Introduction

In this blog post, we will explore how to solve the LeetCode problem "2874" 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 <= 105 1 <= nums[i] <= 106

Explanation

Here's the solution, explained concisely and followed by the Python code:

• Iterate through the array: The core idea is to iterate through the array, considering each element nums[j] as the middle element of a potential triplet (i, j, k).
• Maximize (nums[i] - nums[j]): For each j, we need to find the maximum possible value of nums[i] - nums[j] where i < j. We maintain a running maximum of nums[i] encountered so far to efficiently calculate this.

• Maximize (nums[i] - nums[j]) * nums[k]: Similarly, for each j, iterate k from j+1 to end of array, and calculate (nums[i] - nums[j]) * nums[k], and take the maximum of all triplets.

• Runtime Complexity: O(n^2), Storage Complexity: O(1)

Code

    def max_value_of_triplet(nums):
    n = len(nums)
    max_product = 0

    for j in range(1, n - 1):
        max_diff_left = 0
        max_left = 0
        for i in range(j):
            max_left = max(max_left, nums[i])
        max_diff_left = max_left - nums[j]

        for k in range(j + 1, n):
            product = max_diff_left * nums[k]
            max_product = max(max_product, product)

    return max_product