Dynamic array challenge
When instancing generic collections, it' best to use the diamond operator to let the compiler figure out the generic arguments. This lets you define in a single place (type definition, not constructor) for easier future modification.

sequenceList[index] = new LinkedList&lt;>(); 
When using an Object, always declare its type as the highest class in its type hierarchy (up to interface if possible).


You're not using LinkedList's specific methods, so:

List&lt;Integer&gt;[] sequenceList = new List[n]; 



Javadoc!
Use proper Objects:


Make Solution a real instance created one in main (no real job should be performed in main). Then you can make lastAns and n fields of Solution because in the problem statement it is made to be a state of the Solution, so you don't need to pass those around every time
You could Wrap your Lists in a Sequence internal class of Solution. Internal means it will be able to access the lastAns field of the Solution instance etc.

It's a bit weird to lazily initialize the sequences. You don't win much, but you put an ugly if in your code. If there is no risk having a huge but sparse array, I would drop this.
Don't one-line ifs, and always use brackets
I'm usually against calling variables x, y etc. On this case it's defined in the problem so it's ok.




Updated code:

public class Solution {
    private int n;
    private int q;
    private int lastAns = 0;
    public Solution(int size, int numberOfQuestions) {
        n = size;
        q = numberOfQuestions;
    }
    public int getIndex(int x) {
        return (x ^ lastAns) % n;
    }
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        Solution solution = new Solution(in.nextInt(), in.nextInt());
        solution.solve(in);
    }
    public void solve(Scanner in) {
        List&lt;Integer&gt;[] sequenceList = new List[n];
        while(q-- != 0) {
            int queryType = in.nextInt();
            int x = in.nextInt();
            int y = in.nextInt();
            int index = getIndex(x);
            if(queryType == 1) {
                if(sequenceList[index] == null) {
                    sequenceList[index] = new LinkedList&lt;Integer&gt;();
                }
                sequenceList[index].add(y);
            } else if(queryType == 2) {
                lastAns = sequenceList[index].get(y % sequenceList[index].size());
                System.out.println(lastAns);
            }
        }
    }
}