Add function to find two unique numbers in a list

Implement a function to find two unique numbers in a list where every other number appears twice. The function raises errors for empty lists and non-integer elements.

Author: RamachandraBhardwaj

bit_manipulation/fin_two_unique_numbers.py

@@ -0,0 +1,56 @@
+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 
+    (different bit between the two unique numbers).
+
+    >>> find_two_unique_numbers([1,2,3,4,1,2])
+    (3, 4)
+
+    >>> find_two_unique_numbers([4,5,6,7,4,5])
+    (6, 7)
+
+    >>> find_two_unique_numbers([10,20,30,10,20,40])
+    (30, 40)
+
+    >>> find_two_unique_numbers([2,3])
+    (2, 3)
+
+    >>> find_two_unique_numbers([])
+    Traceback (most recent call last):
+        ...
+    ValueError: input list must not be empty
+    >>> find_two_unique_numbers([1, 'a', 1])
+    Traceback (most recent call last):
+        ...
+    TypeError: all elements must be integers
+    """
+
+    if not arr:
+        raise ValueError("input list must not be empty")
+    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
+
+    righmost_bit=xor_result & -xor_result
+
+    num1=0
+    num2=0
+
+    for number in arr:
+        if number & righmost_bit:
+            num1^=number
+        else:
+            num2^=number
+    return tuple(sorted((num1,num2)))
+
+if __name__=="__main__":
+    import doctest
+
+    doctest.testmod()