2261. K Divisible Elements Subarrays

Description

Given an integer array nums and two integers k and p, return the number of distinct subarrays, which have at most k elements that are 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

Solution 1

Python3JavaC++GoTypeScript

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
            for j in range(i, n):
                cnt += nums[j] % p == 0
                if cnt > k:
                    break
                s.add(tuple(nums[i : j + 1]))
        return len(s)

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

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

func countDistinct(nums []int, k int, p int) int {
    s := map[string]struct{}{}
    for i := range nums {
        cnt, t := 0, ""
        for _, x := range nums[i:] {
            if x%p == 0 {
                cnt++
                if cnt > k {
                    break
                }
            }
            t += string(x) + ","
            s[t] = struct{}{}
        }
    }
    return len(s)
}

function countDistinct(nums: number[], k: number, p: number): number {
    const n = nums.length;
    const s = new Set();
    for (let i = 0; i < n; ++i) {
        let cnt = 0;
        let t = '';
        for (let j = i; j < n; ++j) {
            if (nums[j] % p === 0 && ++cnt > k) {
                break;
            }
            t += nums[j].toString() + ',';
            s.add(t);
        }
    }
    return s.size;
}

Solution 2

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 x in nums[i:]:
                cnt += x % p == 0
                if cnt > k:
                    break
                t += str(x) + ","
                s.add(t)
        return len(s)

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