Introduction

In this blog post, we will explore how to solve the LeetCode problem "2261" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given an integer array nums and two integers k and p, return the number of distinct subarrays, which have at most k elements that are 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 Follow up: Can you solve this problem in O(n2) time complexity?

Explanation

Here's a breakdown of the approach, complexity, and the Python code:

• High-Level Approach:

  • Iterate through all possible subarrays of the input array nums.
  • For each subarray, count the number of elements divisible by p.
  • If the count is less than or equal to k, add the subarray to a set to ensure distinct subarrays are counted only once.

• Complexity:

  • Runtime Complexity: O(n²), where n is the length of the input array nums.
  • Storage Complexity: O(n²) in the worst case, to store distinct subarrays in the set.

Code

    def count_distinct_subarrays(nums, k, p):
    """
    Counts the number of distinct subarrays with at most k elements divisible by p.

    Args:
        nums: The input integer array.
        k: The maximum number of elements divisible by p allowed in a subarray.
        p: The divisor.

    Returns:
        The number of distinct subarrays satisfying the condition.
    """

    distinct_subarrays = set()
    n = len(nums)

    for i in range(n):
        for j in range(i, n):
            subarray = tuple(nums[i:j + 1])  # Convert to tuple for set usage
            divisible_count = 0
            for num in subarray:
                if num % p == 0:
                    divisible_count += 1

            if divisible_count <= k:
                distinct_subarrays.add(subarray)

    return len(distinct_subarrays)