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Question
Formatted question description: https://leetcode.ca/all/301.html
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 <= 25sconsists of lowercase English letters and parentheses'('and')'.- There will be at most
20parentheses ins.
Algorithm
Use BFS to solve, put the given string into the queue, and then take it out to check whether it is legal
- If legal, return directly
- If it is illegal, traverse it, and remove the bracket characters to generate a new string for the characters in the left and right brackets encountered
- If this string has not been encountered before, put it in the queue and use HashSet to record whether a string has appeared.
Perform the same operation on each element in the queue, and return to the empty set if no valid string is found until the queue is empty.
Code
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import java.util.ArrayList; import java.util.HashSet; import java.util.List; import java.util.Set; public class Remove_Invalid_Parentheses { // reference: https://leetcode.com/problems/remove-invalid-parentheses/solution/ class Solution_dfs { private Set<String> validExpressions = new HashSet<String>(); private int minimumRemoved; private void reset() { this.validExpressions.clear(); this.minimumRemoved = Integer.MAX_VALUE; } private void dfs( String s, int index, int leftCount, int rightCount, StringBuilder expression, int removedCount) { // If we have reached the end of string. if (index == s.length()) { // If the current expression is valid. if (leftCount == rightCount) { // If the current count of removed parentheses is <= the current minimum count if (removedCount <= this.minimumRemoved) { // Convert StringBuilder to a String. This is an expensive operation. // So we only perform this when needed. String possibleAnswer = expression.toString(); // If the current count beats the overall minimum we have till now if (removedCount < this.minimumRemoved) { this.validExpressions.clear(); this.minimumRemoved = removedCount; } this.validExpressions.add(possibleAnswer); } } } else { char currentCharacter = s.charAt(index); int length = expression.length(); // If the current character is neither an opening bracket nor a closing one, // simply recurse further by adding it to the expression StringBuilder if (currentCharacter != '(' && currentCharacter != ')') { expression.append(currentCharacter); this.dfs(s, index + 1, leftCount, rightCount, expression, removedCount); expression.deleteCharAt(length); } else { // Recursion where we delete the current character at 'index' and move forward to 'index+1' this.dfs(s, index + 1, leftCount, rightCount, expression, removedCount + 1); expression.append(currentCharacter); // If it's an opening parenthesis, consider it and recurse if (currentCharacter == '(') { this.dfs(s, index + 1, leftCount + 1, rightCount, expression, removedCount); } else if (rightCount < leftCount) { // For a closing parenthesis, only recurse if right < left this.dfs(s, index + 1, leftCount, rightCount + 1, expression, removedCount); } // Undoing the append operation for other recursions. expression.deleteCharAt(length); } } } public List<String> removeInvalidParentheses(String s) { this.reset(); this.dfs(s, 0, 0, 0, new StringBuilder(), 0); return new ArrayList(this.validExpressions); } } public class Solution_bfs { public List<String> removeInvalidParentheses(String s) { List<String> res = new ArrayList<>(); // sanity check if (s == null) return res; Set<String> visited = new HashSet<>(); Queue<String> queue = new LinkedList<>(); // initialize queue.add(s); visited.add(s); boolean found = false; while (!queue.isEmpty()) { s = queue.poll(); if (isValid(s)) { // found an answer, add to the result res.add(s); found = true; } if (found) continue; // continue to check all in queue, but no more new adding to queue // generate all possible states for (int i = 0; i < s.length(); i++) { // we only try to remove left or right parent if (s.charAt(i) != '(' && s.charAt(i) != ')') continue; String t = s.substring(0, i) + s.substring(i + 1); // skip char at index=i if (!visited.contains(t)) { // for each state, if it's not visited, add it to the queue queue.add(t); visited.add(t); } } } return res; } // helper function checks if string s contains valid parantheses boolean isValid(String s) { int count = 0; for (int i = 0; i < s.length(); i++) { char c = s.charAt(i); if (c == '(') count++; if (c == ')') count--; if (count < 0) return false; } return count == 0; } } } ############ 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, t + c); } } -
