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Tested Python solution for LeetCode 301 with 24 pytest cases. Generate a practice environment with lcpy.
lcpy gen -n 301 # by problem number
lcpy gen -s remove_invalid_parentheses # by problem name
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.
Input: s = "()())()"
Output: ["(())()","()()()"]
Input: s = "(a)())()"
Output: ["(a())()","(a)()()"]
Input: s = ")("
Output: [""]
1 <= s.length <= 25s consists of lowercase English letters and parentheses '(' and ')'.20 parentheses in s.class Solution:
# Time: O(2^p) over the p <= 20 parenthesis positions, each kept or removed
# Space: O(n) recursion depth plus the result set
def remove_invalid_parentheses(self, s: str) -> list[str]:
rem_left = rem_right = 0
for ch in s:
if ch == "(":
rem_left += 1
elif ch == ")":
if rem_left:
rem_left -= 1
else:
rem_right += 1
results: set[str] = set()
path: list[str] = []
n = len(s)
def dfs(i: int, open_count: int, left_rem: int, right_rem: int) -> None:
if left_rem + right_rem > n - i:
return
if i == n:
if left_rem == 0 and right_rem == 0 and open_count == 0:
results.add("".join(path))
return
ch = s[i]
if ch == "(" and left_rem > 0:
dfs(i + 1, open_count, left_rem - 1, right_rem)
elif ch == ")" and right_rem > 0:
dfs(i + 1, open_count, left_rem, right_rem - 1)
path.append(ch)
if ch == "(":
dfs(i + 1, open_count + 1, left_rem, right_rem)
elif ch == ")" and open_count > 0:
dfs(i + 1, open_count - 1, left_rem, right_rem)
elif ch not in "()":
dfs(i + 1, open_count, left_rem, right_rem)
path.pop()
dfs(0, 0, rem_left, rem_right)
return list(results)
| Time | Space |
|---|---|
| O(2^p) over the p <= 20 parenthesis positions, each kept or removed | O(n) recursion depth plus the result set |