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package g2201_2300.s2261_k_divisible_elements_subarrays;

// #Medium #Array #Hash_Table #Trie #Enumeration #Hash_Function #Rolling_Hash
// #2022_06_18_Time_73_ms_(92.32%)_Space_67.2_MB_(78.03%)

import java.util.HashSet;

/**
 * 2261 - K Divisible Elements Subarrays\.
 *
 * Medium
 *
 * 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`
**/
public class Solution {
    public int countDistinct(int[] nums, int k, int p) {
        HashSet 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();
    }
}