3371. Identify the Largest Outlier in an Array - Detailed Explanation

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Problem Statement

Definition: We have an integer array nums of length n containing a special structure. Exactly n-2 of the elements are special numbers. Among the remaining two elements, one is the sum of all special numbers, and the other is an outlier. An outlier is a number that is neither one of the special numbers nor the sum-of-specials element. All special numbers, the sum element, and the outlier are at different indices (though they could have the same value). The task is to identify the largest possible outlier value in the array.

Goal: Return the largest value in nums that can serve as an outlier under these conditions.

Examples:

Example 1:
Input: nums = [2, 3, 5, 10]
Output: 10
Explanation: The special numbers could be 2 and 3 (sum = 5), one element 5 represents that sum, and 10 is the outlier. So 10 is identified as the outlier.
Example 2:
Input: nums = [-2, -1, -3, -6, 4]
Output: 4
Explanation: The special numbers could be -2, -1, -3 (sum = -6), one element -6 represents that sum, and 4 is the outlier. Thus 4 is the largest outlier in the array.
Example 3:
Input: nums = [1, 1, 1, 1, 1, 5, 5]
Output: 5
Explanation: The special numbers could be five 1's (sum = 5), one element 5 represents that sum, and the other 5 is the outlier. Even though the sum element and outlier have the same value 5, they are at different indices.

Constraints:

3 <= nums.length <= 10^5 (the array has at least 3 elements and up to 100,000 elements)
-1000 <= nums[i] <= 1000 (elements can be negative, zero, or positive)
The input is guaranteed to have at least one valid outlier under the given conditions.

Hints

Hint 1: Think about the relationship between the sum of all elements in the array and the special structure. If one element is the sum of the special numbers, how does it relate to the total sum of the array? Can you derive a formula involving the outlier, the sum element, and the total sum?
Hint 2: Try identifying which element could represent the sum of the special numbers. If you knew that value, how would you find the corresponding outlier? Consider using a mathematical approach or a lookup structure to test candidates efficiently.
Hint 3: A brute force approach might attempt to check every possible combination of special numbers. That will be too slow given the constraints. Instead, look for a pattern or formula that directly checks each element in a single pass. Using a hash map or set to quickly check for the existence of required values can be very helpful.

Multiple Approaches

We will discuss two approaches: a brute-force idea for understanding, and an optimal approach. The brute-force approach is mainly for conceptual clarity (it will not work efficiently for large inputs but helps us reason about the problem). The optimal approach uses math and hashing to solve the problem in linear time.

Approach 1: Brute Force (Conceptual)

Idea: Try every possible way to split the array into special numbers, a sum element, and an outlier. In other words, choose two distinct elements in the array to play the roles of (1) the outlier and (2) the sum-of-special. Then check if the rest of the elements (the other n-2 numbers) indeed sum up to the chosen sum element.

How to implement naively:

Compute the total sum of all elements in the array.
Loop through each possible element x in the array as a candidate for outlier.
For each x, loop through each other element y in the array as a candidate for the sum-of-special.
Check if the sum of all remaining elements (i.e. totalSum - x - y) equals y (which means those remaining elements sum up correctly to y). If yes, then x could be an outlier.
Track the largest x that satisfies the condition.

Why it works: If y is the sum of the special numbers and x is the outlier, then by definition y = (totalSum - x - y) (the remaining n-2 elements sum to y). This equation simplifies to totalSum - x = 2*y, or x = totalSum - 2*y. So we are basically checking this relationship for every pair of elements.

Drawbacks: This approach is extremely inefficient. It requires nested loops over the array (O(n²) checks), and computing the sum of remaining elements from scratch for each pair could add even more overhead. With n up to 100k, this is not feasible in practice – the number of combinations grows roughly with n² (which could be ~10^10 checks in the worst case!). It will time out for large inputs.

However, this brute-force reasoning gives us a clue: the formula x = totalSum - 2*y must hold for the outlier x and sum element y. We can use this insight to avoid exhaustive search.

Approach 2: Optimal Solution (Using Math + Hash Map)

Idea: Use the formula derived from the problem’s conditions to identify the outlier in one pass. We know that if y is the sum-of-special and x is the outlier, then totalSum = x + 2*y (because totalSum = x + y + (sum of special) and y = sum of special). This rearranges to x = totalSum - 2*y. So for each element y in the array that could be the sum-of-special, the corresponding outlier would have to be x = totalSum - 2*y. We just need to check which such x exists in the array.

