backtracking/power_sum.py - master - xddxdd/Python

backtracking/power_sum.py
91 строка3 KB
Muhammad Umer Farooq
Updated test cases of power_sum.py (#9978)
14 окт 2023, 11:31
14 окт 2023, 11:3137cae3f
"""
Problem source: https://www.hackerrank.com/challenges/the-power-sum/problem
Find the number of ways that a given integer X, can be expressed as the sum
of the Nth powers of unique, natural numbers. For example, if X=13 and N=2.
We have to find all combinations of unique squares adding up to 13.
The only solution is 2^2+3^2. Constraints: 1<=X<=1000, 2<=N<=10.
"""


def backtrack(
    needed_sum: int,
    power: int,
    current_number: int,
    current_sum: int,
    solutions_count: int,
) -> tuple[int, int]:
    """
    >>> backtrack(13, 2, 1, 0, 0)
    (0, 1)
    >>> backtrack(10, 2, 1, 0, 0)
    (0, 1)
    >>> backtrack(10, 3, 1, 0, 0)
    (0, 0)
    >>> backtrack(20, 2, 1, 0, 0)
    (0, 1)
    >>> backtrack(15, 10, 1, 0, 0)
    (0, 0)
    >>> backtrack(16, 2, 1, 0, 0)
    (0, 1)
    >>> backtrack(20, 1, 1, 0, 0)
    (0, 64)
    """
    if current_sum == needed_sum:
        # If the sum of the powers is equal to needed_sum, then we have a solution.
        solutions_count += 1
        return current_sum, solutions_count

    i_to_n = current_number**power
    if current_sum + i_to_n <= needed_sum:
        # If the sum of the powers is less than needed_sum, then continue adding powers.
        current_sum += i_to_n
        current_sum, solutions_count = backtrack(
            needed_sum, power, current_number + 1, current_sum, solutions_count
        )
        current_sum -= i_to_n
    if i_to_n < needed_sum:
        # If the power of i is less than needed_sum, then try with the next power.
        current_sum, solutions_count = backtrack(
            needed_sum, power, current_number + 1, current_sum, solutions_count
        )
    return current_sum, solutions_count


def solve(needed_sum: int, power: int) -> int:
    """
    >>> solve(13, 2)
    1
    >>> solve(10, 2)
    1
    >>> solve(10, 3)
    0
    >>> solve(20, 2)
    1
    >>> solve(15, 10)
    0
    >>> solve(16, 2)
    1
    >>> solve(20, 1)
    Traceback (most recent call last):
        ...
    ValueError: Invalid input
    needed_sum must be between 1 and 1000, power between 2 and 10.
    >>> solve(-10, 5)
    Traceback (most recent call last):
        ...
    ValueError: Invalid input
    needed_sum must be between 1 and 1000, power between 2 and 10.
    """
    if not (1 <= needed_sum <= 1000 and 2 <= power <= 10):
        raise ValueError(
            "Invalid input\n"
            "needed_sum must be between 1 and 1000, power between 2 and 10."
        )

    return backtrack(needed_sum, power, 1, 0, 0)[1]  # Return the solutions_count


if __name__ == "__main__":
    import doctest

    doctest.testmod()