k-divisible-elements-subarrays

K Divisible Elements Subarrays

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countDistinct

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

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

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

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

&nbsp;
Example 1:

<pre>
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.
</pre>

Example 2:

<pre>
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.
</pre>

&nbsp;
Constraints:

<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>

&nbsp;
Follow up:

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

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