diff --git a/DIRECTORY.md b/DIRECTORY.md
index 98b87f2fe..b37bb35ec 100644
--- a/DIRECTORY.md
+++ b/DIRECTORY.md
@@ -856,6 +856,8 @@
     * [Sol1](project_euler/problem_113/sol1.py)
   * Problem 114
     * [Sol1](project_euler/problem_114/sol1.py)
+  * Problem 115
+    * [Sol1](project_euler/problem_115/sol1.py)
   * Problem 119
     * [Sol1](project_euler/problem_119/sol1.py)
   * Problem 120
diff --git a/project_euler/problem_115/__init__.py b/project_euler/problem_115/__init__.py
new file mode 100644
index 000000000..e69de29bb
diff --git a/project_euler/problem_115/sol1.py b/project_euler/problem_115/sol1.py
new file mode 100644
index 000000000..15a13516d
--- /dev/null
+++ b/project_euler/problem_115/sol1.py
@@ -0,0 +1,62 @@
+"""
+Project Euler Problem 115: https://projecteuler.net/problem=115
+
+NOTE: This is a more difficult version of Problem 114
+(https://projecteuler.net/problem=114).
+
+A row measuring n units in length has red blocks
+with a minimum length of m units placed on it, such that any two red blocks
+(which are allowed to be different lengths) are separated by at least one black square.
+
+Let the fill-count function, F(m, n),
+represent the number of ways that a row can be filled.
+
+For example, F(3, 29) = 673135 and F(3, 30) = 1089155.
+
+That is, for m = 3, it can be seen that n = 30 is the smallest value
+for which the fill-count function first exceeds one million.
+
+In the same way, for m = 10, it can be verified that
+F(10, 56) = 880711 and F(10, 57) = 1148904, so n = 57 is the least value
+for which the fill-count function first exceeds one million.
+
+For m = 50, find the least value of n
+for which the fill-count function first exceeds one million.
+"""
+
+from itertools import count
+
+
+def solution(min_block_length: int = 50) -> int:
+    """
+    Returns for given minimum block length the least value of n
+    for which the fill-count function first exceeds one million
+
+    >>> solution(3)
+    30
+
+    >>> solution(10)
+    57
+    """
+
+    fill_count_functions = [1] * min_block_length
+
+    for n in count(min_block_length):
+        fill_count_functions.append(1)
+
+        for block_length in range(min_block_length, n + 1):
+            for block_start in range(n - block_length):
+                fill_count_functions[n] += fill_count_functions[
+                    n - block_start - block_length - 1
+                ]
+
+            fill_count_functions[n] += 1
+
+        if fill_count_functions[n] > 1_000_000:
+            break
+
+    return n
+
+
+if __name__ == "__main__":
+    print(f"{solution() = }")