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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;
}
};
-
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(); } } ############ class Solution { public String minRemoveToMakeValid(String s) { Deque<Character> stk = new ArrayDeque<>(); int x = 0; for (int i = 0; i < s.length(); ++i) { char c = s.charAt(i); if (c == ')' && x == 0) { continue; } if (c == '(') { ++x; } else if (c == ')') { --x; } stk.push(c); } StringBuilder ans = new StringBuilder(); x = 0; while (!stk.isEmpty()) { char c = stk.pop(); if (c == '(' && x == 0) { continue; } if (c == ')') { ++x; } else if (c == '(') { --x; } ans.append(c); } return ans.reverse().toString(); } }
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// 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; } };
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class Solution: def minRemoveToMakeValid(self, s: str) -> str: stk = [] x = 0 for c in s: if c == ')' and x == 0: continue if c == '(': x += 1 elif c == ')': x -= 1 stk.append(c) x = 0 ans = [] for c in stk[::-1]: if c == '(' and x == 0: continue if c == ')': x += 1 elif c == '(': x -= 1 ans.append(c) return ''.join(ans[::-1]) ############ # 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)
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func minRemoveToMakeValid(s string) string { stk := []byte{} x := 0 for i := range s { c := s[i] if c == ')' && x == 0 { continue } if c == '(' { x++ } else if c == ')' { x-- } stk = append(stk, c) } ans := []byte{} x = 0 for i := len(stk) - 1; i >= 0; i-- { c := stk[i] if c == '(' && x == 0 { continue } if c == ')' { x++ } else if c == '(' { x-- } ans = append(ans, c) } for i, j := 0, len(ans)-1; i < j; i, j = i+1, j-1 { ans[i], ans[j] = ans[j], ans[i] } return string(ans) }
-
function minRemoveToMakeValid(s: string): string { let left = 0; let right = 0; for (const c of s) { if (c === '(') { left++; } else if (c === ')') { if (right < left) { right++; } } } let hasLeft = 0; let res = ''; for (const c of s) { if (c === '(') { if (hasLeft < right) { hasLeft++; res += c; } } else if (c === ')') { if (hasLeft != 0 && right !== 0) { right--; hasLeft--; res += c; } } else { res += c; } } return res; }
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impl Solution { pub fn min_remove_to_make_valid(s: String) -> String { let bs = s.as_bytes(); let mut right = { let mut left = 0; let mut right = 0; for c in bs.iter() { match c { &b'(' => left += 1, &b')' if right < left => right += 1, _ => {} } } right }; let mut has_left = 0; let mut res = vec![]; for c in bs.iter() { match c { &b'(' => { if has_left < right { has_left += 1; res.push(*c); } } &b')' => { if has_left != 0 && right != 0 { right -= 1; has_left -= 1; res.push(*c); } } _ => { res.push(*c); } } } String::from_utf8_lossy(&res).to_string() } }