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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

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) {
} 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] + ",";
}
}
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]) + ","
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):

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() + ',';
}
}
return s.size;
}



Explain:

nope.

Complexity:

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