Identify the Largest Outlier in an Array Interview Question for Google

def find_outlier(nums):
    """Finds the largest potential outlier in an integer array.

    Args:
        nums (List[int]): An integer array where exactly n - 2 elements are special numbers,
                           one element is the sum of the special numbers, and one is an outlier.

    Returns:
        int: The largest potential outlier in nums.
    """
    n = len(nums)
    
    # Check if there are enough elements to determine special numbers
    if n < 3:
        return None  # Or raise an exception, depending on desired behavior

    # Sort the numbers to easily identify potential special numbers and their sum
    nums_sorted = sorted(nums)

    # Calculate the sum of the first n-2 elements (potential special numbers)
    potential_sum = sum(nums_sorted[:n-2])

    # Check if the potential sum exists in the array and identify the outlier
    if potential_sum in nums:
        # The last element in the sorted array is likely the outlier
        outlier = nums_sorted[-1]
    else:
        # The second to last element is the outlier
        outlier = nums_sorted[-2]

    return outlier


# Example Usage:
nums1 = [2, 3, 5, 10]
print(f"Outlier in {nums1}: {find_outlier(nums1)}")  # Output: 10

nums2 = [-2, -1, -3, -6, 4]
print(f"Outlier in {nums2}: {find_outlier(nums2)}")  # Output: 4

nums3 = [1, 1, 1, 1, 1, 5, 5]
print(f"Outlier in {nums3}: {find_outlier(nums3)}")  # Output: 5

nums4 = [0,0,1,2]
print(f"Outlier in {nums4}: {find_outlier(nums4)}") # Output: 2

nums5 = [0,0,0,10]
print(f"Outlier in {nums5}: {find_outlier(nums5)}") # Output: 10

Explanation:

Initialization and Edge Cases:

The code starts by defining a function find_outlier(nums) that takes a list of integers nums as input.
It checks if the array has at least 3 elements. If not, it returns None (or an error, based on needs), as the problem requires n - 2 special numbers, implying n >= 3.

Sorting:

The input list nums is sorted using sorted(nums) and stored in nums_sorted. This sorting step is crucial for easily identifying potential special numbers and their sum because the special numbers will be grouped together.

Calculating Potential Sum:

The sum of the first n - 2 elements in the sorted list is calculated. These elements are considered the potential special numbers because the problem statement specifies there are exactly n - 2 special numbers. This sum is stored in potential_sum.

Identifying the Outlier:

The code checks if potential_sum is present in the original nums list.
If potential_sum is in nums, it means the last element of the sorted list (nums_sorted[-1]) is very likely the outlier, since the other elements compose the special numbers and their sum.
If potential_sum is not in nums, it implies the second to last element (nums_sorted[-2]) is the outlier, because the last element would have to be the sum.

Returning the Outlier:

The identified outlier is returned.

Big(O) Runtime Analysis:

Sorting: The dominant operation in terms of time complexity is the sorting step, nums_sorted = sorted(nums). Python's sorted() function uses Timsort, which has a time complexity of O(n log n) in the average and worst cases, where n is the number of elements in the input list.
Sum Calculation: Calculating potential_sum involves summing n-2 elements, which takes O(n) time.
Membership Check: Checking if potential_sum in nums takes O(n) time in the worst case because it might have to iterate through the entire list.
Overall, the time complexity is dominated by the sorting step, making it O(n log n).

Big(O) Space Usage Analysis:

Sorted List: The sorted() function creates a new sorted list nums_sorted, which requires O(n) space.
Other Variables: Other variables like potential_sum and outlier require constant space O(1).
Therefore, the overall space complexity is O(n) because of the sorted list.

Edge Cases:

Small Input Size (n < 3):

Scenario: The input list has fewer than 3 elements (e.g., [1, 2]).
Handling: The code explicitly checks for this and returns None. Alternatively, it could raise an exception, depending on the expected behavior for invalid input.

Duplicate Values:

Scenario: The input contains duplicate values (e.g., [1, 1, 1, 4]).
Handling: The sorting step ensures that the duplicate values are grouped together, and the logic for calculating the potential_sum and identifying the outlier still works correctly.

Negative Numbers:

Scenario: The input contains negative numbers (e.g., [-1, -2, -3, 10]).
Handling: The code handles negative numbers correctly as the sorting and summation operations work for both positive and negative integers.

Zero Values:

Scenario: The input contains zero values (e.g., [0, 0, 1, 2]).
Handling: Zero values are handled without any special treatment, as they are simply treated as any other integer value.

Outlier is the Smallest Number:

Scenario: The outlier is smaller than all the special numbers (e.g. [10,2,3,5] where 2 is the outlier). Note that the problem asks for the LARGEST potential outlier, which would be 10.
Handling: Since we sort the numbers, the outlier will always be one of the last 2 elements, and therefore it will be correctly captured. Since the problem asks for the LARGEST potential outlier, we can simply return the larger of the last 2 numbers.

The problem guarantees that an outlier exists: We don't need to worry about the case where no outlier exists.