Introduction

In this blog post, we will explore how to solve the LeetCode problem "2016" 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 a 0-indexed integer array nums of size n, find the maximum difference between nums[i] and nums[j] (i.e., nums[j] - nums[i]), such that 0 <= i < j < n and nums[i] < nums[j]. Return the maximum difference. If no such i and j exists, return -1. Example 1: Input: nums = [7,1,5,4] Output: 4 Explanation: The maximum difference occurs with i = 1 and j = 2, nums[j] - nums[i] = 5 - 1 = 4. Note that with i = 1 and j = 0, the difference nums[j] - nums[i] = 7 - 1 = 6, but i > j, so it is not valid. Example 2: Input: nums = [9,4,3,2] Output: -1 Explanation: There is no i and j such that i < j and nums[i] < nums[j]. Example 3: Input: nums = [1,5,2,10] Output: 9 Explanation: The maximum difference occurs with i = 0 and j = 3, nums[j] - nums[i] = 10 - 1 = 9. Constraints: n == nums.length 2 <= n <= 1000 1 <= nums[i] <= 109

Explanation

Here's the approach to solve this problem efficiently:

• Maintain Minimum: Iterate through the array, keeping track of the minimum value encountered so far.
• Calculate Difference: At each element, calculate the difference between the current element and the minimum value seen so far.

• Track Maximum: Update the maximum difference found if the current difference is greater.

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

Code

    def maximumDifference(nums):
    """
    Finds the maximum difference between nums[i] and nums[j] such that i < j and nums[i] < nums[j].

    Args:
        nums: A list of integers.

    Returns:
        The maximum difference, or -1 if no such i and j exists.
    """
    min_val = nums[0]
    max_diff = -1

    for i in range(1, len(nums)):
        if nums[i] > min_val:
            max_diff = max(max_diff, nums[i] - min_val)
        else:
            min_val = nums[i]

    return max_diff