K Divisible Elements Subarrays
<p>Given an integer array <code>nums</code> and two integers <code>k</code> and <code>p</code>, return <em>the number of <strong>distinct subarrays,</strong> which have <strong>at most</strong></em> <code>k</code> <em>elements </em>that are <em>divisible by</em> <code>p</code>.</p>

<p>Two arrays <code>nums1</code> and <code>nums2</code> are said to be <strong>distinct</strong> if:</p>

<ul>
	<li>They are of <strong>different</strong> lengths, or</li>
	<li>There exists <strong>at least</strong> one index <code>i</code> where <code>nums1[i] != nums2[i]</code>.</li>
</ul>

<p>A <strong>subarray</strong> is defined as a <strong>non-empty</strong> contiguous sequence of elements in an array.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [<u><strong>2</strong></u>,3,3,<u><strong>2</strong></u>,<u><strong>2</strong></u>], k = 2, p = 2
<strong>Output:</strong> 11
<strong>Explanation:</strong>
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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3,4], k = 4, p = 1
<strong>Output:</strong> 10
<strong>Explanation:</strong>
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.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 200</code></li>
	<li><code>1 &lt;= nums[i], p &lt;= 200</code></li>
	<li><code>1 &lt;= k &lt;= nums.length</code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong></p>

<p>Can you solve this problem in O(n<sup>2</sup>) time complexity?</p>
Trie; Array; Hash Table; Enumeration; Hash Function; Rolling Hash
function countDistinct(nums: number[], k: number, p: number): number {
    const s = new Set<bigint>();
    const [base1, base2] = [131, 13331];
    const [mod1, mod2] = [1000000007, 1000000009];
    for (let i = 0; i < nums.length; i++) {
        let [h1, h2, cnt] = [0, 0, 0];
        for (let j = i; j < nums.length; j++) {
            if (nums[j] % p === 0) {
                cnt++;
                if (cnt > k) {
                    break;
                }
            }
            h1 = (h1 * base1 + nums[j]) % mod1;
            h2 = (h2 * base2 + nums[j]) % mod2;
            s.add((BigInt(h1) << 32n) | BigInt(h2));
        }
    }
    return s.size;
}
