#### Power Sum

m
p
```"""
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()
```