// OJ: https://leetcode.com/problems/remove-invalid-parentheses/ // Time: O(2^N) // Space: O(N) class Solution { public: vector<string> removeInvalidParentheses(string s) { int N = s.size(); vector<int> right(N + 1); for (int i = N - 1; i >= 0; --i) right[i] = max(0, right[i + 1] + (s[i] == '(' || s[i] == ')' ? (s[i] == ')' ? 1 : -1) : 0)); unordered_set<string> ans; string tmp; function<void(int, int)> dfs = [&](int i, int left) { if (i == N) { if (left == 0) ans.insert(tmp); return; } if (s[i] == '(') { if (left + 1 > right[i + 1]) { // we can remove this '(' dfs(i + 1, left); } ++left; } else if (s[i] == ')') { if (right[i] > left) { // we can remove this ')' dfs(i + 1, left); } if (left > 0) { --left; } else return; } tmp.push_back(s[i]); dfs(i + 1, left); tmp.pop_back(); }; dfs(0, 0); return vector<string>(begin(ans), end(ans)); } }; -
# python3 class Solution: ''' - `i` 表示当前处理到字符串 $s$ 的第 $i$ 个字符； - `l` 和 `r` 分别表示剩余待删除的左、右括号的数量； - `t` 表示当前得到的字符串； - `lcnt` 和 `rcnt` 分别表示当前得到的字符串中左、右括号的数量。 ''' 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) # note here is elif, not another if, because, to get min-deletion valid string, always delete from extra '(' first elif s[i] == ')' and r: dfs(i + 1, l, r - 1, lcnt, rcnt, t) # s[i] is a char, or not delete any '(' or ')' so wait for future recursions to do it dfs(i + 1, l, r, lcnt + (s[i] == '('), rcnt + (s[i] == ')'), t + s[i]) l = r = 0 for c in s: # count how many ( or ) to be removed to get valid one if c == '(': l += 1 elif c == ')': if l: l -= 1 else: r += 1 ans = set() n = len(s) dfs(0, l, r, 0, 0, '') # minimum removal, is archieved by strictly applying counted l/r return list(ans) # bfs version class Solution(object): def removeInvalidParentheses(self, s): def isvalid(s): ctr = 0 for c in s: if c == '(': ctr += 1 elif c == ')': ctr -= 1 if ctr < 0: return False return ctr == 0 ''' >>> a = {1,2,3,1,2,3} >>> a {1, 2, 3} >>> list(a) [1, 2, 3] ''' level = {s} while True: valid = filter(isvalid, level) if valid: return valid level = {s[:i] + s[i+1:] for s in level for i in range(len(s))} # bfs version, python3 class Solution: def removeInvalidParentheses(self, s: str) -> List[str]: def isvalid(s): ctr = 0 for c in s: if c == '(': ctr += 1 elif c == ')': ctr -= 1 if ctr < 0: return False return ctr == 0 level = {s} while True: valid = filter(isvalid, level) converted = list(valid) if converted: # in python3, valid from filter() will always be not-None, so need to convert to list return converted level = {s[:i] + s[i+1:] for s in level for i in range(len(s))} # official-solution: Backtracking. https://leetcode.com/problems/remove-invalid-parentheses/editorial/ class Solution(object): def __init__(self): self.valid_expressions = None self.min_removed = None def reset(self): self.valid_expressions = set() self.min_removed = float("inf") """ string: The original string we are recursing on. index: current index in the original string. left: number of left parentheses till now. right: number of right parentheses till now. ans: the resulting expression in this particular recursion. ignored: number of parentheses ignored in this particular recursion. """ def remaining(self, string, index, left_count, right_count, expr, rem_count): # If we have reached the end of string. if index == len(string): # If the current expression is valid. The only scenario where it can be # invalid here is if left > right. The other way around we handled early on in the recursion. if left_count == right_count: if rem_count <= self.min_removed: # This is the resulting expression. # Strings are immutable in Python so we move around a list in the recursion # and eventually join to get the final string. possible_ans = "".join(expr) # If the current count of brackets removed < current minimum, ignore # previous answers and update the current minimum count. if rem_count < self.min_removed: self.valid_expressions = set() self.min_removed = rem_count self.valid_expressions.add(possible_ans) else: current_char = string[index] # If the current character is not a parenthesis, just recurse one step ahead. if current_char != '(' and current_char != ')': expr.append(current_char) self.remaining(string, index + 1, left_count, right_count, expr, rem_count) expr.pop() else: # Else, one recursion is with ignoring the current character. # So, we increment the ignored counter and leave the left and right untouched. self.remaining(string, index + 1, left_count, right_count, expr, rem_count + 1) expr.append(current_char) # If the