XOR Array

Think in prefix XORs. This problem looks like it wants some clever XOR wizardry, but the real trick is to stop staring at the array directly.

Define

$p_0=0,$

and for $i\ge 1$,

$p_i=a_1\oplus a_2\oplus\dots\oplus a_i.$

Then the XOR of a subarray $[x,y]$ is

$a_x\oplus a_{x+1}\oplus\dots\oplus a_y=p_y\oplus p_{x-1}.$

This is $0$ exactly when

$p_y=p_{x-1}.$

So we are not really constructing array values. We are constructing prefix values.

What do we need?

We need exactly one zero-XOR subarray, namely $[l,r]$.

That means exactly one matching pair among prefix values:

$p_r=p_{l-1}.$

And no other pair of prefix values may be equal. If some other $p_j=p_i$ with $i<j$, then subarray $[i+1,j]$ would also have XOR $0$, which would be bad.

So the whole problem becomes:

Construct $p_0,p_1,\dots,p_n$ such that all values are distinct except $p_r=p_{l-1}$.

That is hilariously easier.

Construction

Set

$p_i=i$

for every $i$, except overwrite

$p_r=l-1.$

Now $p_r=p_{l-1}$, so subarray $[l,r]$ has XOR $0$.

Are there any other duplicates? Nope. Initially $p_i=i$ makes everything distinct. Replacing $p_r$ with $l-1$ only creates one duplicate with position $l-1$.

Also, because $l<r$, we have

$r-(l-1)\ge 2,$

so these two equal prefix values are not adjacent. Every adjacent pair $p_{i-1},p_i$ is still different, which matters because

$a_i=p_i\oplus p_{i-1}.$

If adjacent prefixes were equal, $a_i$ would be $0$, and array values must be positive. Here that never happens.

Recovering the array

Once the prefixes are fixed, define

$a_i=p_i\oplus p_{i-1}.$

Then the prefix XOR of the produced array is exactly the $p_i$ sequence we designed. So the duplicate-prefix argument directly proves correctness.

Bounds

The largest prefix value is at most $n$, and $n\le 4\cdot 10^5$. Therefore each

$a_i=p_i\oplus p_{i-1}$

is also small, definitely positive and at most $10^9$. No weird big constants, no randomization, no nonsense.

Proof of correctness

Let $p_0,p_1,\dots,p_n$ be the constructed prefix array, where $p_i=i$ except $p_r=l-1$.

First, $p_r=p_{l-1}$, so the XOR of subarray $[l,r]$ is

$p_r\oplus p_{l-1}=0.$

Now consider any subarray $[x,y]$ whose XOR is $0$. Then

$p_y\oplus p_{x-1}=0,$

so

$p_y=p_{x-1}.$

By construction, the only equal pair of prefix positions is $(l-1,r)$. Therefore

$x-1=l-1\quad\text{and}\quad y=r,$

which gives $x=l$ and $y=r$. Thus no other subarray has XOR $0$.

Finally, for every $i$, the output value is $a_i=p_i\oplus p_{i-1}$. Adjacent prefix values are never equal, so $a_i\ne 0$. Since all prefix values are small, all $a_i$ also satisfy $1\le a_i\le 10^9$.

Therefore the construction is valid. Clean, short, and honestly kind of rude to overthink.