Efficient plan:

First compute totalSum of all elements in nums.
Create a hash map or dictionary (cnt) to count occurrences of each number in nums. This allows quick existence checks and handles duplicates.
Loop through each element x in nums as a candidate outlier (we could also loop through candidates for y; both ways are similar in complexity). For each x:
- Compute t = totalSum - x. This t is the sum of the rest of the array if x were removed.
- Check if t is even. If t is not even, then t/2 would not be an integer, and it’s impossible for the remaining n-1 elements to be split into two equal parts y + y with one part as an element. In that case, skip this x (it cannot satisfy totalSum - x = 2*y if the left side is odd).
- Let half = t / 2. This is the value that would need to be the sum-of-special element y if x is outlier.
- Check if half exists in the array (using the hash map). If not, then no element has the value needed to be the sum-of-special for this x.
- If half exists, we have to ensure that the index of half is distinct from x's index (they can be the same value, but not the same exact element). In practice, this means:
  - If x != half, then we're fine (the outlier value and sum value are different, so they are naturally different elements).
  - If x == half, then that value would have to appear at least twice in the array so that one instance can serve as the sum-of-special and another instance as the outlier. We can check this by seeing if cnt[half] > 1 (more than one occurrence of that value).
- If the above conditions pass, then x is a valid outlier candidate. We record it as a possible answer.
Among all valid outlier candidates found, return the largest value.

This approach smartly avoids checking combinations explicitly. We reduce the problem to checking each element once with O(1) work (thanks to the precomputed sum and hash map lookups).

Why it’s correct: We systematically enforce the required relationship (totalSum - x = 2*y) for each element x. Any x that fulfills this and has a counterpart y = half in the array corresponds to a scenario where removing x leaves an array whose sum is 2*y, and since y itself is in the array, it implies the rest sum to y. This is exactly the condition for x being an outlier and y being the sum-of-special. By checking all x, we consider all possible outlier scenarios and pick the largest.

Optimizations: We use a hash map for O(1) average lookups to check half existence. Computing the total sum and building the map are each O(n). The single loop through elements is O(n). Thus, this approach runs in linear time overall, which is efficient even for n = 100k.

Code Implementation

Python Implementation

Python3

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Java Implementation

Java

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Complexity Analysis

Time Complexity:

Brute Force Approach: O(n²) in the worst case. We would potentially examine every pair of elements (outlier, sum) among n elements. Even if we optimize checking the sum of the rest using a precomputed total, the pairwise checking makes it quadratic. This is infeasible for large n (e.g., 100k elements would lead to billions of checks).
Optimal Approach: O(n) on average. Calculating the total sum is O(n). Building the frequency map is O(n). The loop through each element and constant-time checks (modulus, lookup, comparisons) is O(n). So overall it scales linearly with the array size. This will comfortably handle n up to 100,000. (In the worst case, hash map lookups could degrade if many collisions occur, but with a good hash function and the data size, it's effectively O(n) average.)

Space Complexity:

Brute Force Approach: O(1) (excluding input storage), as it only uses a few extra variables. However, if we consider storing combinations or so (which we didn't explicitly do), it could blow up, but the straightforward brute force just uses the input and a couple of counters.
Optimal Approach: O(n) additional space. We use a hash map to store frequency of each element, which in the worst case holds every element once (if all values are unique). This is proportional to n. The memory trade-off is acceptable for the given constraint (100k elements).

The optimal approach is clearly preferable due to its linear time performance. It leverages extra space for efficiency, which is a common trade-off in algorithm design (using a hash map to reduce time complexity).

Step-by-Step Walkthrough with Visuals

Let's walk through the optimal solution step by step using an example to see how it identifies the outlier.

Consider Example 2: nums = [-2, -1, -3, -6, 4]. The correct output is 4. We will determine this through the algorithm:

Calculate the total sum:
totalSum = (-2) + (-1) + (-3) + (-6) + 4 = -8.
Build the frequency map:
freq = { -6: 1, -3: 1, -2: 1, -1: 1, 4: 1 } (each number appears once in this case).
Iterate over each element as a potential outlier x:
- Candidate x = -2:
  - Compute t = totalSum - x = -8 - (-2) = -6.
  - Check if t is even. -6 is even, so proceed.
  - Compute half = t/2 = -6/2 = -3.
  - Check if half (which is -3) exists in the array. Yes, -3 is present in freq.
  - Are x and half distinct? Here x = -2 and half = -3 are different values, so they're definitely different elements.
  - Conclusion: -2 is a valid outlier candidate (with corresponding sum element -3).
- Candidate x = -1:
  - t = -8 - (-1) = -7.
  - t is odd (-7), so we skip this candidate (if the sum of the rest is -7, it cannot be twice some integer).
  - Conclusion: -1 cannot be an outlier.
- Candidate x = -3:
  - t = -8 - (-3) = -5.
  - t is odd (-5), skip this candidate.
  - Conclusion: -3 cannot be an outlier.
- Candidate x = -6:
  - t = -8 - (-6) = -2.
  - t is even.
  - half = -2/2 = -1.
  - Check existence of half = -1. Yes, -1 is in the array.
  - x = -6 and half = -1 are distinct values.
  - Conclusion: -6 is a valid outlier candidate (with corresponding sum element -1).
- Candidate x = 4:
  - t = -8 - 4 = -12.
  - t is even.
  - half = -12/2 = -6.
  - Check existence of half = -6. Yes, -6 is in the array.
  - x = 4 and half = -6 are distinct values.
  - Conclusion: 4 is a valid outlier candidate (with corresponding sum element -6).
Determine the largest outlier:
Among the candidates we found, the values were -2, -6, and 4. The largest (maximum) value is 4. So, the algorithm would return 4, which matches the expected answer.

