Identify the Largest Outlier in an Array solution and explanation
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The problem
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.
Constraints:
3 <= nums.length <= 105-1000 <= nums[i] <= 1000- The input is generated such that at least one potential outlier exists in
nums.
Approach overview
Problem Overview: You receive an integer array where all elements except one form a special structure: among the remaining numbers, one value equals the sum of several others. The task is to identify the largest number that could be the outlier while the rest of the array still satisfies this condition.
The key observation comes from simple algebra. If o is the outlier and y is the element equal to the sum of the special numbers, then:
totalSum - o - y = y
Rearranging gives y = (totalSum - o) / 2. For any candidate outlier o, the array must contain a valid y. Efficient solutions focus on testing this relationship quickly.
Approach 1: Sorting and Iterative Check (O(n log n) time, O(1) extra space)
Sort the array first, then iterate from the largest value toward the smallest while treating each element as a possible outlier. For each candidate o, compute (totalSum - o) / 2 and check whether that value exists elsewhere in the array. Sorting helps you evaluate larger outliers first and use binary search or pointer checks for existence. The approach is straightforward and works well when the array size is moderate.
This solution relies mainly on Array manipulation and enumeration of candidates. The downside is the O(n log n) cost from sorting.
Approach 2: Hashing for Frequency and Sum (O(n) time, O(n) space)
Use a frequency map to store counts of every value in the array. Compute the total sum once. Then iterate through the array and treat each number as the potential outlier o. Calculate remaining = totalSum - o. If remaining is even, the required element must be y = remaining / 2.
Check the hash map to see if y exists among the remaining elements. Frequency handling matters: if y equals the candidate outlier, ensure its count is greater than one before accepting it. Track the largest valid outlier while scanning the array. Hash lookups keep the validation constant time.
This method uses a Hash Table for frequency counting and efficient lookups. The algorithm performs a single pass after building the map, giving an optimal O(n) time solution with O(n) space.
Recommended for interviews: Start by explaining the equation derived from the sum relationship. A brute reasoning step shows you understand the structure of the problem. Then move to the hash map solution, which interviewers typically expect because it reduces the search to constant-time lookups using Counting and frequency tracking. The O(n) hashing approach is both clean and optimal.
Approach 1: Sorting and Iterative Check
By sorting the array, we can easily identify potential sums and iterate through the list to find the largest value that cannot be a part of the sum. Leveraging the sorted property, we can assess elements from the smallest to the largest, mindful of how they can sum to intermediate values that appear in the array.
def largest_outlier(nums):
nums.sort()
total_sum = sum(nums[:-1])
return nums[-2] if total_sum == nums[-1] else nums[-1]
nums = [2, 3, 5, 10]
print(largest_outlier(nums))import java.util.Arrays;
public class LargestOutlier {
public static int largestOutlier(int[] nums) {
Arrays.sort(nums);
int totalSum = 0;
for (int i = 0; i < nums.length - 1; i++) {
totalSum += nums[i];
}
return totalSum == nums[nums.length - 1] ? nums[nums.length - 2] : nums[nums.length - 1];
}
public static void main(String[] args) {
int[] nums = {2, 3, 5, 10};
System.out.println(largestOutlier(nums));
}
}#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int largestOutlier(vector<int>& nums) {
sort(nums.begin(), nums.end());
int total_sum = 0;
for (int i = 0; i < nums.size() - 1; ++i) {
total_sum += nums[i];
}
return (total_sum == nums[nums.size() - 1]) ? nums[nums.size() - 2] : nums[nums.size() - 1];
}
int main() {
vector<int> nums = {2, 3, 5, 10};
cout << largestOutlier(nums) << endl;
return 0;
}function largestOutlier(nums) {
nums.sort((a, b) => a - b);
let totalSum = 0;
for (let i = 0; i < nums.length - 1; i++) {
totalSum += nums[i];
}
return totalSum === nums[nums.length - 1] ? nums[nums.length - 2] : nums[nums.length - 1];
}
let nums = [2, 3, 5, 10];
console.log(largestOutlier(nums));#include <stdio.h>
#include <stdlib.h>
int cmp (const void * a, const void * b) { return (*(int*)a - *(int*)b); }
int largestOutlier(int* nums, int size) {
qsort(nums, size, sizeof(int), cmp);
int total_sum = 0;
for (int i = 0; i < size - 1; i++) {
total_sum += nums[i];
}
int largest = nums[size - 1];
if (total_sum == nums[size - 1]) {
largest = nums[size - 2];
}
return largest;
}
int main() {
int nums[] = {2, 3, 5, 10};
int size = sizeof(nums) / sizeof(nums[0]);
printf("%d\n", largestOutlier(nums, size));
return 0;
}using System;
public class LargestOutlier {
public static int LargestOutlierFunction(int[] nums) {
Array.Sort(nums);
int totalSum = 0;
for (int i = 0; i < nums.Length - 1; i++) {
totalSum += nums[i];
}
return totalSum == nums[nums.Length - 1] ? nums[nums.Length - 2] : nums[nums.Length - 1];
}
public static void Main() {
int[] nums = {2, 3, 5, 10};
Console.WriteLine(LargestOutlierFunction(nums));
}
}In this Python solution, sort() efficiently orders the data, and a simple iterative sum checks whether the maximum of respective sums equals the list's last value. If so, the respective element behind it is an outlier.
