Addition on a Segment

Key observation

The order of $b$ is a distraction. We only need the final multiset of values, because the statement lets us reorder the array at the end.

So imagine sorting the target values however we want. The cleanest choice is nonincreasing order:

$$b_1 \ge b_2 \ge \dots \ge b_n.$$

Now the final array looks like a histogram. A segment addition is like placing one horizontal strip of height $1$ over some consecutive columns.

Upper bound

Suppose some operation uses a segment of length $L$.

Every position inside that segment is incremented at least once, so every one of those $L$ final values is positive. Therefore:

$$L \le \#\{i : b_i > 0\}.$$

So the answer can never exceed the number of positive elements in $b$. That part is brutally simple: you cannot touch 7 final-zero positions and then pretend nothing happened. Math has receipts.

Let

$$P = \#\{i : b_i > 0\}.$$

We have shown:

$$\text{answer} \le P.$$

Achieving the bound

Now sort $b$ in nonincreasing order. All positive values are in the prefix $1,2,\dots,P$.

Since the input is guaranteed feasible, there exists some valid sequence of exactly $n$ segment additions that produces this sorted array. Look at the very first layer: every positive position must be incremented at least once, and no zero position can be incremented at all.

So the segment covering the first layer can be chosen as exactly:

$$[1,P].$$

That gives us an operation of length $P$, which matches the upper bound.

Another way to see the same thing: for each height $x$, the positions with $b_i \ge x$ form a prefix after sorting, so layers are naturally contiguous. The widest layer is height $1$, and its width is exactly the number of positive values.

Final answer

The maximum possible segment length is:

$$\boxed{\#\{i : b_i > 0\}}$$

So each test case is just counting positives. That's it. Rated 1200, but the core idea is basically: zeros are lava, positives are usable ground.