Difference Pyramid

After every operation, every value is still one of $0$, $1$, and $2$. Therefore, the final value is also one of these three values.

First, determine the parity of the final value. For integers $x,y$,

$$|x-y|\equiv x+y\pmod 2.$$

Thus, modulo $2$, one operation is equivalent to replacing adjacent values by their sum.

After repeating this $N-1$ times, Pascal-triangle coefficients appear. If $P$ is the parity of the final value, then

$$P\equiv\sum_{i=1}^{N}\binom{N-1}{i-1}A_i\pmod 2.$$

By Lucas's theorem,

$$\binom{n}{k}$$

is odd if and only if every set bit of $k$ is also set in $n$. Equivalently,

$$(k\mathbin{\&}n)=k.$$

Therefore, with $n=N-1$, XOR $A_i\bmod2$ over all positions satisfying $(i-1)\mathbin{\&}n=i-1$. This computes $P$ in $O(N)$ time.

If $P=1$, the final value is odd. Since it is one of $0$, $1$, and $2$, the answer must be $1$.

Now suppose $P=0$. The final value is either $0$ or $2$.

If the final value is $2$, the original array cannot contain a $1$. This can be proved by induction on the array length. The claim is trivial for length $1$. For a longer array whose final value is $2$, apply the induction hypothesis to the array after the first operation. Every value after the first operation must be even. Hence every adjacent pair in the original array has the same parity, so all original values have the same parity. If they were all odd, then they would all be $1$, the first operation would produce only $0$s, and the final value would be $0$, a contradiction. Thus all original values are $0$ or $2$.

Consequently, if $P=0$ and at least one $A_i$ equals $1$, the answer is $0$.

It remains to handle the case where every $A_i$ is $0$ or $2$. Let $C_i=A_i/2$, so $C_i$ is binary. Since

$$|2x-2y|=2|x-y|,$$

every value in the entire Difference Pyramid for $A$ is exactly twice the corresponding value for $C$. Thus, the final value for $A$ is twice the final value for $C$.

Because $C$ is binary, its final value is exactly its parity, which is

$$Q\equiv\sum_{i=1}^{N}\binom{N-1}{i-1}\frac{A_i}{2}\pmod2.$$

Compute $Q$ with the same Lucas-theorem test and print $2Q$.

We can compute $P$, $Q$, and whether a $1$ exists in a single pass. The time complexity is $O(N)$ and the memory complexity is $O(1)$.

Solution written by GPT5.6