How about putting all positive numbers in one group and negative in second group

Let $$$S$$$ denotes sum of element of array $$$a$$$.

Claim: Answer is |S|.

Proof: Let sum of all positive elements is $$$S_{pos}$$$ and sum of all negative elements $$$S_{neg}$$$. Put all positive numbers in first group and negative numbers in second group. We get $$$||S_{pos}| - |S_{neg}|| = |S|$$$.

Let's prove that we can not do better than that. Let $$$S_1$$$ denotes sum of elements of first group and $$$S_2$$$ denotes sum of elements of second group. We have $$$|S_1| - |S_2| \leq |S_1 + S_2| = |S|$$$. Hence $$$|S|$$$ is the upperbound for the answer.

Instead of subsequences solve for substrings. That is there should not be any substring $$$\texttt{BAN}$$$ after performing operations.

In one operation you can destroy atmost $$$2$$$ substrings. Find minimum operations to destroy $$$n$$$ substrings.

Congrats, you have solved for subsequences also!

No subsequences of string BAN would also mean no substrings of BAN in original string. Let minimum number of operations to have no substrings of BAN be $$$x$$$, it would be also be the lower bound for having no subsequences of string BAN.

Claim: $$$x = \left \lceil\frac{n}{2}\right \rceil$$$.

Proof: Swap $$$i$$$-th B from start with $$$i$$$-th N from end for $$$1 \leq i \leq \left \lceil\frac{n}{2}\right \rceil$$$. We can see that, no substrings of BAN exists after performing $$$ \left \lceil\frac{n}{2}\right \rceil$$$ operations. Since we can only destroy atmost $$$2$$$ substrings in one operations, $$$\left \lceil\frac{n}{2}\right \rceil$$$ is minimum possible.

Now if you see clearly, after performing above operations, there does not exist any subsequence of string BAN in original string. Hence $$$\left \lceil\frac{n}{2}\right \rceil$$$ is also the answer for the original problem.

Divide problem into two different cases. When $$$a_1 \gt \min(a)$$$ and when $$$a_1 = \min(a)$$$.

You do not need more hints to solve the problem.

Case 1: $$$a_1 \gt \min(a)$$$

$$$\texttt{Alice}$$$ can force the $$$\texttt{Bob}$$$ to always decrease the minimum element by always choosing minimum element of $$$a$$$ in her turn. Where as $$$\texttt{Bob}$$$ can not do much, all other elements he would swap with would be greater than or equal to $$$\min(a)$$$. Even if there exists multiple minimums in $$$a$$$, In first move $$$\texttt{Alice}$$$ would decrease from $$$a_1$$$, hence in this case $$$\texttt{Alice}$$$ would always win.

Case 2: $$$a_1 = \min(a)$$$

In this case optimal startegy for $$$\texttt{Bob}$$$ would be to always chhose minimum element of the array, which is $$$a_1$$$. $$$\texttt{Alice}$$$ would always be swapping the element greater than $$$a_1$$$ in her turn, hence in the case $$$\texttt{Bob}$$$ would always win