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* Updated test cases of power_sum.py * updated * updated. * remove extra comment and used ** instead of pow * [pre-commit.ci] auto fixes from pre-commit.com hooks for more information, see https://pre-commit.ci * Update backtracking/power_sum.py * [pre-commit.ci] auto fixes from pre-commit.com hooks for more information, see https://pre-commit.ci --------- Co-authored-by: pre-commit-ci[bot] <66853113+pre-commit-ci[bot]@users.noreply.github.com> Co-authored-by: Tianyi Zheng <tianyizheng02@gmail.com>
92 lines
2.6 KiB
Python
92 lines
2.6 KiB
Python
"""
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Problem source: https://www.hackerrank.com/challenges/the-power-sum/problem
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Find the number of ways that a given integer X, can be expressed as the sum
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of the Nth powers of unique, natural numbers. For example, if X=13 and N=2.
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We have to find all combinations of unique squares adding up to 13.
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The only solution is 2^2+3^2. Constraints: 1<=X<=1000, 2<=N<=10.
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"""
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def backtrack(
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needed_sum: int,
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power: int,
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current_number: int,
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current_sum: int,
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solutions_count: int,
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) -> tuple[int, int]:
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"""
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>>> backtrack(13, 2, 1, 0, 0)
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(0, 1)
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>>> backtrack(10, 2, 1, 0, 0)
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(0, 1)
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>>> backtrack(10, 3, 1, 0, 0)
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(0, 0)
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>>> backtrack(20, 2, 1, 0, 0)
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(0, 1)
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>>> backtrack(15, 10, 1, 0, 0)
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(0, 0)
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>>> backtrack(16, 2, 1, 0, 0)
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(0, 1)
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>>> backtrack(20, 1, 1, 0, 0)
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(0, 64)
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"""
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if current_sum == needed_sum:
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# If the sum of the powers is equal to needed_sum, then we have a solution.
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solutions_count += 1
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return current_sum, solutions_count
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i_to_n = current_number**power
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if current_sum + i_to_n <= needed_sum:
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# If the sum of the powers is less than needed_sum, then continue adding powers.
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current_sum += i_to_n
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current_sum, solutions_count = backtrack(
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needed_sum, power, current_number + 1, current_sum, solutions_count
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)
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current_sum -= i_to_n
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if i_to_n < needed_sum:
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# If the power of i is less than needed_sum, then try with the next power.
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current_sum, solutions_count = backtrack(
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needed_sum, power, current_number + 1, current_sum, solutions_count
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)
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return current_sum, solutions_count
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def solve(needed_sum: int, power: int) -> int:
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"""
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>>> solve(13, 2)
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1
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>>> solve(10, 2)
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1
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>>> solve(10, 3)
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0
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>>> solve(20, 2)
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1
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>>> solve(15, 10)
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0
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>>> solve(16, 2)
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1
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>>> solve(20, 1)
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Traceback (most recent call last):
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...
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ValueError: Invalid input
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needed_sum must be between 1 and 1000, power between 2 and 10.
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>>> solve(-10, 5)
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Traceback (most recent call last):
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...
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ValueError: Invalid input
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needed_sum must be between 1 and 1000, power between 2 and 10.
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"""
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if not (1 <= needed_sum <= 1000 and 2 <= power <= 10):
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raise ValueError(
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"Invalid input\n"
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"needed_sum must be between 1 and 1000, power between 2 and 10."
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)
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return backtrack(needed_sum, power, 1, 0, 0)[1] # Return the solutions_count
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if __name__ == "__main__":
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import doctest
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doctest.testmod()
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