README_EN.md

2261. K Divisible Elements Subarrays

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Description

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

 

Follow up:

Can you solve this problem in O(n²) time complexity?

Solutions

Python3

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)

Java

class Solution {
    public int countDistinct(int[] nums, int k, int p) {
        Set<String> s = new HashSet<>();
        for (int i = 0, n = nums.length; i < n; ++i) {
            int cnt = 0;
            String t = "";
            for (int j = i; j < n; ++j) {
                if (nums[j] % p == 0) {
                    ++cnt;
                }
                if (cnt > k) {
                    break;
                }
                t += nums[j] + ",";
                s.add(t);
            }
        }
        return s.size();
    }
}

C++

class Solution {
public:
    int countDistinct(vector<int>& nums, int k, int p) {
        unordered_set<string> s;
        for (int i = 0, n = nums.size(); i < n; ++i) {
            int cnt = 0;
            string t = "";
            for (int j = i; j < n; ++j) {
                if (nums[j] % p == 0) ++cnt;
                if (cnt > k) break;
                t += to_string(nums[j]) + ",";
                s.insert(t);
            }
        }
        return s.size();
    }
};

Go

func countDistinct(nums []int, k int, p int) int {
	s := map[string]bool{}
	for i, n := 0, len(nums); i < n; i++ {
		cnt := 0
		t := ""
		for j := i; j < n; j++ {
			if nums[j]%p == 0 {
				cnt++
			}
			if cnt > k {
				break
			}
			t += string(nums[j]) + ","
			s[t] = true
		}
	}
	return len(s)
}

TypeScript

function countDistinct(nums: number[], k: number, p: number): number {
    const n = nums.length;
    const numSet = new Set(nums);
    const verfiedSet = new Set<number>();
    for (let i of numSet) {
        if (i % p != 0) continue;
        verfiedSet.add(i);
    }
    let ans = new Set<string>();
    for (let i = 0; i < n; i++) {
        let sub = [];
        for (let j = i, cnt = 0; j < n; j++) {
            const num = nums[j];
            if (verfiedSet.has(num)) cnt++;
            if (cnt > k) break;
            sub.push(num);
            const str = sub.join(',');
            ans.add(str);
        }
    }
    return ans.size;
}

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