From 62dcbea943e8cc4ea4d83eff115c4e6f6a4808af Mon Sep 17 00:00:00 2001
From: duongoku <nguyenvubinhduong@gmail.com>
Date: Mon, 26 Jun 2023 14:39:18 +0700
Subject: [PATCH]  Add power sum problem (#8832)

* Add powersum problem

* Add doctest

* [pre-commit.ci] auto fixes from pre-commit.com hooks

for more information, see https://pre-commit.ci

* Add more doctests

* [pre-commit.ci] auto fixes from pre-commit.com hooks

for more information, see https://pre-commit.ci

* Add more doctests

* Improve paramater name

* Fix line too long

* Remove global variables

* Apply suggestions from code review

* Apply suggestions from code review

---------

Co-authored-by: pre-commit-ci[bot] <66853113+pre-commit-ci[bot]@users.noreply.github.com>
Co-authored-by: Christian Clauss <cclauss@me.com>
---
 backtracking/power_sum.py | 93 +++++++++++++++++++++++++++++++++++++++
 1 file changed, 93 insertions(+)
 create mode 100644 backtracking/power_sum.py

diff --git a/backtracking/power_sum.py b/backtracking/power_sum.py
new file mode 100644
index 000000000..fcf1429f8
--- /dev/null
+++ b/backtracking/power_sum.py
@@ -0,0 +1,93 @@
+"""
+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.
+"""
+
+from math import pow
+
+
+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(100, 2, 1, 0, 0)
+    (0, 3)
+    >>> backtrack(100, 3, 1, 0, 0)
+    (0, 1)
+    >>> backtrack(800, 2, 1, 0, 0)
+    (0, 561)
+    >>> backtrack(1000, 10, 1, 0, 0)
+    (0, 0)
+    >>> backtrack(400, 2, 1, 0, 0)
+    (0, 55)
+    >>> backtrack(50, 1, 1, 0, 0)
+    (0, 3658)
+    """
+    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 = int(pow(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(100, 2)
+    3
+    >>> solve(100, 3)
+    1
+    >>> solve(800, 2)
+    561
+    >>> solve(1000, 10)
+    0
+    >>> solve(400, 2)
+    55
+    >>> solve(50, 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()