Hints: Magical Tiered Cake

Three pegs hold stacks. Layer $i$ is smaller than layer $j$ when $i < j$, and you may only place a layer on a strictly larger one below it (or on an empty peg).

Because of that rule, what must every peg’s stack look like if you write layers from top to bottom?

Answer to Hint 1: Each peg’s stack is always sorted increasingly from top to bottom: if both $3$ and $7$ are on the same peg, they appear as $[\ldots, 3, 7, \ldots]$ with $3$ above $7$.

So the whole state is: three increasing lists. No arbitrary permutations on a peg.

Now rewrite the magic rule. If peg $S$ contains layer $i$, how many layers on that same peg are smaller than $i$ (and therefore above $i$)?

Answer to Hint 2: Exactly $a_i$ layers on that peg must be smaller than $i$.

Equivalently, with layer set $S$ on a peg, layer $i \in S$ is movable iff

Try small $n$ by hand: when $a = [0,1,2]$, which layer can move from the full kitchen stack $[1,2,3]$?

Answer to Hint 3: Only layer $1$ (needs $0$ above; it is on top). After moving $1$ away, layer $2$ has exactly one smaller layer above it in the kitchen — but wait, $1$ left, so now $2$ is on top with $0$ above…

Actually after removing $1$, the kitchen is $[2,3]$. Layer $2$ has $0$ smaller layers above, but $a_2 = 1$. So not movable yet.

For $a = [0,1,2]$, start from $[1,2,3]$: only layer $3$ is movable first (two smaller layers above). That suggests moving large layers first in this case. What simple impossibility can you spot from $|\{j < i\}| \le i - 1$?

Answer to Hint 4: Always $a_i \le i - 1$ (0-based: $a_i \le i$ in input index $i$ for layer $i+1$… check your indexing).

If $a_i > i - 1$, print NO. Example: $a = [2,2,2]$ for $n = 3$.

Claim (the hard part): this is also sufficient — if every $a_i \le i - 1$, a solution within $2^n$ moves always exists.

Compare two extremes: all $a_i = 0$ vs all $a_i = i - 1$. How many moves do you expect in each case?

Answer to Hint 5:

• All $a_i = i - 1$: from $[1,\ldots,n]$ in the kitchen, layer $n$ is movable immediately (all $n-1$ smaller layers above). Peel $n, n-1, \ldots, 1$ straight to the party — $n$ moves (sample test 2).
• All $a_i = 0$: only the top of each stack moves. Extracting the bottom is Tower of Hanoi — $2^n - 1$ moves (sample test 1).

Both fit in $2^n$. The general case interpolates between “peel the bottom” and “only move the top.”

Focus on a block $l, l+1, \ldots, k$ together on one peg. When is bottom layer $k$ movable?

Answer to Hint 6: Among the block, layer $k$ has exactly $(k - l)$ smaller members above it. It is movable iff

Define

If $d > 0$, you must relocate the $d$ smallest layers of the block ($l, \ldots, l+d-1$) off the peg before moving $k$.

This suggests a recursive function move_range(l, k, src, dst, aux). What are the four phases?

Answer to Hint 7:

1. Stash $l..l+d-1$ from src to aux.
2. Move layer $k$ from src to dst.
3. Finish middle layers $l+d, \ldots, k-1$ (each may need a few smaller layers temporarily brought back or sent away so that exactly $a_m$ smaller layers share its peg when it moves).
4. Unstash $l..l+d-1$ onto dst.

Steps 1 and 4 are smaller instances of the same recursion; step 2 is one move.

For $a = [0,1,2]$, what is $d$ when $l = 1, k = 3$?

Answer to Hint 8: $d = (3 - 1) - a_3 = 2 - 2 = 0$. No stash. Move $3$, then middle layers $2, 1$ to the party — 3 moves total.

For $a = [0,0,0]$, $k = 3$, $l = 1$: $d = 2 - 0 = 2$. Stash layers $1$ and $2$, move $3$, then rebuild on the party. That is exactly $2^3 - 1$ moves.

When processing middle layer $m$, you may need more or fewer smaller layers on its peg than currently present. How do you fix each case?

Answer to Hint 9:

Let $c$ = number of smaller layers on the same peg as $m$.

• If $c < a_m$: pull some layer $x < m$ from another peg (often from the stash) onto this peg.
• If $c > a_m$: move the largest smaller layer on this peg to a spare peg. Choose a spare where the move is legal (empty or top larger than $x$). Do not park $x$ on the party peg if a larger middle layer still has to land there later with $x$ on top.

After adjusting, move $m$ to the party peg.

Why is the answer always YES when $a_i \le i - 1$, and why are $\le 2^n$ moves enough?

Answer to Hint 10: Necessity was $a_i \le i - 1$. For sufficiency, the recursion always makes progress: either a shorter interval or a layer fixed at the party. Move count is sandwiched between the $n$-move peel (all $a_i = i-1$) and the $2^n - 1$ Hanoi (all $a_i = 0$), hence $\le 2^n$.

Implementation sketch:

1. Check $a_i \le i - 1$; else NO.
2. Call move_range(1, n, kitchen, party, warehouse).
3. Simulate moves to verify; if the trace breaks on a corner case, BFS on the state graph with cap $2^n$ (reference code does this).

Output YES, $k$, then $k$ lines id from to.