0mar and Alternating Sums

The non-decreasing order is doing basically all the work here. The actual magnitudes up to $10^9$ are mostly a decoy; what matters is the parity of how many indices we choose from each equal-value group.

Parity choices inside one group

Suppose some value occurs $s$ times. Exactly half of its $2^s$ subsets have even size and half have odd size, so either prescribed parity has

$2^{s-1}$

choices. One way to see this is to fix one occurrence and toggle whether it is selected; this pairs every even-sized subset with a unique odd-sized subset.

This counts actual index subsets, not merely possible counts, so duplicate values are handled correctly.

If the array has $d$ distinct values with group sizes $s_1,\ldots,s_d$, then any fixed assignment of even/odd parity to every group has

$\prod_{j=1}^d 2^{s_j-1}=2^{(s_1+\cdots+s_d)-d}=2^{n-d}$

realizations.

Call this common quantity $B=2^{n-d}$.

First solve the positive-only case

Take any selected positive sequence

$b_1\le b_2\le\cdots\le b_k.$

If $k$ is odd, rewrite its alternating sum as

$b_1+(b_3-b_2)+(b_5-b_4)+\cdots+(b_k-b_{k-1}).$

Every parenthesized difference is non-negative and $b_1>0$, so this sum is strictly positive. It cannot be zero.

If $k$ is even, pair consecutive positions:

$P=(b_1-b_2)+(b_3-b_4)+\cdots+(b_{k-1}-b_k).$

Every term is non-positive because the sequence is sorted. Therefore $P=0$ exactly when every pair contains equal values.

That is equivalent to selecting every distinct value an even number of times:

• If every count is even, all copies of each value form equal pairs.
• If some count is odd, one pair must cross from that value to a larger value, making its difference negative.

Thus, when $-1$ is absent, every group must have even selected size. There are exactly

$B=2^{n-d}$

valid subsequences. The empty subsequence is already included.

Bring back the $-1$s

Because the entire array is sorted, all occurrences of $-1$ come before every positive value. Let $r$ be the number of selected $-1$s.

An even number of $-1$s

Their contribution is

$(-1)-(-1)+(-1)-(-1)+\cdots=0.$

Since $r$ is even, the positive suffix starts with a plus sign. Its alternating sum must therefore be zero, which requires every positive group to be selected evenly.

So every group, including the $-1$ group, has prescribed even parity. This contributes exactly $B$ subsequences.

An odd number of $-1$s

The selected $-1$s contribute $-1$. They also occupy an odd number of positions, so the first selected positive receives a minus sign.

Let $P$ denote the positives' usual alternating sum starting with plus. The complete sum is then

$-1-P.$

For it to be zero, we need

$P=-1.$

We now characterize exactly when that happens.

An odd-length positive sequence has $P>0$, as proved earlier, so its length must be even. Consider the positive value groups selected an odd number of times, and list their distinct values as

$y_1<y_2<\cdots<y_{2q}.$

Groups selected evenly contribute zero and do not change the sign entering the next group. Each odd group contributes one uncancelled copy and flips that sign. Consequently,

$P=(y_1-y_2)+(y_3-y_4)+\cdots+(y_{2q-1}-y_{2q}).$

Each difference is at most $-1$. Their sum can be exactly $-1$ only when

• there is exactly one pair, so $q=1$; and
• its values differ by exactly one, so $y_2=y_1+1$.

Therefore, for some positive value $v$, the groups $v$ and $v+1$ must be selected oddly, while every other positive group is selected evenly. The $-1$ group is also selected oddly.

Let $\ell$ be the number of positive values $v$ for which both $v$ and $v+1$ occur in the array. Every such pair defines one parity assignment, and each assignment has exactly $B$ realizations. Different pairs give different parity assignments, so there is no double-counting.

Final formula

If $-1$ does not occur, only the positive-only case exists:

$\boxed{\text{answer}=2^{n-d}}.$

If $-1$ occurs, there is one all-even parity assignment and $\ell$ valid odd-$-1$ assignments:

$\boxed{\text{answer}=2^{n-d}(\ell+1)}.$

Here $d$ is the number of all distinct array values, including $-1$ when present.

For example, in $[-1,1,1,2,3]$, we have $d=4$, so $B=2^{5-4}=2$. The consecutive positive pairs are $(1,2)$ and $(2,3)$, hence $\ell=2$ and the answer is $2(2+1)=6$.

Algorithm

Scan the already sorted array and count:

• $d$, the number of distinct-value groups;
• whether $-1$ occurs;
• $\ell$, the number of transitions between distinct positive values whose difference is exactly one.

Then apply the formula above using precomputed powers of two modulo $10^9+7$.

Correctness proof

For positive-only selections, the pairing argument proves that alternating sum zero is equivalent to every value group having even selected size.

With an even number of selected $-1$s, those elements cancel and do not reverse the positive signs, so the same all-even condition is necessary and sufficient.

With an odd number of selected $-1$s, the total is zero exactly when the positives have usual alternating sum $-1$. The odd-group expansion proves this happens exactly when one occurring pair $(v,v+1)$ has odd counts and every other positive group has an even count.

Every fixed parity assignment has $2^{n-d}$ index subsets. Hence the algorithm counts one such assignment without $-1$, or $\ell+1$ disjoint assignments when $-1$ exists. Therefore the returned formula counts every valid subsequence exactly once.

Complexity

Across one test case, the scan takes $O(n)$ time and $O(1)$ auxiliary space. The shared powers-of-two table uses $O(2\cdot10^5)$ memory and preprocessing time.