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 {
public:
    int maxProduct(vector<int>& nums) {
        vector<int> pos;
        vector<int> neg;

        for(int num : nums){
            num-=1;
            if(num>=0) pos.push_back(num);
            else neg.push_back(num);
        }

        sort(pos.begin(),pos.end());
        sort(neg.begin(), neg.end());

        if(pos.size() >= 2 && neg.size()>=2){
            return max(pos[pos.size()-1]*pos[pos.size()-2], neg[neg.size()-1] * neg[neg.size()-2]);
        }
        else if(pos.size() >= 2){
            return pos[pos.size()-1]*pos[pos.size()-2];
        }
        else if(neg.size() >= 2){
            return neg[neg.size()-1], neg[neg.size()-2];
        }
        else{
            return pos[0] * neg[neg.size() - 1];
        }
    }
};
13000