Ten Thousand Empty Cases

The constraint block reads:

1 <= t <= 10^4                       (number of test cases)
1 <= n <= 2*10^5                     (array length in one case)
the sum of n over all test cases does not exceed 2*10^5

Your per-case algorithm is $O(n \log n)$ . What is your real budget, and where is the risk?

The sum bound adds nothing, since $n \le 2 \cdot 10^5$ already caps every case.

Total work across the whole run is bounded by $O!\left(\left(\sum n\right) \log n\right) \approx 3.5 \cdot 10^6$ , so the algorithm is fine. The risk is any per-case work that depends on the maximum $n$ rather than this case's $n$ : clearing a fixed $2 \cdot 10^5$ array at the top of each case costs $10^4 \cdot 2 \cdot 10^5 = 2 \cdot 10^9$ .

The budget is per test case, so you get roughly $10^8$ operations for each of the up to $10^4$ cases.

With $t$ as large as $10^4$ the binding constraint is $t$ , so you need an $O(1)$ answer per case.