[pre-commit.ci] auto fixes from pre-commit.com hooks

for more information, see https://pre-commit.ci

Author: pre-commit-ci[bot]

bit_manipulation/fin_two_unique_numbers.py

@@ -1,11 +1,11 @@
-#Reference : https://www.geeksforgeeks.org/dsa/find-two-non-repeating-elements-in-an-array-of-repeating-elements
-def find_two_unique_numbers(arr: list[int]) -> tuple[int,int]:
+# Reference : https://www.geeksforgeeks.org/dsa/find-two-non-repeating-elements-in-an-array-of-repeating-elements
+def find_two_unique_numbers(arr: list[int]) -> tuple[int, int]:
     """
     Given a list of integers where every elemnt appears twice except for two numbers,
     find the two numbers that appear only once.
 
     this method returns the tuple of two numbers that appear only once using bitwise XOR
-    and using the property of x & -x to isolate the rightmost bit 
+    and using the property of x & -x to isolate the rightmost bit
     (different bit between the two unique numbers).
 
     >>> find_two_unique_numbers([1,2,3,4,1,2])
@@ -32,26 +32,27 @@ def find_two_unique_numbers(arr: list[int]) -> tuple[int,int]:
 
     if not arr:
         raise ValueError("input list must not be empty")
-    if not all(isinstance(x,int) for x in arr):
+    if not all(isinstance(x, int) for x in arr):
         raise TypeError("all elements must be integers")
-    
+
     xor_result = 0
     for number in arr:
-        xor_result^=number
+        xor_result ^= number
 
-    righmost_bit=xor_result & -xor_result
+    righmost_bit = xor_result & -xor_result
 
-    num1=0
-    num2=0
+    num1 = 0
+    num2 = 0
 
     for number in arr:
         if number & righmost_bit:
-            num1^=number
+            num1 ^= number
         else:
-            num2^=number
-    return tuple(sorted((num1,num2)))
+            num2 ^= number
+    return tuple(sorted((num1, num2)))
+
 
-if __name__=="__main__":
+if __name__ == "__main__":
     import doctest
 
     doctest.testmod()