Remove the minimum number of invalid parentheses in order to make the input string valid. Return all possible results.
Note: The input string may contain letters other than the parentheses ( and ).
Examples:
"()())()" -> ["()()()", "(())()"]
"(a)())()" -> ["(a)()()", "(a())()"]
")(" -> [""]
Credits:
Special thanks to @hpplayer for adding this problem and creating all test cases.
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思路：
题意是给出一个string，其中的小括号可能不配对，移除不配对的括号，返回所有的解。
首先，如何判断括号是否合法。可以用栈，这也是栈这个数据结构的一个典型应用。也可用一个count计数器，遇到左括号++，右括号--，一旦count小于0，就说明不合法。比较推荐count方式，空间复杂度较低。
BFS: 枚举去除的点，当找到后停止BFS树的扩展（因为要去除最少括号，所以即使有其他的结果，也一定在同一层）
DFS: 统计左右括号能删的个数，进行DFS。
Java: DFS
Calculate the number of invalid parentheses of the original string. Iterate through the string. Remove each character and DFS if the number of invalid parentheses decreases.
This solution is based on the fact that if we‘re on the right path to the optimal string, the number of invalid parentheses must always decrease.
Time complexity:
In the worst case, I could have some input like "))))))))", where I need to search through the entire string. The good thing is duplicates will be pruned by the hash set. Calculating mis-match takes O(n). So the overall time complexity is O(n^2).
class Solution {
public List<String> removeInvalidParentheses(String s) {
List<String> list = new ArrayList<>();
Set<String> visited = new HashSet<>();
removeInvalidParentheses(s, numberOfInvalid(s), list, visited);
return list;
}
private void removeInvalidParentheses(String s, int invalid, List<String> list, Set<String> visited) {
if (invalid == 0) {
list.add(s);
return;
}
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) != ‘(‘ && s.charAt(i) != ‘)‘) {
continue;
}
String child = s.substring(0, i) + s.substring(i + 1, s.length());
if (!visited.contains(child)) {
visited.add(child);
int next = numberOfInvalid(child);
if (next < invalid) {
removeInvalidParentheses(child, next, list, visited);
}
}
}
}
private int numberOfInvalid(String s) {
int open = 0;
int close = 0;
for (char c : s.toCharArray()) {
if (c == ‘(‘) {
open++;
} else if (c == ‘)‘) {
if (open == 0) {
close++;
} else {
open--;
}
}
}
return open + close;
}
}
Java: BFS
Thought process:
BFS:
Graph definition:
Vertex: a candidate string.
Edge: two strings s1 and s2 have an edge if s1 equals s2 with one parenthesis deleted.
Put string into a queue.
For the current size of the queue, poll a string from the queue.
Iterate through the string. For each character, remove it, and check if the parentheses are valid.
If so, iterate over current level and return the result.
If not, offer the new string to the queue.
Time complexity:
Say the string‘s length is n. For every character, the choice is to keep or remove. So there are 2^n total states to check. Check if a string is valid is O(1). So the overall time complexity is O(2^n).
class Solution {
public List<String> removeInvalidParentheses(String s) {
List<String> list = new ArrayList<>();
Queue<String> queue = new LinkedList<>();
queue.offer(s);
Set<String> visited = new HashSet<>();
visited.add(s);
boolean found = false;
while (!found && !queue.isEmpty()) {
int size = queue.size();
for (int i = 0; i < size; i++) {
String str = queue.poll();
if (isValid(str)) {
list.add(str);
found = true;
continue;
}
for (int j = 0; j < str.length(); j++) {
if (str.charAt(j) != ‘(‘ && str.charAt(j) != ‘)‘) {
continue;
}
String child = str.substring(0, j) + str.substring(j + 1);
if (!visited.contains(child)) {
queue.offer(child);
visited.add(child);
}
}
}
}
return list;
}
private boolean isValid(String s) {
int open = 0;
for (char c : s.toCharArray()) {
if (c == ‘(‘) {
open++;
} else if (c == ‘)‘) {
if (open == 0) {
return false;
}
open--;
}
}
return open == 0;
