I'm using the below approach using two-pointers and using HashSet and the complexity is O(n) in both time and space, I don't understand why it should follow that approach in the solution to bring the complexity to O(n^3)

import java.util.*;

class Solution {

  public static Set<List<Integer>> findSubarrays(int[] arr, int target) {
    Set<List<Integer>> result = new HashSet<>();

        int left = 0;
        int right = left + 1;

        while (right < arr.length) {

            int product = arr[right] * arr[left];
            if (product < target) {
                result.add(List.of(arr[left]));
                result.add(List.of(arr[right]));
                result.add(Arrays.asList(arr[left], arr[right]));
            } else {
                result.add(List.of(arr[right]));
            }

            left++;
            right++;
        }

        return result;
  }

  public static void main(String[] args) {
    System.out.println(
          Solution.findSubarrays(new int[] { 2, 5, 3, 10 }, 30));
    System.out.println(
          Solution.findSubarrays(new int[] { 8, 2, 6, 5 }, 50));
  }
}