<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>

class Solution:
    def countDistinct(self, nums: List[int], k: int, p: int) -> int:
        n = len(nums)
        res = 0
        seen = set()

        def count(sub):
            count = 0
            for v in sub:
                if v % p == 0:
                    count += 1
            return count <= k

        for i in range(n):
            for j in range(i+1, n+1):
                sub = nums[i:j]
                # print(sub)
                if tuple(sub) not in seen and count(sub):
                    seen.add(tuple(sub))
                    res += 1
        return res
1000