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 largest{
    public:
    int i,j; 
    largest(){
        this->i = i; 
        this->j = j; 
    }
}; 

class Solution {
public:
    largest is_match(vector<int> vc){
        largest l = largest(); 
        int mx = INT_MIN; 
        int n = vc.size(); 
        for(int i=0; i<n-1; i++){
            for(int j=i+1; j<n; j++){
                if(vc[i]+vc[j]>mx){
                    mx = vc[i]+vc[j]; 
                    l.i = i; 
                    l.j = j;  
                }
            }
        }
        return l; 
    }

    int maxProduct(vector<int>& nums) {
        largest l = is_match(nums); 
        cout<<l.i<<' '<<l.j<<endl; 
        return (nums[l.i]-1)*(nums[l.j]-1); 
    }
};

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