TheAlgorithms · dhruvmanila · Oct 16, 2020 · Oct 8, 2020 · Oct 8, 2020 · Oct 15, 2020
diff --git a/project_euler/problem_174/__init__.py b/project_euler/problem_174/__init__.py
diff --git a/project_euler/problem_174/sol1.py b/project_euler/problem_174/sol1.py
@@ -0,0 +1,52 @@
+"""
+Project Euler Problem 174: https://projecteuler.net/problem=174
+
+We shall define a square lamina to be a square outline with a square "hole" so that
+the shape possesses vertical and horizontal symmetry.
+
+Given eight tiles it is possible to form a lamina in only one way: 3x3 square with a
+1x1 hole in the middle. However, using thirty-two tiles it is possible to form two
+distinct laminae.
+
+If t represents the number of tiles used, we shall say that t = 8 is type L(1) and
+t = 32 is type L(2).
+
+Let N(n) be the number of t ≤ 1000000 such that t is type L(n); for example,
+N(15) = 832.
+
+What is ∑ N(n) for 1 ≤ n ≤ 10?
+"""
+
+from collections import defaultdict
+from math import ceil, sqrt
+
+
+def solution(t_limit: int = 1000000, n_limit: int = 10) -> int:
+    """
+    Return the sum of N(n) for 1 <= n <= n_limit.
+
+    >>> solution(1000,5)
+    249
+    >>> solution(10000,10)
+    2383
+    """
+    count: defaultdict = defaultdict(int)
+
+    for outer_width in range(3, (t_limit // 4) + 2):
+        if outer_width * outer_width > t_limit:
+            hole_width_lower_bound = max(
+                ceil(sqrt(outer_width * outer_width - t_limit)), 1
+            )
+        else:
+            hole_width_lower_bound = 1
+
+        hole_width_lower_bound += (outer_width - hole_width_lower_bound) % 2
+
+        for hole_width in range(hole_width_lower_bound, outer_width - 1, 2):
+            count[outer_width * outer_width - hole_width * hole_width] += 1
+
+    return sum(1 for n in count.values() if 1 <= n <= 10)
+
+
+if __name__ == "__main__":
+    print(f"{solution() = }")