22. Generate Parentheses ¶

Time: $O(2^{2n})$
Space: $O(n)$

C++JavaPythonJavaScript

class Solution {
 public:
  vector<string> generateParenthesis(int n) {
    vector<string> ans;
    dfs(n, n, "", ans);
    return ans;
  }

 private:
  void dfs(int l, int r, string&& path, vector<string>& ans) {
    if (l == 0 && r == 0) {
      ans.push_back(path);
      return;
    }

    if (l > 0) {
      path.push_back('(');
      dfs(l - 1, r, std::move(path), ans);
      path.pop_back();
    }
    if (l < r) {
      path.push_back(')');
      dfs(l, r - 1, std::move(path), ans);
      path.pop_back();
    }
  }
};

class Solution {
  public List<String> generateParenthesis(int n) {
    List<String> ans = new ArrayList<>();
    dfs(n, n, new StringBuilder(), ans);
    return ans;
  }

  private void dfs(int l, int r, StringBuilder sb, List<String> ans) {
    if (l == 0 && r == 0) {
      ans.add(sb.toString());
      return;
    }

    if (l > 0) {
      sb.append("(");
      dfs(l - 1, r, sb, ans);
      sb.deleteCharAt(sb.length() - 1);
    }
    if (l < r) {
      sb.append(")");
      dfs(l, r - 1, sb, ans);
      sb.deleteCharAt(sb.length() - 1);
    }
  }
}

class Solution:
  def generateParenthesis(self, n):
    ans = []

    def dfs(l: int, r: int, s: list[str]) -> None:
      if l == 0 and r == 0:
        ans.append(''.join(s))
      if l > 0:
        s.append('(')
        dfs(l - 1, r, s)
        s.pop()
      if l < r:
        s.append(')')
        dfs(l, r - 1, s)
        s.pop()

    dfs(n, n, [])
    return ans

var generateParenthesis = function(n) {
  const res = [];
  const backtrack = (open, close, cur) => {
    if (cur.length === 2 * n) { res.push(cur); return; }
    if (open < n) backtrack(open + 1, close, cur + "(");
    if (close < open) backtrack(open, close + 1, cur + ")");
  };
  backtrack(0, 0, "");
  return res;
};