2261. K Divisible Elements Subarrays

Description

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

k-divisible-elements-subarrays.py

class Solution:
    def countDistinct(self, nums: List[int], k: int, p: int) -> int:
        BASE, MOD = 401, 8390190857
        n = len(nums)
        h = [1]
        seen = set()

        for _ in range(n):
            h.append((h[-1] * BASE) % MOD)

        for sz in range(1, n + 1):
            curr = 0
            count = 0

            for i in range(sz):
                curr = (curr * BASE) + nums[i]
                curr %= MOD

                if nums[i] % p == 0:
                    count += 1

            if count <= k:
                seen.add(curr)

            for i in range(sz, n):
                if nums[i] % p == 0:
                    count += 1

                if nums[i - sz] % p == 0:
                    count -= 1

                curr = (curr * BASE) + nums[i]
                curr %= MOD
                curr -= nums[i - sz] * h[sz]
                curr %= MOD

                if count <= k:
                    seen.add(curr)

        return len(seen)