Introduction

In this blog post, we will explore how to solve the LeetCode problem "2475" 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

You are given a 0-indexed array of positive integers nums. Find the number of triplets (i, j, k) that meet the following conditions: 0 <= i < j < k < nums.length nums[i], nums[j], and nums[k] are pairwise distinct. In other words, nums[i] != nums[j], nums[i] != nums[k], and nums[j] != nums[k]. Return the number of triplets that meet the conditions. Example 1: Input: nums = [4,4,2,4,3] Output: 3 Explanation: The following triplets meet the conditions: - (0, 2, 4) because 4 != 2 != 3 - (1, 2, 4) because 4 != 2 != 3 - (2, 3, 4) because 2 != 4 != 3 Since there are 3 triplets, we return 3. Note that (2, 0, 4) is not a valid triplet because 2 > 0. Example 2: Input: nums = [1,1,1,1,1] Output: 0 Explanation: No triplets meet the conditions so we return 0. Constraints: 3 <= nums.length <= 100 1 <= nums[i] <= 1000

Explanation

• Iterate through all possible triplets (i, j, k) such that 0 <= i < j < k < nums.length.
  • For each triplet, check if the elements at these indices are pairwise distinct (nums[i] != nums[j], nums[i] != nums[k], and nums[j] != nums[k]).
  • If the triplet satisfies the distinctness condition, increment a counter. Finally, return the counter.

• Runtime Complexity: O(n^3), Storage Complexity: O(1)

Code

    def count_triplets(nums):
    """
    Counts the number of triplets (i, j, k) that meet the conditions:
    0 <= i < j < k < nums.length
    nums[i], nums[j], and nums[k] are pairwise distinct.

    Args:
        nums: A 0-indexed array of positive integers.

    Returns:
        The number of triplets that meet the conditions.
    """

    n = len(nums)
    count = 0
    for i in range(n):
        for j in range(i + 1, n):
            for k in range(j + 1, n):
                if nums[i] != nums[j] and nums[i] != nums[k] and nums[j] != nums[k]:
                    count += 1
    return count