all_combinations.py

"""
In this problem, we want to determine all possible combinations of k
numbers out of 1 ... n. We use backtracking to solve this problem.

Time complexity: O(C(n,k)) which is O(n choose k) = O((n!/(k! * (n - k)!))),
"""

from __future__ import annotations

from itertools import combinations


def combination_lists(n: int, k: int) -> list[list[int]]:
    """
    Generates all possible combinations of k numbers out of 1 ... n using itertools.

    >>> combination_lists(n=4, k=2)
    [[1, 2], [1, 3], [1, 4], [2, 3], [2, 4], [3, 4]]
    >>> combination_lists(n=5, k=3)
    [[1, 2, 3], [1, 2, 4], [1, 2, 5],
     [1, 3, 4], [1, 3, 5], [1, 4, 5],
     [2, 3, 4], [2, 3, 5], [2, 4, 5],
     [3, 4, 5]]
    """
    return [list(x) for x in combinations(range(1, n + 1), k)]


def generate_all_combinations(n: int, k: int) -> list[list[int]]:
    """
    Generates all possible combinations of k numbers out of 1 ... n using backtracking.

    >>> generate_all_combinations(n=4, k=2)
    [[1, 2], [1, 3], [1, 4], [2, 3], [2, 4], [3, 4]]
    >>> generate_all_combinations(n=0, k=0)
    [[]]
    >>> generate_all_combinations(n=10, k=-1)
    Traceback (most recent call last):
        ...
    ValueError: k must not be negative
    >>> generate_all_combinations(n=-1, k=10)
    Traceback (most recent call last):
        ...
    ValueError: n must not be negative
    >>> generate_all_combinations(n=5, k=4)
    [[1, 2, 3, 4], [1, 2, 3, 5], [1, 2, 4, 5], [1, 3, 4, 5], [2, 3, 4, 5]]
    >>> generate_all_combinations(n=3, k=3)
    [[1, 2, 3]]
    >>> generate_all_combinations(n=3, k=1)
    [[1], [2], [3]]
    >>> generate_all_combinations(n=1, k=0)
    [[]]
    >>> generate_all_combinations(n=1, k=1)
    [[1]]
    >>> from itertools import combinations
    >>> all(generate_all_combinations(n, k) == combination_lists(n, k)
    ...     for n in range(1, 6) for k in range(1, 6))
    True
    """
    if k < 0:
        raise ValueError("k must not be negative")
    if n < 0:
        raise ValueError("n must not be negative")

    result: list[list[int]] = []
    create_all_state(1, n, k, [], result)
    return result


def create_all_state(
    increment: int,
    total_number: int,
    level: int,
    current_list: list[int],
    total_list: list[list[int]],
) -> None:
    """
    Helper function to recursively build all combinations.

    >>> create_all_state(1, 4, 2, [], result := [])
    >>> result
    [[1, 2], [1, 3], [1, 4], [2, 3], [2, 4], [3, 4]]
    >>> create_all_state(1, 3, 3, [], result := [])
    >>> result
    [[1, 2, 3]]
    >>> create_all_state(2, 2, 1, [1], result := [])
    >>> result
    [[1, 2]]
    >>> create_all_state(1, 0, 0, [], result := [])
    >>> result
    [[]]
    >>> create_all_state(1, 4, 0, [1, 2], result := [])
    >>> result
    [[1, 2]]
    >>> create_all_state(5, 4, 2, [1, 2], result := [])
    >>> result
    []
    """
    if level == 0:
        total_list.append(current_list[:])
        return

    for i in range(increment, total_number - level + 2):
        current_list.append(i)
        create_all_state(i + 1, total_number, level - 1, current_list, total_list)
        current_list.pop()


if __name__ == "__main__":
    from doctest import testmod

    testmod()
    print(generate_all_combinations(n=4, k=2))
    tests = ((n, k) for n in range(1, 5) for k in range(1, 5))
    for n, k in tests:
        print(n, k, generate_all_combinations(n, k) == combination_lists(n, k))

    print("Benchmark:")
    from timeit import timeit

    for func in ("combination_lists", "generate_all_combinations"):
        print(f"{func:>25}(): {timeit(f'{func}(n=4, k = 2)', globals=globals())}")