One Night At Freddy's

Observation 1

The guard should always flash the current maximum danger level.

Suppose the dangers are sorted as

$$d_1 \ge d_2 \ge \cdots \ge d_m.$$

If we reset $d_p$ for some $p>1$, then $d_1$ stays alive. If we reset $d_1$, then after sorting, the whole state is component-wise no worse for the guard. Since animatronic labels do not matter, resetting the maximum dominates every other choice. Anything else is just bad horror-movie decision making.

Observation 2

Now the guard's strategy is fixed: every flashlight deletes the current maximum.

Let $k$ be the number of flashlight uses still remaining before the current second. A danger pile only matters if it can survive those $k$ future deletions. So the adversary needs multiple simultaneous threats:

$$s = \min(m, k+1).$$

Only the $s$ largest dangers are relevant. If $m>k+1$, anything below the top $k+1$ is no larger than the weakest relevant threat, so it cannot improve the final maximum unless future increments are spent on it. Spending that increment on the weakest relevant threat is at least as good.

Among the $s$ relevant threats, the guard keeps deleting the largest ones. Therefore the adversary should increment the weakest relevant threat. This is the classic balancing idea: if you want one pile to survive deletions, do not make one pile gigantic; make enough piles annoyingly equal.

So with sorted decreasing dangers, the adversary increments position

$$\min(m,k+1)$$

using $1$-based indexing.

Algorithm

Maintain all $m$ danger values sorted decreasing.

For each second $sec=1,2,\ldots,\ell$:

1. Let $k$ be the number of unused flashlight actions.
2. Increment the $\min(m,k+1)$-th largest danger value.
3. Sort the array again.
4. If $sec$ is a flashlight time, reset the largest value to $0$, sort again, and decrease $k$.

At the end, the largest danger value is the minimax answer.

Correctness

First, the guard's greedy move is optimal: resetting the current maximum gives a sorted state no worse than resetting any smaller value, and future play only depends on the multiset of danger values.

Second, with $k$ flashes left, the adversary must build $k+1$ threats if possible, because the guard can erase at most $k$ of them. Thus exactly the top $\min(m,k+1)$ values are the relevant candidates for the final surviving maximum.

Finally, inside those candidates, increasing anything except the weakest candidate cannot be better. The strongest candidates are the ones most likely to be deleted, while the weakest candidate is the bottleneck for what can survive all future maximum-deletions. Therefore incrementing the weakest relevant candidate is optimal for the adversary.

The simulation applies the optimal guard move and the optimal adversary move at every step, so it computes exactly the minimum $x$ the guard can guarantee.

Complexity

For one test case, we sort $m$ values for each of $\ell$ seconds:

$$O(\ell \cdot m\log m).$$

Across all test cases,

$$\sum m\ell \le 2\cdot 10^5,$$

so this comfortably fits. Memory usage is $O(m)$.

Source checked: official Codeforces editorial.