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Formatted question description: https://leetcode.ca/all/2261.html
2261. K Divisible Elements Subarrays
 Difficulty: Medium.
 Related Topics: Array, Hash Table, Trie, Rolling Hash, Hash Function, Enumeration.
 Similar Questions: Subarrays with K Different Integers, Count Number of Nice Subarrays, Subarray With Elements Greater Than Varying Threshold.
Problem
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
wherenums1[i] != nums2[i]
.
A subarray is defined as a nonempty 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 (Java, C++, Python)

class Solution { public int countDistinct(int[] nums, int k, int p) { HashSet<Long> numSubarray = new HashSet<>(); for (int i = 0; i < nums.length; i++) { int countDiv = 0; long hashCode = 1; for (int j = i; j < nums.length; j++) { hashCode = 199L * hashCode + nums[j]; if (nums[j] % p == 0) { countDiv++; } if (countDiv <= k) { numSubarray.add(hashCode); } else { break; } } } return numSubarray.size(); } }

Todo

class Solution: def countDistinct(self, nums: List[int], k: int, p: int) > int: n = len(nums) s = set() for i in range(n): cnt = 0 t = "" for j in range(i, n): if nums[j] % p == 0: cnt += 1 if cnt <= k: t += str(nums[j]) + "," s.add(t) else: break return len(s) ############ # 2261. K Divisible Elements Subarrays # https://leetcode.com/problems/kdivisibleelementssubarrays/ class Solution: def countDistinct(self, nums: List[int], k: int, p: int) > int: s = set() i = 0 count = 0 n = len(nums) def go(arr, start, end): sub_list = [] for i in range(start, end + 1): for j in range(i + 1, end + 1): sub = arr[i:j] sub_list.append(sub) return sub_list for j, x in enumerate(nums): if x % p == 0: count += 1 while count > k: if nums[i] % p == 0: count = 1 i += 1 if j == n  1 or count == k: for sub in go(nums, i, j + 1): s.add(tuple(sub)) return len(s)
Explain:
nope.
Complexity:
 Time complexity : O(n).
 Space complexity : O(n).