Add Project Euler problem 131 solution 1 (TheAlgorithms#8179)

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DIRECTORY.md

@@ -196,11 +196,14 @@
     * [Disjoint Set](data_structures/disjoint_set/disjoint_set.py)
   * Hashing
     * [Double Hash](data_structures/hashing/double_hash.py)
+    * [Hash Map](data_structures/hashing/hash_map.py)
     * [Hash Table](data_structures/hashing/hash_table.py)
     * [Hash Table With Linked List](data_structures/hashing/hash_table_with_linked_list.py)
     * Number Theory
       * [Prime Numbers](data_structures/hashing/number_theory/prime_numbers.py)
     * [Quadratic Probing](data_structures/hashing/quadratic_probing.py)
+    * Tests
+      * [Test Hash Map](data_structures/hashing/tests/test_hash_map.py)
   * Heap
     * [Binomial Heap](data_structures/heap/binomial_heap.py)
     * [Heap](data_structures/heap/heap.py)
@@ -973,6 +976,8 @@
     * [Sol1](project_euler/problem_125/sol1.py)
   * Problem 129
     * [Sol1](project_euler/problem_129/sol1.py)
+  * Problem 131
+    * [Sol1](project_euler/problem_131/sol1.py)
   * Problem 135
     * [Sol1](project_euler/problem_135/sol1.py)
   * Problem 144

project_euler/problem_131/__init__.py

@@ -0,0 +1,56 @@
+"""
+Project Euler Problem 131: https://projecteuler.net/problem=131
+
+There are some prime values, p, for which there exists a positive integer, n,
+such that the expression n^3 + n^2p is a perfect cube.
+
+For example, when p = 19, 8^3 + 8^2 x 19 = 12^3.
+
+What is perhaps most surprising is that for each prime with this property
+the value of n is unique, and there are only four such primes below one-hundred.
+
+How many primes below one million have this remarkable property?
+"""
+
+from math import isqrt
+
+
+def is_prime(number: int) -> bool:
+    """
+    Determines whether number is prime
+
+    >>> is_prime(3)
+    True
+
+    >>> is_prime(4)
+    False
+    """
+
+    for divisor in range(2, isqrt(number) + 1):
+        if number % divisor == 0:
+            return False
+    return True
+
+
+def solution(max_prime: int = 10**6) -> int:
+    """
+    Returns number of primes below max_prime with the property
+
+    >>> solution(100)
+    4
+    """
+
+    primes_count = 0
+    cube_index = 1
+    prime_candidate = 7
+    while prime_candidate < max_prime:
+        primes_count += is_prime(prime_candidate)
+
+        cube_index += 1
+        prime_candidate += 6 * cube_index
+
+    return primes_count
+
+
+if __name__ == "__main__":
+    print(f"{solution() = }")