Time Complexity: O(n log n).
Space Complexity: O(1).
Approach 2: Hashing for Frequency and Sum
To find the largest outlier efficiently without sorting, use a hashmap to count element frequencies and determine potential outliers by verifying if an element could be a sum by considering n-2 special numbers. The goal is to ensure only one potential sum matches an element, allowing easy identification of the outlier.
def largest_outlier(nums):
from collections import Counter
freq = Counter(nums)
total_sum = sum(nums)
for num in nums:
potential_sum = total_sum - num
if potential_sum in freq and potential_sum != num:
return num
return -1
nums = [2, 3, 5, 10]
print(largest_outlier(nums))import java.util.HashMap;
public class LargestOutlier {
public static int largestOutlier(int[] nums) {
HashMap<Integer, Integer> freq = new HashMap<>();
int sum = 0;
for (int num : nums) {
freq.put(num, freq.getOrDefault(num, 0) + 1);
sum += num;
}
for (int num : nums) {
int potential_sum = sum - num;
if (freq.containsKey(potential_sum) && potential_sum != num) {
return num;
}
}
return -1; // logically should never reach.
}
public static void main(String[] args) {
int[] nums = {2, 3, 5, 10};
System.out.println(largestOutlier(nums));
}
}#include <iostream>
#include <unordered_map>
#include <vector>
using namespace std;
int largestOutlier(vector<int>& nums) {
unordered_map<int, int> freq;
int sum = 0;
for (int num : nums) {
freq[num]++;
sum += num;
}
for (int num : nums) {
int potential_sum = sum - num;
if (freq.find(potential_sum) != freq.end() && potential_sum != num) {
return num;
}
}
return -1; // should not reach here because there's at least one outlier.
}
int main() {
vector<int> nums = {2, 3, 5, 10};
cout << largestOutlier(nums) << endl;
return 0;
}function largestOutlier(nums) {
const freq = {};
let totalSum = 0;
nums.forEach(num => {
freq[num] = (freq[num] || 0) + 1;
totalSum += num;
});
for (let num of nums) {
const potentialSum = totalSum - num;
if (freq[potentialSum] && potentialSum !== num) {
return num;
}
}
return -1; // logically it won't happen
}
const nums = [2, 3, 5, 10];
console.log(largestOutlier(nums));#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX_NUM 2001
#define OFFSET 1000
int largestOutlier(int* nums, int size) {
int frequency[MAX_NUM] = {0};
int sum = 0;
for (int i = 0; i < size; ++i) {
frequency[nums[i] + OFFSET]++;
sum += nums[i];
}
for (int i = size - 1; i >= 0; --i) {
int outlier_candidate = nums[i];
int potential_sum = sum - outlier_candidate;
if (frequency[potential_sum + OFFSET] > 0 && potential_sum != outlier_candidate) {
return outlier_candidate;
}
}
return -1;
}
int main() {
int nums[] = {2, 3, 5, 10};
int size = sizeof(nums)/sizeof(nums[0]);
printf("%d\n", largestOutlier(nums, size));
return 0;
}using System;
using System.Collections.Generic;
public class LargestOutlier {
public static int LargestOutlierFunction(int[] nums) {
Dictionary<int, int> frequency = new Dictionary<int, int>();
int sum = 0;
foreach (int num in nums) {
if (frequency.ContainsKey(num))
frequency[num]++;
else
frequency[num] = 1;
sum += num;
}
foreach (int num in nums) {
int potentialSum = sum - num;
if (frequency.ContainsKey(potentialSum) && potentialSum != num) {
return num;
}
}
return -1; // Should never happen.
}
public static void Main() {
int[] nums = {2, 3, 5, 10};
Console.WriteLine(LargestOutlierFunction(nums));
}
}Python's `Counter` from the collections module provides frequency counts. By assessing potential sums with the total minus a given number, it identifies the largest outlier when certain criteria are met.
