Generate parentheses (#10903)

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2024-11-23 21:11:08 +00:00 · 2023-10-29 18:25:31 +05:30 · 2023-10-29 18:25:31 +05:30 · cc22d0b0ba
commit cc22d0b0ba
parent 403d3b8a29
1 changed files with 77 additions and 0 deletions
--- a/backtracking/generate_parentheses.py
+++ 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()