Alright, here’s how I solved the Dynamic Array problem:

You start by creating n empty arrays and setting lastAnswer = 0. Then, you just go through each query one by one.

If the query type is 1, you find which array to use with this formula: idx = (x ^ lastAnswer) % n Then you just push y into that array.

If the query type is 2, you again use the same formula to find the array, but this time you read a value from it: lastAnswer = arr[idx][y % arr[idx].size()] and store or print that value.

Here’s what the code looks like (C++):

include

using namespace std;

int main() { int n, q; cin >> n >> q;

vector<vector<int>> arr(n);
int lastAnswer = 0;

while (q--) {
    int type, x, y;
    cin >> type >> x >> y;

    int idx = (x ^ lastAnswer) % n;

    if (type == 1) {
        arr[idx].push_back(y);
    } else {
        lastAnswer = arr[idx][y % arr[idx].size()];
        cout << lastAnswer << endl;
    }
}

return 0;

}

That’s it. It’s basically about keeping track of the arrays using that XOR trick and knowing when to append or fetch values. Once you get that logic down, it’s pretty straightforward.