Problem Description

The problem provides an integer array named nums, along with two integers k and p. The objective is to find out how many distinct subarrays exist within nums such that each subarray contains at most k elements divisible by p.

Subarrays are contiguous parts of the array, and they have to be non-empty. Distinct subarrays are those that differ in size or have at least one differing element at any position when compared.

It's important to understand:

• What a subarray is: a continuous sequence from the array.
• A distinct subarray: when compared to another, it has a different length or at least one different element at any index.
• The condition to be met: having at most k elements divisible by p in each subarray.

The challenge is to consider all possible subarrays and count only the unique ones matching the divisibility constraint.

Intuition

The intuition behind the solution comprises two main aspects: iteration and uniqueness tracking.

• Iteration: To find all possible subarrays, we iterate through the original array. Starting from each index in nums, we add elements one by one to the current subarray and check if they meet the condition (at most k elements divisible by p).

• Uniqueness Tracking: As we keep extending these subarrays, we convert them into a string representation, by concatenating elements separated by commas, in order to record each unique one. This string acts as a unique key for the particular subarray.

• Break Condition: An important aspect is knowing when to stop extending a subarray. This is determined by the count of elements divisible by p. As soon as we have more than k such elements, we break the inner loop and proceed with the next starting index.

• Set for Uniqueness: We use a set data structure, since it automatically ensures that only unique subarrays are counted. We add the string representation of each valid subarray (up to k divisible elements) to this set.

The key lies in sequential scanning and the use of a set to ensure distinct subarrays are tracked and counted. The iteration ensures that every possible subarray configuration is considered, while the set data structure helps in avoiding duplicates and keeping the count of distinct subarrays only.

Learn more about Trie patterns.

Solution Approach

The solution approach uses a brute force method to find all possible subarrays and utilizes a set to store unique subarrays satisfying the condition. To understand the algorithm, let's break down the code:

• We have a nested loop; the outer loop starts with index i from 0 to the end of the array.
• For each position in the array, the inner loop adds elements to the subarray and checks if it still satisfies our condition.
• We keep track of the number of elements divisible by p using the variable cnt.
• To maintain the uniqueness, we create a string t to represent the current subarray. As we iterate with the inner loop, we append the current element x concatenated with a comma to t.
• After appending each element x, we check the divisibility condition. If cnt exceeds k, we stop extending that subarray and break the inner loop.
• If cnt is still less than or equal to k, we add t to our set s.
• The set 's' maintains all unique subarrays encountered during the loops.
• Finally, we return the size of the set s, which gives us the number of unique subarrays.

Key Components:

• Brute Force Enumeration: The solution uses a brute force approach to generate all possible subarrays starting from each index in the original array.
• Set for Uniqueness: A set is employed to handle the uniqueness of subarrays, which means any duplicate representation is automatically neglected.
• String Representation: Subarrays are represented as strings, which allows for an easy comparison of uniqueness.
• Breaking Condition: The algorithm includes a breaking condition within the inner loop that halts further extension of a subarray once it no longer satisfies the divisibility condition.

The time complexity of this algorithm is O(n^3) in the worst case, where n is the length of nums. This is because we have a nested loop (O(n^2)) and string concatenation within the inner loop (O(n)), which can be quite inefficient for very large arrays. However, for smaller arrays or with constraints on k, this approach is sufficient to find the correct answer.

Example Walkthrough

Let's take a small example to illustrate the solution approach.

Suppose we have the following inputs:

• nums = [3, 1, 2, 3]
• k = 2
• p = 3

Now, we will walk through the algorithm step by step:

1. Initialize a set s to keep track of unique subarrays.

2. Initialize an outer loop that starts with index i at 0 and goes up to the length of nums.

3. For each iteration of the outer loop, we start from the i-th element and initialize an inner loop that creates subarrays by adding one element at a time to the current subarray.

4. Inside the inner loop, we have a cnt variable to count the number of elements divisible by p and a string t for the subarray representation.

5. Starting with index i = 0 and nums[i] = 3:

   • For the first element, cnt is incremented as 3 is divisible by p = 3.
   • The string t becomes "3," (representation of the subarray [3]), and we add it to set s.

6. The inner loop extends the subarray to include the next element 1, making it [3, 1]:

   • cnt remains 1 since 1 is not divisible by 3.
   • The new subarray is represented by "3,1,", and we add it to set s.

7. We continue adding elements to the subarray and update cnt and t accordingly:

   • Next, we include element 2, resulting in subarray [3, 1, 2]. As 2 is not divisible by 3, cnt stays the same. The string t is updated to "3,1,2,", and it is added to s.
   • Then, we add the last element 3, forming subarray [3, 1, 2, 3]. The cnt now becomes 2 since 3 is divisible by 3. The string t is "3,1,2,3,", and it is also added to s.

8. At this point, if there were more elements to add, and if adding another element divisible by p made cnt exceed k, the inner loop would break, and we would not add the new subarray representation to s.

9. The outer loop moves to the next starting index (i = 1) and repeats the process, generating all subarrays starting from nums[1].

10. The algorithm continues iterating this way, ensuring that all subarrays are considered, with only unique subarrays that have at most k divisible elements are added to s.

11. Once the loops are finished, the distinct subarrays in set s will be:

• "3,"
• "3,1,"
• "3,1,2,"
• "3,1,2,3,"
• "1,"
• "1,2,"
• "1,2,3,"
• "2,"
• "2,3,"
• "3," (which already exists and isn't counted again)

Thus, the size of set s gives us the count of distinct subarrays, which is 9 in this example.

This walkthrough demonstrates how the brute force approach and set data structure come together to solve the given problem by considering every single possible subarray and maintaining their uniqueness.

Solution Implementation

Time and Space Complexity

Time Complexity

The given Python code has a nested loop structure where the outer loop runs n times (where n is the length of the nums list) and the inner loop can run up to n times in the worst case. For each iteration of the inner loop, the code checks whether the current number is divisible by p and breaks the loop if the count of numbers divisible by p exceeds k.

The += operator on a string inside the inner loop has a time complexity of O(m) each time it is executed, where m is the current length of the string t. Since t grows linearly with the number of iterations, the string concatenation inside the loop can have a quadratic effect in terms of the number of elements encountered.

Considering all these factors, the overall worst-case time complexity is O(n^3) - this happens when all numbers are less than p or when k is large, allowing the inner loop to run for all elements without breaking early.

Space Complexity

The space complexity mainly comes from the set s that stores unique strings constructed from the list elements. In the worst case, it can store all subarrays, and the size of the subarrays are incremental starting from 1 to n. The space required for these strings could be considered as summing an arithmetic progression from 1 to n, giving a space complexity of O(n*(n+1)/2), which simplifies to O(n^2).

Additionally, there is a minor space cost for the variables cnt and t, but they don't grow with n in a way that would affect the overall space complexity.

Therefore, the overall space complexity of the code is O(n^2).

Learn more about how to find time and space complexity quickly using problem constraints.