < s.size(); ++i) { if (s[i] == '(') ++left; else if (s[i] == ')') { if (left == 0) continue; --left; } s[j++] = s[i]; } s.resize(j); reverse(begin(s), end(s)); j = 0, left = 0; for (int i = 0; i < s.size(); ++i) { if (s[i] == ')') ++left; else if (s[i] == '(') { if (left == 0) continue; --left; } s[j++] = s[i]; } s.resize(j); reverse(begin(s), end(s)); return s; } };
-
# 1249. Minimum Remove to Make Valid Parentheses # https://leetcode.com/problems/minimum-remove-to-make-valid-parentheses/ class Solution: def minRemoveToMakeValid(self, s: str): stack = [] s = list(s) for i,c in enumerate(s): if c == ")": if stack: stack.pop() else: s[i] = "" elif c == "(": stack.append(i) while stack: s[stack.pop()] = "" return "".join(s) def minRemoveToMakeValid(self, s: str): stack = [] res = [""] * len(s) forbidden = set() for i,c in enumerate(s): if c == ")": if stack and stack[-1][1] == "(": stack.pop() res[i] = c else: forbidden.add(i) elif c == "(": res[i] = c stack.append((i,c)) else: res[i] = c if len(stack) > 0: forbidden |= set([i for i,_ in stack]) for i in forbidden: res[i] = "" return "".join(res) def minRemoveToMakeValid(self, s: str): stack = [] res = [] forbidden = set() for i,c in enumerate(s): if c != "(" and c != ")": res.append(c) else: if c == ")": if stack and stack[-1][1] == "(": stack.pop() res.append(c) else: forbidden.add(i) else: res.append(c) stack.append((i,c)) if len(stack) > 0: forbidden |= set([i for i,_ in stack]) ans = [] for i,x in enumerate(s): if i not in forbidden: ans.append(x) return "".join(ans) return "".join(res)