Algorithm Notes
Summary: Generate Parentheses — notes not yet curated.
Time: Estimate via loops/recurrences; common classes: O(1), O(log n), O(n), O(n log n), O(n^2)
Space: Count auxiliary structures and recursion depth.
Tip: See the Big-O Guide for how to derive bounds and compare trade-offs.
Big-O Guide
Source
"""
Generate Parentheses
Given `n` pairs of parentheses, write a function to generate all
combinations of well-formed parentheses.
Example:
Input: n = 3
Output: ["((()))","(()())","(())()","()(())","()()()"]
Constraints:
- 1 <= n <= 8
"""
from interview_workbook.leetcode._registry import register_problem
from interview_workbook.leetcode._types import Category, Difficulty
class Solution:
def solve(self, n: int) -> list[str]:
"""Return all combinations of n pairs of valid parentheses."""
res = []
def backtrack(s, left, right):
if len(s) == 2 * n:
res.append(s)
return
if left < n:
backtrack(s + "(", left + 1, right)
if right < left:
backtrack(s + ")", left, right + 1)
backtrack("", 0, 0)
return res
def demo():
return str(Solution().solve(3))
register_problem(
id=22,
slug="generate_parentheses",
title="Generate Parentheses",
category=Category.STACK,
difficulty=Difficulty.MEDIUM,
tags=["stack", "backtracking"],
url="https://leetcode.com/problems/generate-parentheses/",
notes="Backtracking with stack constraints. Ensure left>=right always.",
)