K Divisible Elements Subarrays - LeetCode solutions

Problem

Given an integer array nums and two integers k and p, return **the number of *distinct subarrays* which have at most** k elements divisible by p.

Two arrays nums1 and nums2 are said to be distinct if:

They are of different lengths, or
There exists at least one index i where nums1[i] != nums2[i].

A subarray is defined as a non-empty contiguous sequence of elements in an array.

Example 1:

Input: nums = [2,3,3,2,2], k = 2, p = 2
Output: 11
Explanation:
The elements at indices 0, 3, and 4 are divisible by p = 2.
The 11 distinct subarrays which have at most k = 2 elements divisible by 2 are:
[2], [2,3], [2,3,3], [2,3,3,2], [3], [3,3], [3,3,2], [3,3,2,2], [3,2], [3,2,2], and [2,2].
Note that the subarrays [2] and [3] occur more than once in nums, but they should each be counted only once.
The subarray [2,3,3,2,2] should not be counted because it has 3 elements that are divisible by 2.

Example 2:

Input: nums = [1,2,3,4], k = 4, p = 1
Output: 10
Explanation:
All element of nums are divisible by p = 1.
Also, every subarray of nums will have at most 4 elements that are divisible by 1.
Since all subarrays are distinct, the total number of subarrays satisfying all the constraints is 10.

Constraints:

1 <= nums.length <= 200
1 <= nums[i], p <= 200
1 <= k <= nums.length

Follow up:

Can you solve this problem in O(n2) time complexity?

Solution (Java)

class Solution {
    public int countDistinct(int[] nums, int k, int p) {
        HashSet<Long> numSubarray = new HashSet<>();
        for (int i = 0; i < nums.length; i++) {
            int countDiv = 0;
            long hashCode = 1;
            for (int j = i; j < nums.length; j++) {
                hashCode = 199L * hashCode + nums[j];
                if (nums[j] % p == 0) {
                    countDiv++;
                }
                if (countDiv <= k) {
                    numSubarray.add(hashCode);
                } else {
                    break;
                }
            }
        }
        return numSubarray.size();
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).