Since $$$1 ≤ a_i ≤ 26$$$, we can assign each value from $$$1$$$ to $$$26$$$ to a letter a-z.

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Using $$$7$$$ characters, we can encode $$$26^7 \approx 8 \cdot 10^9$$$ possibilities. For each number in the array, encode into a $$$7$$$-digit base-$$$26$$$ number. It is important to make sure each number takes up the same number of characters.

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Let's try to solve the problem player B has to solve.

Let the range of a subarray be $$$max(p_l, p_{l+1}, ..., p_r) - min(p_l, p_{l+1}, ..., p_r)$$$. A subarray has range $$$n-1$$$ iff it contains both $$$1$$$ and $$$n$$$. We can binary search on the shortest prefix with range $$$n-1$$$, this will find the latter position of $$$1$$$ or $$$n$$$. Also binary search on the shortest suffix with range $$$n-1$$$, this will find the former position of $$$1$$$ or $$$n$$$. Now we've found the position of $$$1$$$ and $$$n$$$, but we don't know which is which.

This is where player A comes in. If $$$1$$$ appears before $$$n$$$, he will send $$$0$$$; otherwise, he will send $$$1$$$. Now player B can figure out which is which.

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Let's phrase the problem slightly differently.

• Player A has a binary string of length $$$15$$$, and he will send a binary string of length $$$20$$$.
• On the way, none or one of the bits get flipped.
• Given the altered string, $$$B$$$ must find the original binary string of length $$$15$$$.

The first $$$15$$$ bits ($$$s_{1..15}$$$) will be allocated for the original string, and the remaining $$$5$$$ bits will be used to reconstruct where the flip happened.

The next $$$4$$$ ($$$s_{15..19}$$$) will house the bitwise XOR of the all the positions in the original string that are 1. The last bit ($$$s_{20}$$$) will be the number of bits in previous $$$4$$$ bits mod $$$2$$$.

Now there are $$$2$$$ cases:

Case 1: A flip happens in $$$s_{16..20}$$$

We can detect this case by checking if the last bit matches what it's supposed to be. If it doesn't, a flip has happened somewhere in the last $$$5$$$ bits. In this case, the original $$$15$$$ are completely untouched.

Case 2: A flip doesn't happen or happens in $$$s_{1..15}$$$

Here, we calculate the bitwise XOR of all the 1 in $$$s_{1..15}$$$. Let $$$x$$$ be this value and $$$y$$$ be the number stored in $$$s_{16..19}$$$. The cool thing is: $$$x \bigoplus y$$$ will be the position of the flipped bit, or $$$0$$$ if no flip occurred. This is because if a flip occured at position $$$k$$$, $$$x$$$ becomes $$$x \bigoplus k$$$.

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