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Given an integer array `nums`

and two integers `k`

and `p`

, return *the number of distinct subarrays which have at most*

`k`

`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`

```
class Solution {
public:
int countDistinct(vector<int>& nums, int k, int p) {
int n=nums.size();
set<vector<int>>ans;
int i,j;
for(i=0;i<n;i++)
{
vector<int>tt;
int ct=0;
for(j=i;j<n;j++)
{
tt.push_back(nums[j]);
if(nums[j]%p==0)
++ct;
if(ct>k)
break;
ans.insert(tt);
}
}
return ans.size();
}
};
```

```
class Solution:
def countDistinct(self, nums: List[int], k: int, p: int) -> int:
n = len(nums)
sub_arrays = set()
# generate all combinations of subarray
for start in range(n):
cnt = 0
temp = ''
for i in range(start, n):
if nums[i]%p == 0:
cnt+=1
temp+=str(nums[i]) + ',' # build the sequence subarray in CSV format
if cnt>k: # check for termination
break
sub_arrays.add(temp)
return len(sub_arrays)
```

```
class Solution {
public int countDistinct(int[] nums, int k, int p) {
int n = nums.length;
// we are storing hashcode for all the substrings so that we can compare them faster.
// main goal is to avoid entire sub array comparision using hashcode.
Set<Long> ways = new HashSet<>();
for(int i = 0; i < n; i++) {
int cnt = 0;
long hc = 1; // this is the running hashcode for sub array [i...j]
for(int j = i; j < n; j++) {
hc = 199L * hc + nums[j]; // updating running hashcode, since we nums are <=200, we shall consider a prime near 200 to avoid collision
if(nums[j] % p == 0)
cnt++;
if(cnt <= k) { // if current subarray [i...j] is valid, add its hashcode in our storage.
ways.add(hc);
}
}
}
return ways.size();
}
}
```

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