diff --git a/backtracking/generate_parentheses.py b/backtracking/generate_parentheses.py new file mode 100644 index 000000000..18c21e2a9 --- /dev/null +++ b/backtracking/generate_parentheses.py @@ -0,0 +1,77 @@ +""" +author: Aayush Soni +Given n pairs of parentheses, write a function to generate all +combinations of well-formed parentheses. +Input: n = 2 +Output: ["(())","()()"] +Leetcode link: https://leetcode.com/problems/generate-parentheses/description/ +""" + + +def backtrack( + partial: str, open_count: int, close_count: int, n: int, result: list[str] +) -> None: + """ + Generate valid combinations of balanced parentheses using recursion. + + :param partial: A string representing the current combination. + :param open_count: An integer representing the count of open parentheses. + :param close_count: An integer representing the count of close parentheses. + :param n: An integer representing the total number of pairs. + :param result: A list to store valid combinations. + :return: None + + This function uses recursion to explore all possible combinations, + ensuring that at each step, the parentheses remain balanced. + + Example: + >>> result = [] + >>> backtrack("", 0, 0, 2, result) + >>> result + ['(())', '()()'] + """ + if len(partial) == 2 * n: + # When the combination is complete, add it to the result. + result.append(partial) + return + + if open_count < n: + # If we can add an open parenthesis, do so, and recurse. + backtrack(partial + "(", open_count + 1, close_count, n, result) + + if close_count < open_count: + # If we can add a close parenthesis (it won't make the combination invalid), + # do so, and recurse. + backtrack(partial + ")", open_count, close_count + 1, n, result) + + +def generate_parenthesis(n: int) -> list[str]: + """ + Generate valid combinations of balanced parentheses for a given n. + + :param n: An integer representing the number of pairs of parentheses. + :return: A list of strings with valid combinations. + + This function uses a recursive approach to generate the combinations. + + Time Complexity: O(2^(2n)) - In the worst case, we have 2^(2n) combinations. + Space Complexity: O(n) - where 'n' is the number of pairs. + + Example 1: + >>> generate_parenthesis(3) + ['((()))', '(()())', '(())()', '()(())', '()()()'] + + Example 2: + >>> generate_parenthesis(1) + ['()'] + """ + + result: list[str] = [] + backtrack("", 0, 0, n, result) + return result + + +if __name__ == "__main__": + import doctest + + doctest.testmod()