Generate Parentheses LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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In this post, you will find the solution for the Generate Parentheses in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Link for the Problem – Generate Parentheses– LeetCode Problem

Generate Parentheses– LeetCode Problem

Problem:

Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.

Example 1:

Input: n = 3
Output: ["((()))","(()())","(())()","()(())","()()()"]

Example 2:

Input: n = 1
Output: ["()"]

Constraints:

• 1 <= n <= 8

Generate Parentheses– LeetCode Solutions

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, move(path), ans);
      path.pop_back();
    }
    if (l < r) {
      path.push_back(')');
      dfs(l, r - 1, 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, final 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: str) -> None:
      if l == 0 and r == 0:
        ans.append(s)
      if l > 0:
        dfs(l - 1, r, s + '(')
      if l < r:
        dfs(l, r - 1, s + ')')

    dfs(n, n, '')

    return ans