current parenthesis is an opening bracket, we consider it # and increment left and move forward if string[index] == '(': self.remaining(string, index + 1, left_count + 1, right_count, expr, rem_count) elif right_count < left_count: # If the current parenthesis is a closing bracket, we consider it only if we # have more number of opening brackets and increment right and move forward. self.remaining(string, index + 1, left_count, right_count + 1, expr, rem_count) expr.pop() def removeInvalidParentheses(self, s): """ :type s: str :rtype: List[str] """ # Reset the class level variables that we use for every test case. self.reset() # Recursive call self.remaining(s, 0, 0, 0, [], 0) return list(self.valid_expressions) # official-solution: Limited Backtracking. https://leetcode.com/problems/remove-invalid-parentheses/editorial/ class Solution: def removeInvalidParentheses(self, s): """ :type s: str :rtype: List[str] """ left = 0 right = 0 # First, we find out the number of misplaced left and right parentheses. for char in s: # Simply record the left one. if char == '(': left += 1 elif char == ')': # If we don't have a matching left, then this is a misplaced right, record it. right = right + 1 if left == 0 else right # Decrement count of left parentheses because we have found a right # which CAN be a matching one for a left. left = left - 1 if left > 0 else left result = {} def recurse(s, index, left_count, right_count, left_rem, right_rem, expr): # If we reached the end of the string, just check if the resulting expression is # valid or not and also if we have removed the total number of left and right # parentheses that we should have removed. if index == len(s): if left_rem == 0 and right_rem == 0: ans = "".join(expr) result[ans] = 1 else: # The discard case. Note that here we have our pruning condition. # We don't recurse if the remaining count for that parenthesis is == 0. if (s[index] == '(' and left_rem > 0) or (s[index] == ')' and right_rem > 0): recurse(s, index + 1, left_count, right_count, left_rem - (s[index] == '('), right_rem - (s[index] == ')'), expr) expr.append(s[index]) # Simply recurse one step further if the current character is not a parenthesis. if s[index] != '(' and s[index] != ')': recurse(s, index + 1, left_count, right_count, left_rem, right_rem, expr) elif s[index] == '(': # Consider an opening bracket. recurse(s, index + 1, left_count + 1, right_count, left_rem, right_rem, expr) elif s[index] == ')' and left_count > right_count: # Consider a closing bracket. recurse(s, index + 1, left_count, right_count + 1, left_rem, right_rem, expr) # Pop for backtracking. expr.pop() # Now, the left and right variables tell us the number of misplaced left and # right parentheses and that greatly helps pruning the recursion. recurse(s, 0, 0, 0, left, right, []) return list(result.keys()) ############ class Solution(object): def removeInvalidParentheses(self, s): """ :type s: str :rtype: List[str] """ def isValid(s): stack = [] for c in s: if c == "(": stack.append("(") elif c == ")": stack.append(")") if len(stack) >= 2 and stack[-2] + stack[-1] == "()": stack.pop() stack.pop() return len(stack) def dfs(s, res, cache, length): if s in cache: return if len(s) == length: if isValid(s) == 0: res.append(s) return for i in range(0, len(s)): if s[i] == "(" or s[i] == ")" and len(s) - 1 >= length: dfs(s[:i] + s[i + 1:], res, cache, length) cache.add(s[:i] + s[i + 1:]) prepLeft = "" for i in range(0, len(s)): if s[i] == "(": prepLeft += s[i:] break elif s[i] != ")": prepLeft += s[i] prepRight = "" for i in reversed(range(0, len(prepLeft))): if prepLeft[i] == ")": prepRight += prepLeft[:i + 1][::-1] break elif prepLeft[i] != "(": prepRight += prepLeft[i] s = prepRight[::-1] length = len(s) - isValid(s) res = [] cache = set() dfs(s, res, cache, length) return res -
func removeInvalidParentheses(s string) []string { vis := map[string]bool{} l, r, n := 0, 0, len(s) for _, c := range s { if c == '(' { l++ } else if c == ')' { if l > 0 { l-- } else { r++ } } } var dfs func(i, l, r, lcnt, rcnt int, t string) dfs = func(i, l, r, lcnt, rcnt int, t string) { if i == n { if l == 0 && r == 0 { vis[t] = true } return } if n-i < l+r || lcnt < rcnt { return } if s[i] == '(' && l > 0 { dfs(i+1, l-1, r, lcnt, rcnt, t) } if s[i] == ')' && r > 0 { dfs(i+1, l, r-1, lcnt, rcnt, t) } x, y := 0, 0 if s[i] == '(' { x = 1 } else if s[i] == ')' { y = 1 } dfs(i+1, l, r, lcnt+x, rcnt+y, t+string(s[i])) } dfs(0, l, r, 0, 0, "") ans := make([]string, 0, len(vis)) for v := range vis { ans = append(ans, v) } return ans }