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It's not clear how to formally define worst-case (or near-worst-case) instances, but here is something you could try. The idea is to combine the following two tidbits: We know some hard instances f …

Your problem is not really well-defined, since it's not clear what is the parameter $n$. That said, under any reasonable interpretation, your problem is still NP-complete, as can be shown by padding. …

Suppose that you could solve this in $T(n)$. Given a list of positive integers $a_1,\ldots,a_n$ and a target $T$, consider the two instances $a_1,\ldots,a_n,-T$ and $a_1,\ldots,a_n,-T+1$. Denoting by …

This is A053632. The prefixes of this sequence convergence to the more well-known A000009, the number of partitions into distinct (or odd) parts. You shouldn't expect a clean formula, though it's of c …

Suppose that there are $n$ groups, that group $i$ has the $m_i$ elements $x_{ij}$. It is a nice exercise to show that the sum of weights of all valid vectors is $$ \prod_{i=1}^n \sum_{j=1}^{m_i} x_{ij …

This answer assumes that a Yes instance of your problem with elements $x_0,\ldots,x_{n-1}$ and targets $[X,X+1]$ is one in which there is a subset summing to either $X$ or $X+1$. Given an instance of …

Suppose that the value at location $i$ is $j$. According to the continuity of $k$-sums (mentioned in an answer to the linked question), the range of possible sums of values at locations $1,\ldots,i-1$ …

Your problem is known as optimal alphabetic binary tree (or various similar names). This is an ordered version of Huffman coding, in which the two numbers being added don't have to be adjacent. The pr …

The idea is that two sorted lists of length $n$ can be merged into one sorted list of length $2n$ in time $O(n)$. This is a standard procedure used in the mergesort algorithm. Given a list of integers …

If you have $n$ non-negative integers $x_1,\ldots,x_n$ in the range $0,\ldots,T$, then their susbets can sum up only to integers in the range $0,\ldots,nT$. This allows us to solve SUBSET-SUM efficien …

Given an instance of the second problem, we can easily reduce it to an instance of the (decision version of) the first problem: simply take $W = k$. There is a subset of sum at least $k$ and at most $ …

The observation is that the denominator of the reduced fraction $1/p_{i_1} + \cdots + 1/p_{i_m}$ is $p_{i_1} \cdots p_{i_m}$. To see this, it suffices to notice that the (unreduced) numerator isn't di …

Showing that this your problem is in NP is easy. To show that your problem is NP-hard, reduce from PARTITION. The reduction simply chooses a large enough modulus $k$. Details left to you.

For every $i \in \{0,\ldots,n\}$ (where $n$ is the length of the array) and for every $a,b,c \in \{0,\ldots,15\}$, we determine whether it is possible to partition the first $i$ elements of the array …