public static List circularArrayRotation(List a, int k, List queries) { // Write your code here LinkedList list=new LinkedList<>(a); ArrayList ans=new ArrayList<>(); if(k>a.size()){ k=k%a.size(); for(int i=1;i<=k;i++){ int n=list.removeLast(); list.addFirst(n); System.out.println(list); } } else if(a.size()>k){ for(int i=0;i

public static List circularArrayRotation(List a, int k, List queries) { // Write your code here LinkedList list=new LinkedList<>(a); ArrayList ans=new ArrayList<>(); if(k>a.size()){ k=k%a.size(); for(int i=1;i<=k;i++){ int n=list.removeLast(); list.addFirst(n); System.out.println(list); } } else if(a.size()>k){ for(int i=0;i

Python Solution

def circularArrayRotation(a, k, queries):
    for i in range(k):
        a.insert(0, a.pop())
    return (a[i] for i in queries)

Here is my c++ solution, you can watch the explanation here : https://youtu.be/MaT-4dnozJI

Solution 1 O(n + m) :

vector<int> circularArrayRotation(vector<int> a, int k, vector<int> queries) {
    vector<int> newArray(a.size()), result;
    for(int i = 0; i < a.size(); i++){
        newArray[(i + k) % a.size()] = a[i];
    }
    for(int i = 0; i < queries.size(); i++) result.push_back(newArray[queries[i]]);
    return result;
}

Solution 2 O(m) :

vector<int> circularArrayRotation(vector<int> a, int k, vector<int> queries) {
    vector<int> result;
    for(int i = 0; i < queries.size(); i++) {
        int previousIndex = (queries[i] - k + a.size() * 100000 ) % a.size();
        result.push_back(a[previousIndex]);
    }
    return result;
}

public static List circularArrayRotation(List a, int k, List queries) {

    for(int i=0 ; i<k ; i++){
        rotateArray(a);
    }
    List<Integer> res = new ArrayList<>();
    for(int q : queries){
        res.add(a.get(q));
    }
    return res;
}

public static void rotateArray(List<Integer> a){

    a.add(0,a.get(a.size()-1));
    a.remove(a.size()-1);   

}