[Hacker Rank]: Between Two Sets solved ✓

Author: Gonzalo Diaz

src/hackerrank/implementation/between_two_sets.md

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+# [Between Two Sets](https://www.hackerrank.com/challenges/between-two-sets)
+
+- Difficulty: #easy
+- Category: #implementation
+
+There will be two arrays of integers. Determine all integers that satisfy
+the following two conditions:
+
+1. The elements of the first array are all factors of the integer being
+considered
+2. The integer being considered is a factor of all elements of the second
+array
+
+These numbers are referred to as being between the two arrays. Determine
+how many such numbers exist.
+
+## Example
+
+$ a = [2, 6] $
+
+$ b = [24, 36] $
+
+There are two numbers between the arrays: $ 6 $ and $ 12 $.
+$ 6 \bmod 2 = 0 $, $ 6 \bmod 6 = 0 $, $ 24 \bmod 6 = 0 $ and $ 36 \bmod 6 = 0 $
+ for the first value.
+$ 12 \bmod 2 = 0 $, $ 12 \bmod 6 = 0 $, and $ 24 \bmod 12 = 0 $,
+$ 36 \bmod 12 = 0 $ for the second value. Return $ 2 $.
+
+## Function Description
+
+Complete the getTotalX function in the editor below. It should return the
+number of integers that are betwen the sets.
+
+getTotalX has the following parameter(s):
+
+- int a[n]: an array of integers
+- int b[m]: an array of integers
+
+## Returns
+
+- int: the number of integers that are between the sets
+
+## Input Format
+
+The first line contains two space-separated integers, n and m, the number
+of elements in arrays a and b.
+The second line contains n distinct space-separated integers $ a[i] $ where
+$ 0 \leq i < n $.
+The third line contains m distinct space-separated integers $ b[j] $ where
+$ 0 \leq j < m $.
+
+## Constraints
+
+- $ 1 \leq n, m < 10 $
+- $ 1 \leq a[i] \leq 100 $
+- $ 1 \leq b[j] \leq 100 $
+
+## Sample Input
+
+```text
+2 3
+2 4
+12 32 96
+```
+
+## Sample Output
+
+```text
+3
+```
+
+## Explanation
+
+2 and 4 divide evenly into 4, 8, 12 and 16.
+
+4, 8 and 16 divide evenly into 16, 32, 96.
+
+4, 8 and 16 are the only three numbers for which each element of a is a factor
+and each is a factor of all elements of b.

src/hackerrank/implementation/between_two_sets.py

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+import logging
+
+LOGGER = logging.getLogger(__name__)
+
+
+def is_factor(_n: int, group: list[int]) -> bool:
+    result: bool = True
+    i: int = 0
+
+    if len(group) == 0:
+        return False
+
+    while (i < len(group) and result):
+        if _n % group[i] != 0:
+            result = False
+
+        i += 1
+
+    return result
+
+
+def factor_of(_n: int, group: list[int]):
+    result: bool = True
+    i: int = 0
+
+    if len(group) == 0:
+        return False
+
+    while (i < len(group) and result):
+        if group[i] % _n != 0:
+            result = False
+
+        i += 1
+
+    return result
+
+
+def get_total_x(_a: list[int], _b: list[int]) -> int:
+    _max_ = 0
+    for _, j in enumerate(_b):
+        if j > _max_:
+            _max_ = j
+
+    result: list[int] = []
+
+    for i in range(2, _max_):
+        if is_factor(i, _a) and factor_of(i, _b):
+            result.append(i)
+
+    output = len(result)
+    LOGGER.info('Between Two Sets result: %i', output)
+
+    return output

src/hackerrank/implementation/between_two_sets_test.py

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+import unittest
+from .between_two_sets import get_total_x, is_factor, factor_of
+
+
+class TestProblemBetweenTwoSets(unittest.TestCase):
+
+    def test_get_total_x(self):
+
+        tinput: list[int] = [16, 32, 96]
+        solution_found: int = 0
+
+        calculated_a = get_total_x([], tinput)
+
+        self.assertEqual(
+            calculated_a, solution_found,
+            f"get_total_x([], tinput) must be "
+            f"=> {solution_found}")
+
+        calculated_b = get_total_x(tinput, [])
+
+        self.assertEqual(
+            calculated_b, solution_found,
+            f"get_total_x({tinput}, {[]}) must be "
+            f"=> {solution_found}")
+
+        calculated_c = get_total_x([], [])
+
+        self.assertEqual(
+            calculated_c, solution_found,
+            f"get_total_x({[]}, {[]}) must be "
+            f"=> {solution_found}")
+
+        calculated_d = is_factor(1, [])
+
+        self.assertEqual(
+            calculated_d, solution_found,
+            f"is_factor({1}, {[]}) must be "
+            f"=> {False}")
+
+        calculated_e = factor_of(1, [])
+
+        self.assertEqual(
+            calculated_e, solution_found,
+            f"factor_of({1}, {[]}) must be "
+            f"=> {False}")
+
+    def test_get_total_x_case_0(self):
+
+        _a_ = [2, 4]
+        _b_ = [16, 32, 96]
+        _b_reverse_ = [96, 32, 16]
+
+        solution_found = 3
+
+        self.assertEqual(
+            get_total_x(_a_, _b_), solution_found,
+            f"get_total_x({_a_}, {_b_}) must be "
+            f"=> {solution_found}")
+
+        self.assertEqual(
+            get_total_x(_a_, _b_reverse_), solution_found,
+            f"get_total_x({_a_}, {_b_reverse_}) must be "
+            f"=> {solution_found}")