Class Solution
- java.lang.Object
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- g2201_2300.s2261_k_divisible_elements_subarrays.Solution
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public class Solution extends Object
2261 - K Divisible Elements Subarrays.Medium
Given an integer array
nums
and two integersk
andp
, return the number of distinct subarrays which have at mostk
elements divisible byp
.Two arrays
nums1
andnums2
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 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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Constructor Summary
Constructors Constructor Description Solution()
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Method Summary
All Methods Instance Methods Concrete Methods Modifier and Type Method Description int
countDistinct(int[] nums, int k, int p)
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