Description
Given a string s that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.
Return a list of unique strings that are valid with the minimum number of removals. You may return the answer in any order.
Example 1:
Input: s = "()())()"
Output: ["(())()","()()()"]
Example 2:
Input: s = "(a)())()"
Output: ["(a())()","(a)()()"]
Example 3:
Input: s = ")("
Output: [""]
Constraints:
1 <= s.length <= 25s consists of lowercase English letters and parentheses '(' and ')'.- There will be at most
20 parentheses in s.
Solutions
Solution 1
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28 | class Solution:
def removeInvalidParentheses(self, s: str) -> List[str]:
def dfs(i, l, r, lcnt, rcnt, t):
if i == n:
if l == 0 and r == 0:
ans.add(t)
return
if n - i < l + r or lcnt < rcnt:
return
if s[i] == '(' and l:
dfs(i + 1, l - 1, r, lcnt, rcnt, t)
elif s[i] == ')' and r:
dfs(i + 1, l, r - 1, lcnt, rcnt, t)
dfs(i + 1, l, r, lcnt + (s[i] == '('), rcnt + (s[i] == ')'), t + s[i])
l = r = 0
for c in s:
if c == '(':
l += 1
elif c == ')':
if l:
l -= 1
else:
r += 1
ans = set()
n = len(s)
dfs(0, l, r, 0, 0, '')
return list(ans)
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46 | class Solution {
private String s;
private int n;
private Set<String> ans = new HashSet<>();
public List<String> removeInvalidParentheses(String s) {
this.s = s;
this.n = s.length();
int l = 0, r = 0;
for (char c : s.toCharArray()) {
if (c == '(') {
++l;
} else if (c == ')') {
if (l > 0) {
--l;
} else {
++r;
}
}
}
dfs(0, l, r, 0, 0, "");
return new ArrayList<>(ans);
}
private void dfs(int i, int l, int r, int lcnt, int rcnt, String t) {
if (i == n) {
if (l == 0 && r == 0) {
ans.add(t);
}
return;
}
if (n - i < l + r || lcnt < rcnt) {
return;
}
char c = s.charAt(i);
if (c == '(' && l > 0) {
dfs(i + 1, l - 1, r, lcnt, rcnt, t);
}
if (c == ')' && r > 0) {
dfs(i + 1, l, r - 1, lcnt, rcnt, t);
}
int x = c == '(' ? 1 : 0;
int y = c == ')' ? 1 : 0;
dfs(i + 1, l, r, lcnt + x, rcnt + y
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