Any subsequence that includes both $$$i$$$ and $$$j$$$ should be counted, with only one condition that the number of items in the subsequence to the left of $$$i$$$ should be equal the number of items in the subsequence to the right of $$$j$$$ (to keep $$$i$$$, $$$j$$$ mirrored to each other).

Let's denote the number of available items to the left of $$$i$$$ as $$$L$$$, the number of available items to the right of $$$j$$$ as $$$R$$$. Let's also denote the number of items inbetween $$$i$$$ and $$$j$$$ as $$$M$$$.

It should be clear that $$$n = L + R + M + 2$$$.

Given the above condition only applies to $$$L$$$ and $$$R$$$, the subsequence can include any of the $$$M$$$ items inbetween $$$i$$$ and $$$j$$$, and it would still be valid. This gives us at least $$$W1 = 2^M$$$ options.

There are also options to choose $$$x$$$ items to the left and to the right of $$$i$$$ and $$$j$$$, which gives us $$$\displaystyle W2 = \sum_{X=1}^{X=n} C(L, X)*C(R, X)$$$ more options.

As the case with many other summations involving binomial coefficients, $$$W2$$$ can be reduced to $$$\displaystyle C(L+R, L)$$$.

Thus, $$$\displaystyle f(i, j) = W1 * W2 = 2^M * C(L+R, L)$$$.

where $$$L = i - 1$$$, $$$R = n - j$$$, $$$M = n - L - R - 2$$$.

Which can be calculated fast enough in O(1).

Let's try solving the problem in a reversed way. Let's assume that all pairs $$$(i, j)$$$ are bad initially, and then subtract $$$f(i, j)$$$ for pairs that are not gonna need to change (i.e. $$$arr_i = arr_j$$$).

More specifically the answer should be

$$$\displaystyle (n - 1) * 2^{n - 2} - \sum_{i=1}^{i=n} \sum_{j=i+1}^{j=n} f(i, j)*(arr_i == arr_j)$$$.

calculating $$$f(i, j)$$$ only for pairs where $$$arr_i = arr_j$$$.

The name of the problem gives us some hint that sqrt analysis must be involved.

We can observe that for values that occurr only once in the array, we don't really need to compute anything. (Remember we are now only concerned with pair $$$(i, j)$$$ where $$$arr_i = arr_j$$$).

Similarly, for any index $$$i$$$, where $$$arr_i$$$ occurr a few number of times $$$< sqrt(n)$$$, we need to iterate over a few number of pairs $$$(i, j)$$$ which can be done easily if we keep the occurrences of each value separately.

However, for indices with more repeating values, we need another way.

We are now concerned with values that occurr more than sqrt(n) times in the array. Obviously, there can't be more than sqrt(n) of those values.

Let's have another look at the formula for $$$f(i, j)$$$ that we have compiled above.

$$$\displaystyle f(i, j) = 2^M * C(L+R, L)$$$.

where $$$L = i - 1$$$, $$$R = n - j$$$, $$$M = n - L - R - 2$$$.

Let's write $$$f(i, j)$$$ purely in terms of just i, and j.

$$$\displaystyle f(i, j) = 2^{j - i - 1} * C(n + i - j - 1, i - 1)$$$.

Let's also expand the binomial coefficient

$$$\displaystyle f(i, j) = 2^{j - i - 1} * \frac{factorial(n + i - j - 1)}{factorial(i - 1) * factorial(n - j)}$$$.

Let's organize things a little bit.

$$$\displaystyle f(i, j) = 2^{i - 1} * fact^{-1}(i - 1) * 2^{j} * fact^{-1}(n - j) * fact(n + i - j - 1)$$$.

Now we can notice the terms in the formula are either dependent on $$$i$$$ only, or $$$j$$$ only, or $$$i - j$$$. This is a big hint that we can evaluate the summation of $$$f(i, j)$$$ using multiplying polynomials.

Let's imagine we have two polynomials P1 including terms that depend on i only, and P2 that includes terms that depend on j only.

$$$\displaystyle P1(x) = \sum_{i=1}^{i=n} 2^{i-1} * factorial^{-1}(i-1) * x^{i}$$$

$$$\displaystyle P2(x) = \sum_{j=1}^{j=n} 2^{j} * factorial^{-1}(n - j) * x^{-j}$$$

We can represent the multiplication of them as

$$$\displaystyle P(x) = P1(x) * P2(x)$$$

$$$\displaystyle P(x) = \sum_{i=1}^{i=n} \sum_{j=1}^{j=n} (2^{i-1} * fact^{-1}(i-1) * 2^{j} * fact^{-1}(n - j)) * x^{i-j}$$$

The resulting polynomial $$$P(x)$$$, is kind of the same as summation of $$$f(i, j)$$$ over all possible $$$(i, j)$$$, except that for each term we need to multiply it by $$$fact(n + i - j - 1)$$$ (notice that $$$i - j$$$ is the exponent of $$$x$$$ in the resulting polynomial).

Luckily, we have got number theoretic transform to do the magic for us and multiply polynomials in $$$O(nlog(n))$$$.

We can construct polynomials $$$P1$$$ and $$$P2$$$ for each value $$$V$$$, that occurrs more than $$$sqrt(n)$$$ times in the array.

For example $$$P1(x)$$$ should look like

$$$\displaystyle P1(x) = \sum_{i=1}^{i=n} 2^{i-1} * factorial^{-1}(i-1) * (arr_i == V) * x^{i}$$$.

Doing the polynomial multiplication should get us summation of $$$f(i, j)$$$ for all $$$(i, j)$$$ where $$$arr_i = arr_j = V$$$.

This adds up to $$$O(nsqrt(n)logn)$$$.