Subtask 1

Due to the cyclic nature of the algorithm, it suffices to simulate the algorithm until we encounter the 0 at the front of the string, at which point it can be removed and the remaining characters outputted in order.

Time Complexity: $\mathcal{O}(N)$

Subtask 2

If $M=2$, note that we care only about the 0 that is removed last. This can be done by comparing $a_i$ values, breaking ties by comparing the indices at which the 0's appear in the input string. We can then remove the 0 that is removed first without any consequence.

From this point onwards, we can perform the same sort of simulation as outlined in Subtask 1.

Time Complexity: $\mathcal{O}(N)$

Subtask 3

We still only care about the last 0 that is removed. We can find the index of this 0 by sorting the 0's based on their $a_i$ value and breaking ties by selecting the 0 that is further back. Let's call the index of this last 0 $i$. We now separate the number into two strings $A$ and $B$, where $A$ contains all non-0 characters in the input order before $i$, and $B$ contains all the non-0 characters in the input order after $i$. When we reach $i$ for the last time, the string can be shown to be $B+A$, where $+$ represents concatenation.

A slightly faster solution involves maintaining $i$ by iterating through the string instead of sorting.

Time Complexity: $\mathcal{O}(N\log N)$ or $\mathcal{O}(N)$