Formatted question description: https://leetcode.ca/all/2261.html

2261. K Divisible Elements Subarrays

Difficulty: Medium.
Related Topics: Array, Hash Table, Trie, Rolling Hash, Hash Function, Enumeration.
Similar Questions: Subarrays with K Different Integers, Count Number of Nice Subarrays, Subarray With Elements Greater Than Varying Threshold.

Problem

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?

Solution (Java, C++, Python)

class Solution {
    public int countDistinct(int[] nums, int k, int p) {
        HashSet<Long> numSubarray = new HashSet<>();
        for (int i = 0; i < nums.length; i++) {
            int countDiv = 0;
            long hashCode = 1;
            for (int j = i; j < nums.length; j++) {
                hashCode = 199L * hashCode + nums[j];
                if (nums[j] % p == 0) {
                    countDiv++;
                }
                if (countDiv <= k) {
                    numSubarray.add(hashCode);
                } else {
                    break;
                }
            }
        }
        return numSubarray.size();
    }
}

############

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:
    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)

############

# 2261. K Divisible Elements Subarrays
# https://leetcode.com/problems/k-divisible-elements-subarrays/

class Solution:
    def countDistinct(self, nums: List[int], k: int, p: int) -> int:
        s = set()
        i = 0
        count = 0
        n = len(nums)
        
        def go(arr, start, end): 
            sub_list = [] 
            
            for i in range(start, end + 1): 
                for j in range(i + 1, end + 1): 
                    sub = arr[i:j] 
                    sub_list.append(sub) 

            return sub_list
    
        for j, x in enumerate(nums):
            if x % p == 0:
                count += 1
            
            while count > k:
                if nums[i] % p == 0:
                    count -= 1
                i += 1
            
            
            if j == n - 1 or count == k:
                for sub in go(nums, i, j + 1):
                    s.add(tuple(sub))

        return len(s)

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;
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).

2261 - K Divisible Elements Subarrays

2261. K Divisible Elements Subarrays

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Solution (Java, C++, Python)

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2261. K Divisible Elements Subarrays

Problem

Solution (Java, C++, Python)

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All Solutions