Sure, observe that parity of initial index and final index should remain same as i <-> i+2. But there can be duplicates, so for all values the number of odd/even indices in initial array must be equal to number of odd/even indices in the final sorted array.

We have a solution which we believe is correct but we were getting WA.

Apply meet in the middle. Divide the people into two groups left_part and right_part, calculate dp[mask] for all subsets in left_part and right_part separately in $$$O(3^{10})$$$. Now consider a pair of subsets from left_part and right_part i.e {m1, m2}, Calculate dp[m1 | m2] by iterating over all subsets of m1 i.e. dp[m1 | m2] = min(dp[m1 | m2], dp[m1 ^ b]+dp[m2 | b], where b is the subset of m1.

Overall complexity: $$$(O(T*(3^{10} * 2^{10})) = O(T*(6^{10}))$$$

The total number of operations are roughly $$$6e8$$$ and 4 sec is a little strict but it should pass.

okwedook test cases for E are weak. Many $$$O(N^2)$$$ have been passed. So please consider rejudging all submissions.

A much simpler approach : Let $$$dp(N,K)$$$ be the number of multi sets with sum $$$k$$$ and size $$$N$$$ . You can recursively formulate $$$ dp(N,K) = dp(N,2*K) + dp(N-1,K-1) $$$.The first term $$$dp(N,2*K)$$$ is the case when you don't take 1 and so the sequence starts from $$$\frac{1}{2}$$$ so it is equivalent of taking sum to $$$ 2*k$$$ and let the sequence to start from 1,The other term ,$$$ dp(N-1,K-1) $$$ is the case when you have taken 1 in the multiset,so your sum reduces to $$$k-1$$$ and length to $$$N-1$$$ and you can still start from 1.

How to solve E ?

how to find the coefficient of x^N in that product.I also arrived at this point.

Using min-cost max-flow, take n nodes to the left side ( represents books), m nodes to the right side ( represents algorithms), connect a source to all the n nodes in the left with capacity = total understanding it provides for all algorithms, and cost = cost of the ith book. connect sink with all the m nodes to the right with capacity x and cost 0, also for all the n nodes in the left, connect to each of the m nodes with a capacity equal to a[i][j] and cost 0. Now perform min cost maxflow algorithm and for the maximum flow calculate the minimum cost, if the maximum flow is not equal to m*x, then the answer is -1.(maximum flow cannot be greater than m*x since the capacity of all the edges to sink is x).

For F: let's say some blocks have been placed giving total sum ( by total sum, I mean number of '(' minus number of ')' ), equal to prev. Now for all the blocks left, which have a minimum value of prefix sum greater than or equal to prev , select a block having a maximum total sum.This can be done by using binary search and seg-tree.

Is it wrong ? since I am getting WA verdict , not sure if the implementation is wrong or the logic is wrong.

any hints for C ?