Time Complexity: O(n).
Space Complexity: O(n) due to counter storage
Approach 3: Hash Table + Enumeration
We use a hash table cnt to record the frequency of each element in the array nums.
Next, we enumerate each element x in the array nums as a possible outlier. For each x, we calculate the sum t of all elements in the array nums except x. If t is not even, or half of t is not in cnt, then x does not meet the condition, and we skip this x. Otherwise, if x is not equal to half of t, or x appears more than once in cnt, then x is a possible outlier, and we update the answer.
After enumerating all elements, we return the answer.
The time complexity is O(n), and the space complexity is O(n). Here, n is the length of the array nums.
function getLargestOutlier(nums: number[]): number {
let s = 0;
const cnt: Record<number, number> = {};
for (const x of nums) {
s += x;
cnt[x] = (cnt[x] || 0) + 1;
}
let ans = -Infinity;
for (const [x, v] of Object.entries(cnt)) {
const t = s - +x;
if (t % 2 || !cnt.hasOwnProperty((t / 2) | 0)) {
continue;
}
if (+x != ((t / 2) | 0) || v > 1) {
ans = Math.max(ans, +x);
}
}
return ans;
}func getLargestOutlier(nums []int) int {
s := 0
cnt := map[int]int{}
for _, x := range nums {
s += x
cnt[x]++
}
ans := math.MinInt32
for x, v := range cnt {
t := s - x
if t%2 != 0 || cnt[t/2] == 0 {
continue
}
if x != t/2 || v > 1 {
ans = max(ans, x)
}
}
return ans
}use std::collections::HashMap;
impl Solution {
pub fn get_largest_outlier(nums: Vec<i32>) -> i32 {
let mut s = 0;
let mut cnt = HashMap::new();
for &x in &nums {
s += x;
*cnt.entry(x).or_insert(0) += 1;
}
let mut ans = i32::MIN;
for (&x, &v) in &cnt {
let t = s - x;
if t % 2 != 0 {
continue;
}
let y = t / 2;
if let Some(&count_y) = cnt.get(&y) {
if x != y || v > 1 {
ans = ans.max(x);
}
}
}
ans
}
}Complexity comparison
| Approach | Complexity |
|---|---|
| Sorting and Iterative Check | Time Complexity: O(n log n). |
| Hashing for Frequency and Sum | Time Complexity: O(n). |
| Hash Table + Enumeration | – |
Complexity analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Sorting and Iterative Check | O(n log n) | O(1) | Useful when avoiding extra memory or when the array is already sorted. |
| Hashing for Frequency and Sum | O(n) | O(n) | Best general solution. Fast lookups make candidate validation constant time. |
Frequently asked questions
Is Identify the Largest Outlier in an Array easy or hard?
The problem is rated Medium because the main challenge is recognizing the mathematical relationship between the outlier, the sum element, and the remaining numbers. Once the equation is derived, implementing the O(n) hash map solution is straightforward.
Identify the Largest Outlier in an Array Python/Java solution
Python and Java implementations typically use a dictionary or HashMap to store element counts. After computing the total sum, iterate through the array, compute y = (totalSum - o) / 2, and check the map for that value. This approach runs in O(n) time and O(n) space in both languages.
How to solve Identify the Largest Outlier in an Array in O(n)?
Compute the total sum of the array and build a hash map of value frequencies. For each element treated as the outlier o, calculate remaining = totalSum - o. If remaining is even and remaining / 2 exists in the map (with valid frequency), the condition holds. Track the largest such o while scanning the array once.
What is the best approach for Identify the Largest Outlier in an Array?
The most efficient approach uses a hash map to store the frequency of each number and the total sum of the array. For every candidate outlier o, compute y = (totalSum - o) / 2 and check if that value exists in the map. This reduces validation to constant-time lookups, giving an overall time complexity of O(n) with O(n) space.
Is Identify the Largest Outlier in an Array asked at Google/Amazon/Meta?
Problems involving hash maps, array sums, and frequency counting are common in interviews at companies like Amazon, Google, and Meta. Variants of this problem appear in coding rounds because they test algebraic reasoning combined with efficient data structure use.
What data structure is used in Identify the Largest Outlier in an Array?
The optimal solution uses a hash table (hash map) to maintain frequency counts of numbers in the array. This allows constant-time existence checks when verifying whether the required sum element y is present.
What is the time complexity of Identify the Largest Outlier in an Array?
The optimal solution runs in O(n) time using a hash map for frequency counting and a single scan of the array. A simpler alternative that sorts the array first takes O(n log n) time due to the sorting step.