Pairs Flipping (Easy Version)

Work with 1-indexed positions. If we flip positions $a<b$, then this is operation $b-a$ and we output $l=a$ for that length.

First remove the annoying parity issue. If the original $n$ is even, we solve only the prefix of length $n-1$. The operation of length $n/2$ is unused by that prefix. After the construction, if $s_n=1$, flip positions $(n/2,n)$. This turns $s_n$ into 0; the other endpoint either turns 0 into 1 or 1 into 0, so the total number of 1s does not increase. Nice little cleanup, no drama.

So assume the processed length is odd: $N=2m+1$. Since $N\le 1999999$, we have $m\le 999999$.

Split the processed string into

$$[1,m]\quad\text{and}\quad[m+1,N].$$

The left half has endpoint distances $m-1,m-3,m-5,\ldots$, while the right half has distances $m,m-2,m-4,\ldots$. These parities are different, so every operation length is reserved for exactly one sweep.

For the left half, process from the center outward. Maintain this invariant:

In the already processed part of the left half, all positions are 0 except possibly one pending position $p$.

When adding symmetric endpoints $l<r$, let $d=r-l$.

If $s_l=s_r$, they are easy. If both are 1, flip $(l,r)$ and both become 0; if both are 0, do nothing.

If $s_l\ne s_r$, exactly one endpoint contains a 1. Before making that endpoint the new pending position, kill the old pending position if it is currently 1: flip $(p,p+d)$. This is valid because $p$ lies inside the previously processed interval, so $p+d$ is to the right of the current pair and is still within the processed odd-length prefix. It may mess with the right half, but that is fine because the right half has not been cleaned yet. After that, the endpoint containing 1 becomes the new pending position.

After the left sweep, the whole left half contains at most one active unresolved 1.

Now process the right half from the outside inward. Maintain another invariant:

Outside the current right interval, apart from frozen leftovers we have already given up on, there is at most one active pending position $p$.

For endpoints $l<r$ with $d=r-l$:

If $s_l=s_r$, handle them the same way as before: flip $(l,r)$ only when both are 1.

If $s_l\ne s_r$, one endpoint will become the new active pending position after we shrink the interval. Before that, try to kill the old active $p$. If $p<l$, the only useful partner at distance $d$ is $p+d$; if $p>r$, it is $p-d$. If that partner lies strictly inside $(l,r)$, flip the pair and old $p$ is gone. If not, old $p$ is too far away, so we freeze it as a final leftover. Then the endpoint containing 1 becomes the new active pending position.

The whole proof is in the freeze bound. Suppose an active pending position is created while the current span is $D$. Immediately after shrinking, it is distance $1$ from the remaining interval. Later, after several inward steps, suppose it freezes when the span is $d$ and its gap from the interval is $g$. Each inward step decreases the span by $2$ and increases the gap by $1$, so

$$D=d+2g.$$

A freeze means the pending position is too far to pair into the interval, i.e. $g\ge d$. Therefore

$$D=d+2g\ge 3d.$$

So consecutive freeze spans shrink by a factor of at least $3$.

The first freeze span is at most $m\le 999999$. Hence there can be at most $13$ freezes, since

$$3^{12}=531441\le 999999<3^{13}=1594323.$$

After the right sweep, besides frozen positions, there can be at most one active pending position and at most one unprocessed center position. Therefore the final number of 1s is at most

$$13+2=15.$$

All unused operation lengths simply output $0$.

The construction touches each position $O(1)$ times, so the complexity is $O(n)$ per test case and $O(\sum n)$ overall. Memory usage is $O(n)$.