Identify the Largest Outlier in an Array

You are given an integer array nums. This array contains n elements, where exactly n - 2 elements are special numbers. One of the remaining two elements is the sum of these special numbers, and the other is an outlier.

An outlier is defined as a number that is neither one of the original special numbers nor the element representing the sum of those numbers.

Note that special numbers, the sum element, and the outlier must have distinct indices, but may share the same value.

Return the largest potential outlier in nums.

Example 1:

Input: nums = [2,3,5,10] Output: 10 Explanation: The special numbers could be 2 and 3, thus making their sum 5 and the outlier 10.

Example 2:

Input: nums = [-2,-1,-3,-6,4] Output: 4 Explanation: The special numbers could be -2, -1, and -3, thus making their sum -6 and the outlier 4.

Example 3:

Input: nums = [1,1,1,1,1,5,5] Output: 5 Explanation: The special numbers could be 1, 1, 1, 1, and 1, thus making their sum 5 and the other 5 as the outlier.

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

    Args:
        nums (list of int): An array of integers containing n elements, where exactly n - 2 elements are special numbers.
                             One of the remaining two elements is the sum of these special numbers, and the other is an outlier.

    Returns:
        int: The largest potential outlier in nums.
    """

    n = len(nums)

    # Brute-force approach: Try all possible combinations of n-2 special numbers.
    max_outlier = float('-inf')

    for i in range(n):
        for j in range(i + 1, n):
            # Consider nums[i] and nums[j] as the potential non-special numbers (sum and outlier).
            potential_special = []
            for k in range(n):
                if k != i and k != j:
                    potential_special.append(nums[k])

            special_sum = sum(potential_special)

            # Check if nums[i] is the sum and nums[j] is the outlier, or vice versa.
            if nums[i] == special_sum:
                outlier = nums[j]
                max_outlier = max(max_outlier, outlier)
            elif nums[j] == special_sum:
                outlier = nums[i]
                max_outlier = max(max_outlier, outlier)

    return max_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

Explanation:

1. Brute-Force Approach:

   • The find_outlier function iterates through all possible pairs of elements in the input array nums. It considers each pair as potential non-special numbers (the sum and the outlier).
   • For each pair, it identifies the remaining elements as potential special numbers.
   • It calculates the sum of the potential special numbers.
   • It checks if either of the initially selected numbers matches the calculated sum. If it does, the other number is considered the outlier, and the max_outlier is updated accordingly.
   • Finally, the function returns the largest potential outlier found.

2. Time Complexity:

   • The outer loop iterates n times, and the inner loop iterates n-1 times. Inside the nested loops, there's another loop that iterates n times to create the list of potential_special numbers and calculate their sum. Therefore, the time complexity is O(n^3).

3. Space Complexity:

   • The space complexity is O(n) because, in the worst case, the list potential_special can contain n-2 elements, where n is the number of elements in the input array nums.

Optimized Solution (O(n^2) time complexity):

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

    Args:
        nums (list of int): An array of integers.

    Returns:
        int: The largest potential outlier in nums.
    """
    n = len(nums)
    max_outlier = float('-inf')

    for i in range(n):
        # Consider nums[i] as a potential outlier
        for j in range(n):
            if i == j:
                continue

            # Calculate the sum of all elements except nums[i] and nums[j]
            special_sum = 0
            for k in range(n):
                if k != i and k != j:
                    special_sum += nums[k]

            # If nums[j] is the sum of the special numbers, then nums[i] is an outlier
            if nums[j] == special_sum:
                max_outlier = max(max_outlier, nums[i])

    return max_outlier


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

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

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

Explanation of Optimized Solution:

1. Optimized Approach:

   • The find_outlier_optimized function iterates through the nums array, considering each element as a potential outlier.
   • For each potential outlier, it iterates through the rest of the array to find a potential sum.
   • It computes the sum of all numbers excluding the potential outlier and the potential sum.
   • If the potential sum equals the calculated sum of the remaining numbers, it updates the max_outlier.

2. Time Complexity:

   • The optimized solution reduces the time complexity to O(n^2) because it has two nested loops, each iterating up to n times. The inner loop includes a linear-time calculation of the sum.

3. Space Complexity:

   • The space complexity of the optimized solution is O(1) because it uses a fixed amount of extra space regardless of the input size. Only a few variables (n, max_outlier, i, j, k, special_sum) are used.