## DEV Community

Abhishek Chaudhary

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# K Divisible Elements Subarrays

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`

SOLUTION:

``````class Solution:
def countDistinct(self, nums: List[int], k: int, p: int) -> int:
subs = set()
left, right, divs = 0, 0, 0
while right < len(nums):
if nums[right] % p == 0:
divs += 1
while divs > k and left < right:
if nums[left] % p == 0:
divs -= 1
left += 1
for l in range(left, right + 1):