During the iteration, the algorithm continuously tracks the largest valid outlier found. By the end, 4 is correctly identified as the largest outlier.

We can double-check the result by interpreting it back in terms of the problem: If 4 is the outlier, the remaining array [-2, -1, -3, -6] should contain exactly one sum-of-special and the rest special numbers. Indeed, in those remaining numbers, -6 can be seen as the sum of the special numbers -2, -1, -3 (since -2 + -1 + -3 = -6). Everything fits the definition, confirming our answer.

For a visual summary of this example, consider the array split into three parts for the winning scenario:

Special numbers: [-2, -1, -3] (these sum up to -6)
Sum element: -6 (represents the sum of the special numbers)
Outlier: 4 (stands apart from the above group)

This partitioning is exactly what the algorithm discovered by checking the conditions for each candidate.

Common Mistakes & Edge Cases

Mixing up roles: A common confusion is swapping the outlier and sum-of-special roles. Remember that the outlier is not part of the summation relationship, whereas the sum element is directly equal to the sum of the special numbers. Ensure you apply the formula in the correct order (either check x = totalSum - 2*y for some candidate pair, or as in our approach, treat each x as outlier and find y = (totalSum - x)/2).
Not handling duplicates properly: If the outlier value could be the same as the sum value (like in Example 3 where both were 5), you must verify that the value occurs at least twice in the array. A mistake would be to assume different values for outlier and sum always. The condition “distinct indices” allows the same number to serve as outlier and sum as long as they are two separate elements.
Forgetting negative values and zero: The array can have negative numbers and zero. The logic totalSum - x = 2*y still holds universally. Just ensure your solution doesn’t impose any incorrect assumptions (for instance, dividing an odd negative t by 2 in integer math implicitly floors toward zero, which isn’t what we want – we explicitly needed t to be even to proceed).
Using an inefficient approach for large input: Trying to generate all combinations or test every subset of n-2 special numbers is impractical. A brute force approach will not finish for large n. The key is to use the mathematical relationship and quick lookups. A common pitfall is nested loops without realizing the input size is too big.
Edge case – smallest array (n=3): With 3 elements, there’s only 1 special number (since n-2 = 1), 1 sum element, and 1 outlier. For example, nums = [5, 3, 2]. Here either 5 could be the sum of special (3) and outlier 2, or 3 could be the sum of special (2) and outlier 5. Our algorithm would find both 2 and 5 as candidates and correctly return 5 as the largest outlier. Make sure your solution handles such minimal scenarios correctly.
Edge case – values repeated many times: If many numbers are the same, ensure the logic doesn’t break. For instance, nums = [0, 0, 0] (all zeros) technically fits the definition: special number = 0 (one of them), sum element = 0 (another one, sum of one special which is 0), outlier = 0 (the remaining one). The largest outlier is 0. The algorithm’s duplicate-handling covers this because it would see half = 0 exists and requires at least two occurrences (which there are). Always consider if your checks handle multiple identical values gracefully.

Good Array (Codeforces 1077C): A problem where you need to check if an array has an element equal to the sum of all other elements. This is a simpler variant of the concept (essentially the case where the outlier would be 0 or not needed). It uses similar reasoning (sums and checking existence) to find such an element.
Parity Outlier (Codewars): A challenge where you have to find the single even number in an array of odds or vice versa. While the rule for being an "outlier" is different (parity rather than a sum relationship), it’s also about identifying the one element that doesn’t fit with the rest.
LeetCode 136: Single Number: In this problem, every number except one appears twice, and you must find the one that appears once. The theme of finding an element that stands out among others is similar, though the method (XOR bit manipulation) is very different. It’s a good exercise in identifying a unique element under different conditions.
LeetCode 747: Largest Number At Least Twice of Others: This asks to check if the largest element in an array is at least twice as large as every other element, and return its index if so. Conceptually, it’s about one element being dominant or “standing out” from the rest (though not in terms of a sum). Solving it involves scanning for the largest and second-largest values, somewhat akin to considering outlier vs. others.
Subarray Sum Problems (Two-sum, Three-sum, etc.): These involve finding numbers that sum to a target. While not directly related to outliers, they sharpen the skill of using hashing and two-pointer techniques to find relationships between array elements (like we used hashing and a derived formula here). They help build a mindset for checking conditions efficiently in an array.