Let $S=\{1,2,3,4\}, a_{1}, a_{2}, \cdots$, be any permutation ending with 1, i.e., for any permutation $\left(b_{1}, b_{2}, b_{3}, b_{4}\right)$ of the four numbers in $S$ that does not end with 1, $\left(b_{4} \neq 1\right)$, there exist $i_{1}, i_{2}, i_{3}, i_{4}$, such that $1 \leqslant i_{1}<i_{2}<i_{3}<i_{4} \leqslant k$, and $\left(a_{i_{1}}, a_{i_{2}}, a_{i_{3}}, a_{i_{4}}\right)=\left(b_{1}, b_{2}, b_{3}, b_{4}\right)$. Find the minimum value of the number of terms $k$ in such a sequence.
II. Hint: First consider the minimum value of the number of terms $k$ in any sequence containing a permutation of $S$.
1. For $S=\{1,2,3\}$, it can be proven that a sequence with only 6 terms cannot contain any permutation of $\{1,2,3\}$.
2. For $S=\{1,2,3,4\}$, prove that a sequence with only 11 terms cannot contain any permutation of $\{1,2,3,4\}$.

Direct verification shows that the sequence 1, 2, 3, 4, 1, 2, 3, 1, 4, 2, 3, 1 contains any permutation of $S$, thus 12 is the minimum value.

Returning to the problem itself. If there is a sequence with fewer than 11 terms that contains any permutation of $S$ not ending in 1, then adding one more positive term to this sequence would result in a sequence containing all permutations of $S$, but with fewer than 12 terms, which is a contradiction. Additionally, 1, 2, 3, 4, 1, 2, 3, 1, 4, 2, 3 contains any permutation of $S$ not ending in 1. Therefore, the minimum value of $k$ is 11.