I Will Definitely Make It

The trick is to stop thinking about tower indices. They barely matter. Teleport duration is based only on height difference, so towers with the same height are interchangeable.

Let

$s = h_k$

be the height of the tower where you start.

The Timing Rule

Suppose you are on a tower of height $a$ at time $t$, and you want to teleport upward to height $b \ge a$.

The teleport takes $b-a$ seconds. During that whole time, you are still on height $a$. From the examples, leaving exactly when the old tower gets covered is allowed, so the move is legal iff

$t + (b-a) \le a.$

Now assume we have only moved upward so far. If we started at height $s$ and are currently at height $a$, then the total time spent is

$a-s,$

because all the upward height differences telescope:

$(x_1-s) + (x_2-x_1) + \cdots + (a-x_m) = a-s.$

Plug $t=a-s$ into the teleport condition:

$a-s + (b-a) \le a.$

Simplify:

$b-a \le s.$

That is the whole problem. Your maximum allowed upward height gap is your starting height $s$.

Why Going Down Is Useless

Could dropping to a lower tower ever help? No.

Moving downward costs time too, but it does not get you closer to the maximum height. Even worse, it puts you on a lower tower while the water keeps rising. So any path with downward moves can only be worse than a path that skips that nonsense. We can assume the useful path is nondecreasing in height.

Greedy Reachability

Sort all heights.

Start with cur = h_k. Look at the available tower heights in increasing order. If the next height $x$ satisfies

$x - cur \le h_k,$

then you can reach it, so set cur = x.

If instead

$x - cur > h_k,$

then you cannot reach $x$, and you also cannot reach anything taller than $x$, because those are even farther away. You are stuck. No secret anime teleport arc is coming.

So the algorithm is:

1. Store $s = h_k$.
2. Sort the heights.
3. Scan upward from height $s$.
4. If any needed gap is bigger than $s$, answer NO.
5. Otherwise, answer YES.

Correctness Proof

Lemma 1: There is always an optimal successful path that never moves to a lower height.

A downward move spends positive time unless the height is equal, and it ends at a height no larger than before. Since the goal is to reach a maximum-height tower and water only rises, such a move cannot improve future possibilities. Removing downward movement never makes the situation worse.

Lemma 2: If we have reached height $a$ using only upward moves from starting height $s$, then the elapsed time is $a-s$.

Each upward teleport contributes exactly the increase in height. Summing all increases from $s$ to $a$ telescopes to $a-s$.

Lemma 3: From a reached height $a$, an upward move to height $b \ge a$ is possible iff $b-a \le s$.

By Lemma 2, the current time is $a-s$. The move takes $b-a$ seconds and must finish before or exactly when height $a$ becomes unusable:

$a-s + (b-a) \le a.$

This is equivalent to $b-a \le s$.

Lemma 4: If the sorted scan finds a gap larger than $s$, the maximum height is unreachable.

Let cur be the largest reachable height so far, and let $x$ be the next available height above it. If $x-cur>s$, then by Lemma 3 we cannot move from cur to $x$. Every later height is at least $x$, so it is also unreachable from cur. There is no intermediate tower to bridge the gap.

Lemma 5: If every scanned gap is at most $s$, the maximum height is reachable.

The scan constructs a valid sequence of upward moves. Each move has height difference at most $s$, so by Lemma 3 every move is legal. Eventually the sequence reaches the maximum height.

Therefore, the greedy sorted scan answers YES exactly when reaching a maximum-height tower is possible.

Complexity

Sorting dominates:

$O(n \log n)$

per test case, with total $n \le 10^5$, so this easily fits.