}
}
Python: DFS
class Solution(object):
def removeInvalidParentheses(self, s):
"""
:type s: str
:rtype: List[str]
"""
def dfs(s):
mi = calc(s)
if mi == 0:
return [s]
ans = []
for x in range(len(s)):
if s[x] in (‘(‘, ‘)‘):
ns = s[:x] + s[x+1:]
if ns not in visited and calc(ns) < mi:
visited.add(ns)
ans.extend(dfs(ns))
return ans
def calc(s):
a = b = 0
for c in s:
a += {‘(‘ : 1, ‘)‘ : -1}.get(c, 0)
b += a < 0
a = max(a, 0)
return a + b
visited = set([s])
return dfs(s)
Python: BFS
class Solution(object):
def removeInvalidParentheses(self, s):
"""
:type s: str
:rtype: List[str]
"""
def calc(s):
a = b = 0
for c in s:
a += {‘(‘ : 1, ‘)‘ : -1}.get(c, 0)
b += a < 0
a = max(a, 0)
return a + b
visited = set([s])
ans = []
queue = collections.deque([s])
done = False
while queue:
t = queue.popleft()
mi = calc(t)
if mi == 0:
done = True
ans.append(t)
if done:
continue
for x in range(len(t)):
if t[x] not in (‘(‘, ‘)‘):
continue
ns = t[:x] + t[x+1:]
if ns not in visited and calc(ns) < mi:
visited.add(ns)
queue.append(ns)
return ans
Python: DFS
class Solution(object):
def removeInvalidParentheses(self, s):
"""
:type s: str
:rtype: List[str]
"""
if not s: return [‘‘]
left_remove = right_remove = 0
for c in s:
if c == ‘(‘:
left_remove += 1
elif c == ‘)‘:
if left_remove:
left_remove -= 1
else:
right_remove += 1
ans = set()
self.dfs(0, left_remove, right_remove, 0, ‘‘, s, ans)
return list(ans)
def dfs(self, index, left_remove, right_remove, left_pare, cur, s, ans):
if left_remove < 0 or right_remove < 0 or left_pare < 0: return
if index == len(s):
if left_remove == right_remove == left_pare == 0:
ans.add(cur)
return
if s[index] == ‘(‘:
self.dfs(index + 1, left_remove - 1, right_remove, left_pare, cur, s, ans)
self.dfs(index + 1, left_remove, right_remove, left_pare + 1, cur + s[index], s, ans)
elif s[index] == ‘)‘:
self.dfs(index + 1, left_remove, right_remove - 1, left_pare, cur, s, ans)
self.dfs(index + 1, left_remove, right_remove, left_pare - 1, cur + s[index], s, ans)
else:
self.dfs(index + 1, left_remove, right_remove, left_pare, cur + s[index], s, ans)
Python: BFS
class Solution(object):
def removeInvalidParentheses(self, s):
"""
:type s: str
:rtype: List[str]
"""
if not s: return [‘‘]
q, ans, vis = [s], [], set([s])
found = False
while q:
cur = q.pop(0)
if self.isValidParentheses(cur):
found = True
ans.append(cur)
elif not found:
for i in xrange(len(cur)):
if cur[i] == ‘(‘ or cur[i] == ‘)‘:
t = cur[:i] + cur[i + 1:]
if t not in vis:
q.append(t)
vis.add(t)
return ans
def isValidParentheses(self, s):
cnt = 0
for c in s:
if c == ‘(‘:
cnt += 1
elif c == ‘)‘:
if cnt == 0: return False
cnt -= 1
return cnt == 0
Python: BFS, concise
class Solution(object):
def removeInvalidParentheses(self, s):
"""
:type s: str
:rtype: List[str]
"""
if not s: return [‘‘]
q = {s}
while q:
ans = filter(self.isValidParentheses,q)
if ans: return ans
q = {cur[:i] + cur[i + 1:] for cur in q for i in xrange(len(cur))}
def isValidParentheses(self, s):
cnt = 0
for c in s:
if c == ‘(‘:
cnt += 1
elif c == ‘)‘:
if cnt == 0: return False
cnt -= 1
return cnt == 0
C++:
class Solution {
public:
vector<string> removeInvalidParentheses(string s) {
vector<string> res;
unordered_set<string> visited{{s}};
queue<string> q{{s}};
bool found = false;
while (!q.empty()) {
string t = q.front(); q.pop();
if (isValid(t)) {
res.push_back(t);
found = true;
}
if (found) continue;
for (int i = 0; i < t.size(); ++i) {
if (t[i] != ‘(‘ && t[i] != ‘)‘) continue;
string str = t.substr(0, i) + t.substr(i + 1);
if (!visited.count(str)) {
q.push(str);
visited.insert(str);
}
}
}
return res;
}
bool isValid(string t) {
int cnt = 0;
for (int i = 0; i < t.size(); ++i) {
if (t[i] == ‘(‘) ++cnt;
else if (t[i] == ‘)‘ && --cnt < 0) return false;
}
return cnt == 0;
}
};