Assuming that the $a_i$ are non-negative, you can solve your problem in linear time using a "sliding window" algorithm. I will also assume $T>0$ (otherwise the problem is trivial).

Let $i,j$ be two pointers initialized to $1$, and denote by $\sigma(i,j)$ the sum $a_i + a_{i+1} + \dots + a_{j}$. As long as $i < n$ do the following:

• If $\sigma(i,j) <T$ and $j<n$, increment $j$ by $1$.
• Otherwise, increment $i$ by $1$.

Consider now all pairs of values $(i,j)$ attained by $i$ and $j$ during the previous procedure and, among those that satisfy $\sigma(i,j) \ge T$, return one that minimizes $j-i+1$.

Notice that the above algorithm can be implemented in time $O(n)$ since there are at most $2n-1$ considered pairs $(i,j)$ (each index can be incremented at most $n-1$ times) and you can update the value of $\sigma(i,j)$ in constant time whenever $i$ or $j$ changes (subtract $a_{i}$ just before incrementing $i$, and add $a_j$ immediately after incrementing $j$).

We only need to show that some pair considered $(i,j)$ corresponds to the endpoints $i^*, j^*$ of an optimal subarray $a_{i^*}, a_{i^*+1}, \dots, a_{j^*}$. At some point during the execution of the algorithm we must either have $i = i^*$ or $j=j^*$. Consider the first iteration when this happens.

• If $i=i^*$ then $j \le j^*$. Moreover, for all $j' \in \{j, j+1, \dots, j^*-1\}$ (this set might be empty), we have $\sigma(i, j') < T$ and hence $j$ gets incremented until $j=j^*$ while $i$ remains unchanged. Therefore $(i^*, j^*)$ is considered.

• If $j=j^*$ then $i \le i^*$. Moreover, for all $i' \in \{i, i+1,\dots, i^*-1\}$ (this set might be empty), we have $\sigma(i', j) \ge T$ and hence $i$ gets incremented until $i=i^*$ while $j$ remains unchanged. Therefore $(i^*, j^*)$ is considered.

You can solve your problem in linear time using a "sliding window" algorithm.

Let $i,j$ be two pointers initialized to $1$, and denote by $\sigma(i,j)$ the sum $a_i + a_{i+1} + \dots + a_{j}$. As long as $i < n$ do the following:

• If $\sigma(i,j) <T$ and $j<n$, increment $j$ by $1$.
• Otherwise, increment $i$ by $1$.

Consider now all pairs of values $(i,j)$ attained by $i$ and $j$ during the previous procedure and, among those that satisfy $\sigma(i,j) \ge T$, return one that minimizes $j-i+1$.

Notice that the above algorithm can be implemented in time $O(n)$ since there are at most $2n-1$ considered pairs $(i,j)$ (each index can be incremented at most $n-1$ times) and you can update the value of $\sigma(i,j)$ in constant time whenever $i$ or $j$ changes (subtract $a_{i}$ just before incrementing $i$, and add $a_j$ immediately after incrementing $j$).

We only need to show that some pair considered $(i,j)$ corresponds to the endpoints $i^*, j^*$ of an optimal subarray $a_{i^*}, a_{i^*+1}, \dots, a_{j^*}$. At some point during the execution of the algorithm we must either have $i = i^*$ or $j=j^*$. Consider the first iteration when this happens.

• If $i=i^*$ then $j \le j^*$. Moreover, for all $j' \in \{j, j+1, \dots, j^*-1\}$ (this set might be empty), we have $\sigma(i, j') < T$ and hence $j$ gets incremented until $j=j^*$ while $i$ remains unchanged. Therefore $(i^*, j^*)$ is considered.

• If $j=j^*$ then $i \le i^*$. Moreover, for all $i' \in \{i, i+1,\dots, i^*-1\}$ (this set might be empty), we have $\sigma(i', j) \ge T$ and hence $i$ gets incremented until $i=i^*$ while $j$ remains unchanged. Therefore $(i^*, j^*)$ is considered.