2023-06-26 07:39:18 +00:00
|
|
|
"""
|
|
|
|
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)
|
2023-10-14 08:31:43 +00:00
|
|
|
>>> backtrack(10, 2, 1, 0, 0)
|
|
|
|
(0, 1)
|
|
|
|
>>> backtrack(10, 3, 1, 0, 0)
|
|
|
|
(0, 0)
|
|
|
|
>>> backtrack(20, 2, 1, 0, 0)
|
2023-06-26 07:39:18 +00:00
|
|
|
(0, 1)
|
2023-10-14 08:31:43 +00:00
|
|
|
>>> backtrack(15, 10, 1, 0, 0)
|
2023-06-26 07:39:18 +00:00
|
|
|
(0, 0)
|
2023-10-14 08:31:43 +00:00
|
|
|
>>> backtrack(16, 2, 1, 0, 0)
|
|
|
|
(0, 1)
|
|
|
|
>>> backtrack(20, 1, 1, 0, 0)
|
|
|
|
(0, 64)
|
2023-06-26 07:39:18 +00:00
|
|
|
"""
|
|
|
|
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
|
|
|
|
|
2023-10-14 08:31:43 +00:00
|
|
|
i_to_n = current_number**power
|
2023-06-26 07:39:18 +00:00
|
|
|
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
|
2023-10-14 08:31:43 +00:00
|
|
|
>>> solve(10, 2)
|
2023-06-26 07:39:18 +00:00
|
|
|
1
|
2023-10-14 08:31:43 +00:00
|
|
|
>>> solve(10, 3)
|
2023-06-26 07:39:18 +00:00
|
|
|
0
|
2023-10-14 08:31:43 +00:00
|
|
|
>>> solve(20, 2)
|
|
|
|
1
|
|
|
|
>>> solve(15, 10)
|
|
|
|
0
|
|
|
|
>>> solve(16, 2)
|
|
|
|
1
|
|
|
|
>>> solve(20, 1)
|
2023-06-26 07:39:18 +00:00
|
|
|
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()
|