0940EasyArrayCounting

Is the Middle Reading One of a Kind

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Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

Readings are given as nums, and there is an odd number of them, so exactly one sits in the middle.

Return true when that middle reading appears nowhere else in the list.

Examples

Example 1

Input: nums = [4, 9, 27, 9, 4]
Output: true

The middle reading is 27, which appears nowhere else.

Example 2

Input: nums = [1, 2, 1]
Output: true

The middle reading is 2 and it is the only 2 on the list, even though the reading 1 is repeated.

Example 3

Input: nums = [8, 8, 3, 8, 8]
Output: true

The middle reading is 3, on its own, so the four eights around it make no difference.

Constraints

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

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def is_middle_element_unique(nums: list[int]) -> bool:

Java

public boolean isMiddleElementUnique(int[] nums)

The intended approach

Time O(n) for n readings, one pass.Space O(1).

With an odd count the middle sits at the position given by halving the length and rounding down. Only that one value matters, so there is nothing to tally over the whole list: read the middle value, then walk the list counting how often it turns up, and report whether that count is one. Counting can stop as soon as a second sighting appears.

Pythonentry point: is_middle_element_unique

def is_middle_element_unique(nums):
    middle = nums[len(nums) // 2]
    seen = 0
    for v in nums:
        if v == middle:
            seen += 1
            if seen > 1:
                return False
    return True

The obvious version

Tally every reading in the list, then look the middle one up in that tally. Builds a count for values that are never asked about.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_is_middle_element_unique(nums):
    tally = {}
    for v in nums:
        tally[v] = tally.get(v, 0) + 1
    middle = nums[len(nums) // 2]
    return tally[middle] == 1

How this problem was checked

Reference solution executed against 13 test cases — 13 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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