Class Solution
java.lang.Object
g2201_2300.s2261_k_divisible_elements_subarrays.Solution
2261 - K Divisible Elements Subarrays.<p>Medium</p>
<p>Given an integer array <code>nums</code> and two integers <code>k</code> and <code>p</code>, return <em>the number of <strong>distinct subarrays</strong> which have <strong>at most</strong></em> <code>k</code> <em>elements divisible by</em> <code>p</code>.</p>
<p>Two arrays <code>nums1</code> and <code>nums2</code> are said to be <strong>distinct</strong> if:</p>
<ul>
<li>They are of <strong>different</strong> lengths, or</li>
<li>There exists <strong>at least</strong> one index <code>i</code> where <code>nums1[i] != nums2[i]</code>.</li>
</ul>
<p>A <strong>subarray</strong> is defined as a <strong>non-empty</strong> contiguous sequence of elements in an array.</p>
<p><strong>Example 1:</strong></p>
<p><strong>Input:</strong> nums = [<strong>2</strong> ,3,3, <strong>2</strong> , <strong>2</strong> ], k = 2, p = 2</p>
<p><strong>Output:</strong> 11</p>
<p><strong>Explanation:</strong></p>
<p>The elements at indices 0, 3, and 4 are divisible by p = 2.</p>
<p>The 11 distinct subarrays which have at most k = 2 elements divisible by 2 are:</p>
<p>[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].</p>
<p>Note that the subarrays [2] and [3] occur more than once in nums, but they should each be counted only once.</p>
<p>The subarray [2,3,3,2,2] should not be counted because it has 3 elements that are divisible by 2.</p>
<p><strong>Example 2:</strong></p>
<p><strong>Input:</strong> nums = [1,2,3,4], k = 4, p = 1</p>
<p><strong>Output:</strong> 10</p>
<p><strong>Explanation:</strong></p>
<p>All element of nums are divisible by p = 1.</p>
<p>Also, every subarray of nums will have at most 4 elements that are divisible by 1.</p>
<p>Since all subarrays are distinct, the total number of subarrays satisfying all the constraints is 10.</p>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>1 <= nums.length <= 200</code></li>
<li><code>1 <= nums[i], p <= 200</code></li>
<li><code>1 <= k <= nums.length</code></li>
</ul>
-
Constructor Summary
Constructors -
Method Summary
-
Constructor Details
-
Solution
public Solution()
-
-
Method Details
-
countDistinct
public int countDistinct(int[] nums, int k, int p)
-