Introduction

In this blog post, we will explore how to solve the LeetCode problem "2873" 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 integer array nums. Return the maximum value over all triplets of indices (i, j, k) such that i < j < k. If all such triplets have a negative value, return 0. The value of a triplet of indices (i, j, k) is equal to (nums[i] - nums[j]) nums[k]. Example 1: Input: nums = [12,6,1,2,7] Output: 77 Explanation: The value of the triplet (0, 2, 4) is (nums[0] - nums[2]) nums[4] = 77. It can be shown that there are no ordered triplets of indices with a value greater than 77. Example 2: Input: nums = [1,10,3,4,19] Output: 133 Explanation: The value of the triplet (1, 2, 4) is (nums[1] - nums[2]) nums[4] = 133. It can be shown that there are no ordered triplets of indices with a value greater than 133. Example 3: Input: nums = [1,2,3] Output: 0 Explanation: The only ordered triplet of indices (0, 1, 2) has a negative value of (nums[0] - nums[1]) nums[2] = -3. Hence, the answer would be 0. Constraints: 3 <= nums.length <= 100 1 <= nums[i] <= 106

Explanation

Here's the breakdown of the solution:

• Iterate and Calculate: The solution iterates through all possible triplets of indices (i, j, k) where i < j < k. For each triplet, it calculates the value (nums[i] - nums[j]) * nums[k].
• Maximize: It keeps track of the maximum value encountered so far and updates it whenever a larger triplet value is found.

• Handle Negative Values: If the maximum value remains negative after checking all triplets (meaning all triplets have negative values), the solution returns 0.

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

Code

    def max_value_of_triplet(nums):
    """
    Calculates the maximum value over all triplets of indices (i, j, k) such that i < j < k.

    Args:
        nums: A list of integers.

    Returns:
        The maximum value over all triplets, or 0 if all triplets have a negative value.
    """

    max_val = float('-inf')
    n = len(nums)

    for i in range(n - 2):
        for j in range(i + 1, n - 1):
            for k in range(j + 1, n):
                val = (nums[i] - nums[j]) * nums[k]
                max_val = max(max_val, val)

    if max_val < 0:
        return 0
    else:
        return max_val