 $\Sigma_{\alpha} (r,\mathcal{A})=\Sigma^{\alpha} (r,\mathcal{A})=\Sigma (r,\mathcal{A})$, where $\Sigma (A)$ and $\Sigma (r,\mathcal{A})$ are the classical set of subset and subsequence sums, respectively. 


The set of subset sums $\Sigma_{\alpha} (A)$ and $\Sigma^{\alpha} (A)$ and the set of subsequence sums $\Sigma_{\alpha} (r, \mathcal{A})$ and $\Sigma^{\alpha} (r,\mathcal{A})$ may also be written as unions of sumsets: 


For a finite set $A$ of $k$ integers and for positive integers $h, r$, the \emph{$h$-fold sumset} $hA$ is the collection of all sums of $h$ not-necessarily-distinct elements of $A$, the \emph{$h$-fold restricted sumset} $h~\hat{}A$ is the collection of all sums of $h$ distinct elements of $A$, and the \emph{generalized sumset} $h^{(r)} A$ is the collection of all sums of $h$ elements of $A$ with at most $r$ repetitions for each element (see \cite{mistri14}). Then $\Sigma_{\alpha} (A) = \bigcup_{h=\alpha}^{k} h~\hat{}A$,  $\Sigma^{\alpha} (A) = \bigcup_{h=0}^{k-\alpha} h~\hat{}A$, $\Sigma_{\alpha} (r, \mathcal{A}) = \bigcup_{h=\alpha}^{rk} h^{(r)}A$, and  $\Sigma^{\alpha} (r, \mathcal{A}) = \bigcup_{h=0}^{rk-\alpha} h^{(r)}A$, where $\mathcal{A} = (A)_{r}$ and $0~\hat{}A = 0^{(r)}A =\{0\}$. 
	

An important problem in additive number theory is to find the best possible lower bound for $|\Sigma_{\alpha} (A)|$ and  $|\Sigma_{\alpha} (r,\mathcal{A})|$. Such problems are very useful in some other combinatorial problems such as the zero-sum problems (see \cite{devos09, erdos, grynkiewicz1, peng}). Nathanson \cite{nathu95} studied this problem on $\Sigma_{0} (A)$ in the group of integers. Later, Mistri and Pandey \cite{mistri15} and Jiang and Li\cite{jiang} generalized these results to $\Sigma_{0} (r,\mathcal{A})$, where $\mathcal{A}$ is a finite sequence of integers. Note that these sums may also be studied in any abelian group (for earlier works, in case $\alpha=0$, see \cite{eric12, eric13, devos07, erdos, freeze, griffiths, hamidoune98, hamidoune08, olson}). Recently, Balandraud \cite{eric17} studied the set $\Sigma_{\alpha} (A)$ in the finite prime field $\mathbb{F}_{p}$, where $p$ is a prime number.


A similar problem for sumsets also have been extensively studied in the past (see \cite{alon95, alon96, bajnok, cauchy, dav35, dav47, silva94, eliahou, erdos, nathu, nathu95} and the references therein). For some recent works along this line, one can also see \cite{bajnok15, bajnok16, jagannath, bhanja21, bhanja19, jagannath21, mistri14, mistri18, monopoli}. 



In this paper, using some combinatorial arguments, we obtain the best possible lower bounds for $|\Sigma_{\alpha} (A)|$ in Section \ref{S2} and $|\Sigma_{\alpha} (r,\mathcal{A})|$ in Section \ref{S3}, when the set $A$ (or sequence $\mathcal{A}$) is a finite set (or sequence) of integers. In \cite{bhanja20}, we have already settled this problem when the set $A$ (or sequence $\mathcal{A}$) contains nonnegative (or nonpositive) integers. So, in this paper, we consider the sets which may contain both positive and negative integers.


