Is Unique Middle Element easy or hard?

Unique Middle Element is generally considered an easy problem because the core idea is straightforward frequency counting. The main interview expectation is recognizing when to replace nested loops with hashing to improve performance from O(n²) to O(n).

Unique Middle Element Python/Java solution

Python solutions typically use collections.Counter or a standard dictionary for frequency tracking. Java implementations usually rely on HashMap . Both approaches achieve O(n) time complexity with clean and readable code.

How to solve Unique Middle Element in O(n)?

Use a hash map to count how many times each value appears in the array. After building the frequency table, iterate through the elements and return the value whose frequency is exactly one and satisfies the middle-element condition. Hash lookups work in constant average time, giving an overall O(n) solution.

What is the best approach for Unique Middle Element?

The hash map frequency counting approach is the best overall solution for Unique Middle Element. It processes the array in linear time by storing frequencies in a dictionary or hash map, then identifies the unique middle value in a second pass. The total complexity is O(n) time and O(n) space.

Is Unique Middle Element asked at Google/Amazon/Meta?

Frequency-counting and unique-element problems are common screening questions at companies such as Amazon, Google, and Meta because they test array traversal, hashing, and complexity optimization. The exact problem title may vary, but the underlying pattern appears frequently in coding interviews.

What data structure is used in Unique Middle Element?

The most common data structure used is a hash map or dictionary for storing element frequencies. Some alternative solutions use sorting with array traversal, but hashing gives the best runtime for unsorted inputs.

What is the time complexity of Unique Middle Element?

The optimal solution runs in O(n) time using a hash map for frequency counting. A brute force implementation that counts occurrences for every element separately takes O(n²) time. Sorting-based solutions usually run in O(n log n).

Description Hints 2 Solution Video 2:30

3978. Unique Middle Element

Easy72.3% acceptance

You are given an integer array nums of odd length n.

Return true if the middle element of nums appears exactly once in the array. Otherwise return false.

Example 1:

Input: nums = [1,2,3]

Output: true

Explanation:

The middle element of nums is 2, which appears exactly once.

Thus, the answer is true.

Example 2:

Input: nums = [1,2,2]

Output: false

Explanation:

The middle element of nums is 2, which appears twice.

Thus, the answer is false.

Constraints:

1 <= n == nums.length <= 100
n is odd.
1 <= nums[i] <= 100

Topics

Array Counting

Hints open one at a time

Hint 1

The middle index is nums.length / 2.

Hint 2

Count how many times nums[nums.length / 2] appears in nums.

2:30

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1class Solution:
2    def isMiddleElementUnique(self, nums: list[int]) -> bool:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2

Input

[1,2,3]

Input

[1,2,2]

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