||A|B|C|D|E|F|G|H|I| |:--:|:--:|:--:|:--:|:--:|:--:|:--:|:--:|:--:| |Error_Yuan|800|1200|1600|2100|2300|2600|3000|3400|3500| |sszcdjr|800|1100|1600|2000|2300|2600|2900|3300|3500|

Over all the triples, which is the most important one?

Note that, after sorting, if the triple $$$(1,2,n)$$$ is valid (i.e. $$$a_1+a_2 \gt a_n$$$), then the whole array is valid. Thus, we are going to maximize the value of $$$a_1+a_2-a_n$$$ and check if it is greater than $$$0$$$. To do this, the optimal solution is to choose $$$a=[m-n+1,m-n+2,\ldots,m]$$$, and if $$$(m-n+1)+(m-n+2)\le m$$$, answer is NO.

for _ in range(int(input())):
    n, m = map(int, input().split())
    if m - n + 1 + m - n + 2 > m:
        print("YES")
        print(*range(m - n + 1, m + 1))
    else:
        print("NO")

What condition should $$$s$$$ satisfy to make sure that there is at least one valid index?

Think about the count of $$$\texttt{0}$$$-s and the count of $$$\texttt{1}$$$-s.

Each time we do an operation, if $$$s$$$ consists only of $$$\texttt{0}$$$ or $$$\texttt{1}$$$, we surely cannot find any valid indices. Otherwise, we can always perform the operation successfully. In the $$$i$$$-th operation, if $$$t_i=\texttt{0}$$$, we actually decrease the number of $$$\texttt{1}$$$-s by $$$1$$$, and vice versa. Thus, we only need to maintain the number of $$$\texttt{0}$$$-s and $$$\texttt{1}$$$-s in $$$s$$$. If any of them falls to $$$0$$$ before the last operation, the answer is NO, otherwise answer is YES.

Consider binary search.

In the check function, try to do something backwards (from the end of the array to the start).

Do not think too much. Go straight for a dp solution.

Does your time complexity of dp contain $$$n$$$? Actually, both $$$n$$$ and $$$m$$$ are useless.

Use some data structures to optimize your solution. In fact, fenwick trees are powerful.