Watch 4 video solutions for Unique Middle Element, a easy level problem. This walkthrough by Programming Live with Larry has 59 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
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 <= 100n is odd.1 <= nums[i] <= 100Problem Overview: You need to identify the middle element that appears exactly once while the remaining values may repeat multiple times. The challenge is separating the unique value efficiently without scanning the array repeatedly.
Approach 1: Brute Force Frequency Scan (O(n²) time, O(1) space)
Iterate through every element and count its occurrences by scanning the entire array again. After computing the frequency for the current value, check whether it is unique and whether it satisfies the middle-position condition from the problem statement. This approach avoids extra memory and works well for very small inputs, but repeated comparisons make it slow for larger arrays. Problems involving repeated counting are common in arrays practice sets.
Approach 2: Hash Map Frequency Counting (O(n) time, O(n) space)
Store element frequencies in a hash map using one linear pass. Once frequencies are available, iterate through the array again and return the element whose count equals 1 and matches the required middle index logic. The key insight is that hash lookups reduce repeated counting from linear time to constant average time. This is the standard solution interviewers expect for easy frequency problems involving hash tables.
Approach 3: Sorting and Neighbor Comparison (O(n log n) time, O(1) or O(n) space)
Sort the array first, then scan linearly while comparing neighboring values. A value is unique if it differs from both adjacent elements after sorting. This method is useful when the input is already sorted or when hash-based structures are restricted. The tradeoff is the sorting overhead, but the implementation stays simple and deterministic. Similar techniques appear in sorting and duplicate-removal problems.
Recommended for interviews: The hash map solution is the strongest choice because it gives linear performance with straightforward logic. Interviewers usually expect you to recognize the frequency-counting pattern quickly and optimize away nested loops. Showing the brute force approach first demonstrates baseline reasoning, while moving to the O(n) hash map version shows optimization skill and awareness of time complexity tradeoffs.
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Brute Force Frequency Scan | O(n²) | O(1) | Small inputs or memory-constrained environments |
| Hash Map Frequency Counting | O(n) | O(n) | General case and interview settings |
| Sorting and Neighbor Comparison | O(n log n) | O(1) to O(n) | When the array is already sorted or hashing is unavailable |