O(1) time vector rotateLeft(int d, vector arr) { vector result; int step = d % arr.size();

vector<int> left(arr.begin() + step, arr.end());
vector<int> right(arr.begin(), arr.begin() + step);
result.insert(result.end(), left.begin(), left.end());
result.insert(result.end(), right.begin(), right.end());

return result;

}

public static List<Integer> rotateLeft(int d, List<Integer> arr) {
            //maybe it might be an unnecessarily long solution
    Queue<Integer> q = new LinkedList<>();
    List<Integer> list = new ArrayList<>();
    for(int arrItem : arr){
        q.add(arrItem);
    }
    for(int i = 0; i < d; i++){
        int temp = q.remove();
        q.add(temp);

    }
    for(int i = 0; i < arr.size(); i++ ){
        list.add(q.element());
        q.remove();
    }

    return list;
}

Time Complexity: O(n) <\br>

Space Complexity: O(n)

int* rotateLeft(int d, int arr_count, int* arr, int* result_count) {
    int* res_arr = (int* )malloc(arr_count*sizeof(int));
    *result_count = 0;
    for (int i = 0; i<arr_count; i++){
        res_arr[i] = arr[(i+d)%arr_count];
        (*result_count) ++;
    }
    return res_arr;
}

Here is my one line Python solution!

def rotateLeft(d, arr):
    return arr[d % len(arr):] + arr[:d % len(arr)]

Here is my c++ solution, you can have the video explanation here : https://youtu.be/lyUDSB4Sg7Y

vector<int> rotateLeft(int d, vector<int> arr) {
    int n = arr.size();
    vector<int> result(n);
    for(int i = 0; i < n; i++){
        int newIndex = (i - d + n) % n;
        result[newIndex] = arr[i];
    }
    return result;
}