Added solution for Project Euler problem 109

DIRECTORY.md

@@ -29,6 +29,8 @@
   * [Binary Count Setbits](https://github.com/TheAlgorithms/Python/blob/master/bit_manipulation/binary_count_setbits.py)
   * [Binary Count Trailing Zeros](https://github.com/TheAlgorithms/Python/blob/master/bit_manipulation/binary_count_trailing_zeros.py)
   * [Binary Or Operator](https://github.com/TheAlgorithms/Python/blob/master/bit_manipulation/binary_or_operator.py)
+  * [Binary Shifts](https://github.com/TheAlgorithms/Python/blob/master/bit_manipulation/binary_shifts.py)
+  * [Binary Twos Complement](https://github.com/TheAlgorithms/Python/blob/master/bit_manipulation/binary_twos_complement.py)
   * [Binary Xor Operator](https://github.com/TheAlgorithms/Python/blob/master/bit_manipulation/binary_xor_operator.py)
   * [Count Number Of One Bits](https://github.com/TheAlgorithms/Python/blob/master/bit_manipulation/count_number_of_one_bits.py)
   * [Reverse Bits](https://github.com/TheAlgorithms/Python/blob/master/bit_manipulation/reverse_bits.py)
@@ -278,6 +280,7 @@
 ## Graphics
   * [Bezier Curve](https://github.com/TheAlgorithms/Python/blob/master/graphics/bezier_curve.py)
   * [Koch Snowflake](https://github.com/TheAlgorithms/Python/blob/master/graphics/koch_snowflake.py)
+  * [Mandelbrot](https://github.com/TheAlgorithms/Python/blob/master/graphics/mandelbrot.py)
   * [Vector3 For 2D Rendering](https://github.com/TheAlgorithms/Python/blob/master/graphics/vector3_for_2d_rendering.py)
 
 ## Graphs
@@ -469,6 +472,7 @@
   * [Runge Kutta](https://github.com/TheAlgorithms/Python/blob/master/maths/runge_kutta.py)
   * [Segmented Sieve](https://github.com/TheAlgorithms/Python/blob/master/maths/segmented_sieve.py)
   * Series
+    * [Geometric Mean](https://github.com/TheAlgorithms/Python/blob/master/maths/series/geometric_mean.py)
     * [Geometric Series](https://github.com/TheAlgorithms/Python/blob/master/maths/series/geometric_series.py)
     * [Harmonic Series](https://github.com/TheAlgorithms/Python/blob/master/maths/series/harmonic_series.py)
     * [P Series](https://github.com/TheAlgorithms/Python/blob/master/maths/series/p_series.py)
@@ -757,6 +761,8 @@
     * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_102/sol1.py)
   * Problem 107
     * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_107/sol1.py)
+  * Problem 109
+    * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_109/sol1.py)
   * Problem 112
     * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_112/sol1.py)
   * Problem 113

project_euler/problem_109/sol1.py

@@ -0,0 +1,89 @@
+"""
+In the game of darts a player throws three darts at a target board which is
+split into twenty equal sized sections numbered one to twenty.
+
+The score of a dart is determined by the number of the region that the dart
+lands in. A dart landing outside the red/green outer ring scores zero. The black
+and cream regions inside this ring represent single scores. However, the red/green
+outer ring and middle ring score double and treble scores respectively.
+
+At the centre of the board are two concentric circles called the bull region, or
+bulls-eye. The outer bull is worth 25 points and the inner bull is a double,
+worth 50 points.
+
+There are many variations of rules but in the most popular game the players will
+begin with a score 301 or 501 and the first player to reduce their running total
+to zero is a winner. However, it is normal to play a "doubles out" system, which
+means that the player must land a double (including the double bulls-eye at the
+centre of the board) on their final dart to win; any other dart that would reduce
+their running total to one or lower means the score for that set of three darts
+is "bust".
+
+When a player is able to finish on their current score it is called a "checkout"
+and the highest checkout is 170: T20 T20 D25 (two treble 20s and double bull).
+
+There are exactly eleven distinct ways to checkout on a score of 6:
+
+D3
+D1  D2
+S2  D2
+D2  D1
+S4  D1
+S1  S1  D2
+S1  T1  D1
+S1  S3  D1
+D1  D1  D1
+D1  S2  D1
+S2  S2  D1
+
+Note that D1 D2 is considered different to D2 D1 as they finish on different
+doubles. However, the combination S1 T1 D1 is considered the same as T1 S1 D1.
+
+In addition we shall not include misses in considering combinations; for example,
+D3 is the same as 0 D3 and 0 0 D3.
+
+Incredibly there are 42336 distinct ways of checking out in total.
+
+How many distinct ways can a player checkout with a score less than 100?
+
+Solution:
+    We first construct a list of the possible dart values, separated by type.
+    We then iterate through the doubles, followed by the possible 2 following throws.
+    If the total of these three darts is less than the given limit, we increment
+    the counter.
+"""
+
+from itertools import combinations_with_replacement
+
+
+def solution(limit: int = 100) -> int:
+    """
+    Count the number of distinct ways a player can checkout with a score
+    less than limit.
+    >>> solution(171)
+    42336
+    >>> solution(50)
+    12577
+    """
+    singles: list[int] = [x for x in range(1, 21)] + [25]
+    doubles: list[int] = [2 * x for x in range(1, 21)] + [50]
+    triples: list[int] = [3 * x for x in range(1, 21)]
+    all_values: list[int] = singles + doubles + triples + [0]
+
+    num_checkouts: int = 0
+    double: int
+    throw1: int
+    throw2: int
+    checkout_total: int
+
+    for double in doubles:
+        for throw1, throw2 in combinations_with_replacement(all_values, 2):
+            checkout_total = double + throw1 + throw2
+            if checkout_total < limit:
+                num_checkouts += 1
+
+    return num_checkouts
+
+
+if __name__ == "__main__":
+    print(f"{solution() = }")