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Medium
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Weekly Contest 291 Q3
Trie
Array
Hash Table
Enumeration
Hash Function
Rolling Hash

2261. K Divisible Elements Subarrays

中文文档

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.

Python3

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)

Java

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();
    }
}

C++

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();
    }
};

Go

func countDistinct(nums []int, k int, p int) int {
	s := map[int]bool{}
	base1, base2 := 131, 13331
	mod1, mod2 := 1000000007, 1000000009
	for i := range nums {
		h1, h2, cnt := 0, 0, 0
		for j := i; j < len(nums); j++ {
			if nums[j]%p == 0 {
				cnt++
				if cnt > k {
					break
				}
			}
			h1 = (h1*base1 + nums[j]) % mod1
			h2 = (h2*base2 + nums[j]) % mod2
			s[h1<<32|h2] = true
		}
	}
	return len(s)
}

TypeScript

function countDistinct(nums: number[], k: number, p: number): number {
    const s = new Set<bigint>();
    const [base1, base2] = [131, 13331];
    const [mod1, mod2] = [1000000007, 1000000009];
    for (let i = 0; i < nums.length; i++) {
        let [h1, h2, cnt] = [0, 0, 0];
        for (let j = i; j < nums.length; j++) {
            if (nums[j] % p === 0) {
                cnt++;
                if (cnt > k) {
                    break;
                }
            }
            h1 = (h1 * base1 + nums[j]) % mod1;
            h2 = (h2 * base2 + nums[j]) % mod2;
            s.add((BigInt(h1) << 32n) | BigInt(h2));
        }
    }
    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)