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

Note that, after sorting, if the triple $$$(1,2,n)$$$ is valid (i.e. $$$a_1+a_2>a_n$$$), 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.

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.

The strategy for each one is simple.

Claim. Everyone will keep his potato in his hands (if he has) until the last round.

Proof. For everyone, the person previous and next to him both belong to the opposite team. If he passes his potato to the next person before the last round, the initiative will be passed to the two persons adjacent to him. This would't be better for him than keeping the potato until the last round.

And, in the last round, everyone will pass his potato to the next person (if he can). Thus, according to the claim you can solve the problem by simulating the last round. Time complexity is $$$\mathcal{O}(n)$$$.

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.

The array $$$a$$$ is constructed with the operations. How can we use this property? (In G1 we do not need it)

Suppose we can't undo operations. If we want to make all $$$a_i=x$$$, what is the minimum value of $$$x$$$? Use the property mentioned in Hint 1.

(For the subproblem in Hint 2), try to find an algorithm related to the positions of $$$\texttt{L/R}$$$-s directly.

(For the subproblem in Hint 2), the conclusion is that, the minimum $$$x$$$ equals $$$( \text{# of }\texttt{L}\text{-s} + \text{# of }\texttt{R}\text{-s} - \text{# of adjacent }\texttt{LR}\text{-s} )$$$. (Think why)