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: Enumeration + String Hashing

We can enumerate the left endpoint $i$ of the subarray, and then enumerate the right endpoint $j$ in the range $[i, n)$. During the enumeration of the right endpoint, we use double hashing to store the hash value of the subarray into a set. Finally, we return the size of the set.

The time complexity is $O(n^2)$, and the space complexity is $O(n^2)$. Here, $n$ is the length of the array.

Python3JavaC++GoTypeScript

class Solution:
    def countDistinct(self, nums: List[int], k: int, p: int) -> int:
        s = set()
        n = len(nums)
        base1, base2 = 131, 13331
        mod1, mod2 = 10**9 + 7, 10**9 + 9
        for i in range(n):
            h1 = h2 = cnt = 0
            for j in range(i, n):
                cnt += nums[j] % p == 0
                if cnt > k:
                    break
                h1 = (h1 * base1 + nums[j]) % mod1
                h2 = (h2 * base2 + nums[j]) % mod2
                s.add(h1 << 32 | h2)
        return len(s)

class Solution {
    public int countDistinct(int[] nums, int k, int p) {
        Set<Long> s = new HashSet<>();
        int n = nums.length;
        int base1 = 131, base2 = 13331;
        int mod1 = (int) 1e9 + 7, mod2 = (int) 1e9 + 9;
        for (int i = 0; i < n; ++i) {
            long h1 = 0, h2 = 0;
            int cnt = 0;
            for (int j = i; j < n; ++j) {
                cnt += nums[j] % p == 0 ? 1 : 0;
                if (cnt > k) {
                    break;
                }
                h1 = (h1 * base1 + nums[j]) % mod1;
                h2 = (h2 * base2 + nums[j]) % mod2;
                s.add(h1 << 32 | h2);
            }
        }
        return s.size();
    }
}

class Solution {
public:
    int countDistinct(vector<int>& nums, int k, int p) {
        unordered_set<long long> s;
        int n = nums.size();
        int base1 = 131, base2 = 13331;
        int mod1 = 1e9 + 7, mod2 = 1e9 + 9;
        for (int i = 0; i < n; ++i) {
            long long h1 = 0, h2 = 0;
            int cnt = 0;
            for (int j = i; j < n; ++j) {
                cnt += nums[j] % p == 0;
                if (cnt > k) {
                    break;
                }
                h1 = (h1 * base1 + nums[j]) % mod1;
                h2 = (h2 * base2 + nums[j]) % mod2;
                s.insert(h1 << 32 | h2);
            }
        }
        return s.size();
    }
};