g2201_2300.s2261_k_divisible_elements_subarrays

Class Solution

java.lang.Object

g2201_2300.s2261_k_divisible_elements_subarrays.Solution

public class Solution
extends Object

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

Constructor Summary

Constructors

Constructor and Description
Solution()

Method Summary

Modifier and Type	Method and Description
int	countDistinct(int[] nums, int k, int p)

Methods inherited from class java.lang.Object

clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait

Constructor Detail

Solution

public Solution()

Method Detail

countDistinct

public int countDistinct(int[] nums,
                         int k,
                         int p)