# Solving Leetcode Interviews in Seconds with AI: Maximum Product of Two Elements in an Array


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1464" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), 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 the array of integers nums, you will choose two different indices i and j of that array. Return the maximum value of (nums[i]-1)*(nums[j]-1).   Example 1:  Input: nums = [3,4,5,2] Output: 12  Explanation: If you choose the indices i=1 and j=2 (indexed from 0), you will get the maximum value, that is, (nums[1]-1)*(nums[2]-1) = (4-1)*(5-1) = 3*4 = 12.   Example 2:  Input: nums = [1,5,4,5] Output: 16 Explanation: Choosing the indices i=1 and j=3 (indexed from 0), you will get the maximum value of (5-1)*(5-1) = 16.  Example 3:  Input: nums = [3,7] Output: 12    Constraints:  2 <= nums.length <= 500 1 <= nums[i] <= 10^3  

	# Explanation
	Here's the breakdown of the solution:

*   **Find the two largest numbers:** The problem asks to maximize `(nums[i]-1)*(nums[j]-1)`. To achieve this, we should select the two largest numbers in the array for `nums[i]` and `nums[j]`.
*   **Calculate the product:** Once we have the two largest numbers, we simply compute `(largest - 1) * (second_largest - 1)`.

*   **Time Complexity:** O(n), **Space Complexity:** O(1)

	
	# Code
	```python
	def maxProduct(nums: list[int]) -> int:
    """
    Given the array of integers nums, you will choose two different indices i and j of that array.
    Return the maximum value of (nums[i]-1)*(nums[j]-1).
    """
    largest = 0
    second_largest = 0

    for num in nums:
        if num > largest:
            second_largest = largest
            largest = num
        elif num > second_largest:
            second_largest = num

    return (largest - 1) * (second_largest - 1)
	```