Update beautiful_arrangement.py

Author: zirea3l

dynamic_programming/beautiful_arrangement.py

@@ -1,6 +1,6 @@
 """
-Suppose you have n integers labeled 1 through n. A permutation of those n integers 
-perm (1-indexed) is considered a "beautiful arrangement" if for every i (1 <= i <= n), 
+Suppose you have n integers labeled 1 through n. A permutation of those n integers
+perm (1-indexed) is considered a "beautiful arrangement" if for every i (1 <= i <= n),
 either of the following is true:
 
 -> perm[i] is divisible by i.
@@ -25,35 +25,32 @@ def countarrangement(self, n: int) -> int:
         A boolean list to mark which numbers have 
         already been placed in the permutation.
         """
-        
+
         def backtrack(pos):
             """
             Define the recursive backtracking function.
-            pos is the current position in the permutation we are filling (1-indexed). 
+            pos is the current position in the permutation we are filling (1-indexed).
             We try to assign a number to position pos.
             """
             if pos > n:
-                
-                self.count += 1 
+                self.count += 1  
                 # We found a complete valid arrangement, so increment the total count.
                 return
-            for num in range(1, n + 1): 
-                # Try every candidate number num for the current position pos.
-                
-                
+            for num in range(
+                1, n + 1
+            ):  # Try every candidate number num for the current position pos.
                 """
                 Two checks in one:
                 1. not used[num] — the number num has not been placed yet (we can use it).
-                2. (num % pos == 0 or pos % num == 0) — the 
-                    beautiful-arrangement condition: 
-                    either num divides pos or pos divides num.
+                2. (num % pos == 0 or pos % num == 0) — the beautiful-arrangement condition:
+                number num is compatible with position pos (either num divides pos or pos divides num).
                 If both are true, num is a valid choice for pos.
-            
+
                 """
                 if not used[num] and (num % pos == 0 or pos % num == 0):
                     used[num] = True
                     backtrack(pos + 1)
                     used[num] = False
-        
+
         backtrack(1)
         return self.count