Class Solution
java.lang.Object
g2201_2300.s2261_k_divisible_elements_subarrays.Solution
2261 - K Divisible Elements Subarrays\.
Medium
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`
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Solution
public Solution()
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Method Details
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countDistinct
public int countDistinct(int[] nums, int k, int p)
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