Solution Explanation for LeetCode 2261: K Divisible Elements Subarrays

This problem asks to find the number of distinct subarrays containing at most k elements divisible by p. The key challenge lies in efficiently handling the distinctness constraint and the condition on divisible elements.

Solution 1: Enumeration + String Hashing

This approach uses a brute-force enumeration of all possible subarrays and a string hashing technique to efficiently check for distinctness.

Approach:

1. Nested Loops for Subarray Generation: The outer loop iterates through each possible starting index i of a subarray. The inner loop iterates through all possible ending indices j (from i to the end of the array).

2. Counting Divisible Elements: Inside the inner loop, we count the number of elements in the current subarray (nums[i:j+1]) that are divisible by p.

3. Distinctness Check with Hashing: To avoid redundant counting of identical subarrays, we use double hashing. This involves using two different hash functions (here, using two different bases and modulo operations) to create a unique hash value for each subarray. These hashes are stored in a set. Because of the probability of collision in hash function , the usage of two hashing function will reduce the probability of collision. The set automatically handles the distinctness constraint.

4. Return the Count: Finally, the size of the set (which represents the number of distinct subarrays satisfying the condition) is returned.

Time Complexity: O(n^2), where n is the length of the input array. This is because of the nested loops that enumerate all possible subarrays.

Space Complexity: O(n^2) in the worst case. This is because the set storing the hash values of distinct subarrays could potentially store up to O(n^2) elements if all subarrays are distinct.

Code (Python):

class Solution:
    def countDistinct(self, nums: List[int], k: int, p: int) -> int:
        s = set()
        n = len(nums)
        base1, base2 = 131, 13331 # Two different bases for double hashing
        mod1, mod2 = 10**9 + 7, 10**9 + 9 # Two different modulo values
 
        for i in range(n):
            h1 = h2 = cnt = 0  # Initialize hash values and divisible element count
            for j in range(i, n):
                cnt += nums[j] % p == 0 #Increment if divisible by p
                if cnt > k:
                    break #If greater than k then break from the loop 
                h1 = (h1 * base1 + nums[j]) % mod1 #First hash function
                h2 = (h2 * base2 + nums[j]) % mod2 #Second hash function
                s.add(h1 << 32 | h2) #Combine hashes and add to set
        return len(s)
 

Solution 2 (Simpler, less efficient):

This solution is conceptually simpler but less efficient. It uses string concatenation to represent subarrays and a set to check for distinctness. It is less efficient because string concatenation has a higher time complexity than the hash function in solution 1.

Approach:

1. Nested loops to generate subarrays: Same as in Solution 1.

2. Counting divisible elements: Same as in Solution 1.

3. Subarray representation: The subarray is converted into a string using commas as separators. This is then added to the set to ensure distinctness.

Time Complexity: O(n^3) due to string concatenation which takes O(n) time complexity inside the nested loop

Space Complexity: O(n^2) because of the set that could potentially store all the distinct subarrays in the worst case.

Code (Python):

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 = "" # String to represent the subarray
            for x in nums[i:]:
                cnt += x % p == 0
                if cnt > k:
                    break
                t += str(x) + ","
                s.add(t)
        return len(s)

In summary: Solution 1 (using double hashing) is the preferred approach due to its better time complexity. Solution 2 provides a simpler, but less efficient alternative. Both solutions correctly address the distinct subarray requirement. Remember that the hash function used in solution 1 is probabilistic, meaning there's a tiny chance of hash collisions; however, the use of double hashing mitigates this risk significantly.