Given the array of integers <code>nums</code>, you will choose two different indices <code>i</code> and <code>j</code> of that array. <em>Return the maximum value of</em> <code>(nums[i]-1)*(nums[j]-1)</code>.
<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,4,5,2]
<strong>Output:</strong> 12 
<strong>Explanation:</strong> 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. 
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,5,4,5]
<strong>Output:</strong> 16
<strong>Explanation:</strong> Choosing the indices i=1 and j=3 (indexed from 0), you will get the maximum value of (5-1)*(5-1) = 16.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,7]
<strong>Output:</strong> 12
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= nums.length &lt;= 500</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10^3</code></li>
</ul>

class Solution:
    def maxProduct(self, nums: List[int]) -> int:
        res = 0
        n = len(nums)

        for i in range(n - 1):
            for j in range(i + 1, n):
                res = max((nums[i] - 1) * (nums[j] - 1), res)
        return res

16600