The following already known results have been used to prove the results in this paper.
\begin{theorem}\textup{\cite[Theorem 1.4]{nathu}\label{sumset-thm}}
Let $A$, $B$ be nonempty finite sets of integers. Let $A+B = \{a+b: a \in A, b \in B\}$. Then
	\[|A+B| \geq |A|+|B|-1.\]
This lower bound is optimal.
\end{theorem}

\begin{theorem}\textup{\cite[Theorem 1.3]{nathu}\label{multifold-sumset-thm}}
Let $A$ be a nonempty finite set of integers and let $h$ be a positive integer. Then
\[|hA| \geq h|A|-h+1.\]
This lower bound is optimal.
\end{theorem}

\begin{theorem}\textup{\cite[Theorem 4]{eric17} \label{balandraud-thm}}
Let $A \subset \mathbb{F}_{p}$ such that $A\cap (-A)=\emptyset$. For any integer $\alpha \in [0,|A|]$, we have
\begin{equation*}
|\Sigma_{\alpha} (A)| \geq \min \left\{p, \frac{|A|(|A|+1)}{2}-\frac{\alpha(\alpha+1)}{2}+1 \right\}.
\end{equation*}
This lower bound is optimal.
\end{theorem}



\section{Subset sums}\label{S2}

In Theorem \ref{emptyset-thm} and Corollary \ref{zero-cor}, we present the uniform lower bounds for $|\Sigma_{\alpha} (A)|$ under the conditions $A\cap (-A)=\emptyset$ and $A\cap (-A)=\{0\}$, respectively. These bounds are also seen to be the best.



\begin{theorem}\label{emptyset-thm}
Let $A$ be a set of $k$ integers such that $A\cap (-A)=\emptyset$. For any integer $\alpha \in [0,k]$, we have
\begin{equation}\label{emptyset-eqn}
|\Sigma_{\alpha} (A)| \geq \frac{k(k+1)}{2}-\frac{\alpha(\alpha+1)}{2}+1.
\end{equation}
This lower bound is optimal.
\end{theorem}


\begin{proof}
Let $p$ be a prime number such that $p>\max \left\{ 2\max^{\ast}(A),\frac{k(k+1)}{2}-\frac{\alpha(\alpha+1)}{2}+1 \right\}$, where $\max^{\ast}(A)=\max\{|a|: a\in A\}$. Now, the elements of $A$ can be thought of residue classes modulo $p$. Since $p>2\max^{\ast}(A)$, any two elements of $A$ are different modulo $p$. Furthermore $ A \cap (-A)=\emptyset $. Hence, applying Theorem \ref{balandraud-thm} to the set $A$, we get
\[|\Sigma_{\alpha} (A)| \geq \frac{k(k+1)}{2}-\frac{\alpha(\alpha+1)}{2}+1.\]


Next, to verify that the lower bound in (\ref{emptyset-eqn}) is best possible, let $A=[1,k]$. Then $A\cap (-A)=\emptyset$ and
\begin{align*}
\Sigma_{\alpha} (A) & \subset \left[1+2+\cdots+\alpha, k+(k-1)+\cdots+1\right]
= \left[\frac{\alpha(\alpha+1)}{2}, \frac{k(k+1)}{2}\right].
\end{align*}
Therefore
\[|\Sigma_{\alpha} (A)|\leq \frac{k(k+1)}{2}-\frac{\alpha(\alpha+1)}{2}+1.\]
This together with (\ref{emptyset-eqn}) gives
\[|\Sigma_{\alpha} (A)|
=\frac{k(k+1)}{2}-\frac{\alpha(\alpha+1)}{2}+1.\]
This proves (\ref{emptyset-eqn}) is optimal, and hence completes the proof of the theorem.
\end{proof}



\begin{corollary}\label{zero-cor}
Let $A$ be a set of $k$ integers such that $A\cap (-A)=\{0\}$. For any integer $\alpha\in[0,k]$, we have
\begin{equation}\label{zero-eqn}
|\Sigma_{\alpha} (A)| \geq \frac{k(k-1)}{2}-\frac{\alpha(\alpha-1)}{2}+1.
\end{equation}
