<p>You have an <code>inventory</code> of different colored balls, and there is a customer that wants <code>orders</code> balls of <strong>any</strong> color.</p>

<p>The customer weirdly values the colored balls. Each colored ball&#39;s value is the number of balls <strong>of that color&nbsp;</strong>you currently have in your <code>inventory</code>. For example, if you own <code>6</code> yellow balls, the customer would pay <code>6</code> for the first yellow ball. After the transaction, there are only <code>5</code> yellow balls left, so the next yellow ball is then valued at <code>5</code> (i.e., the value of the balls decreases as you sell more to the customer).</p>

<p>You are given an integer array, <code>inventory</code>, where <code>inventory[i]</code> represents the number of balls of the <code>i<sup>th</sup></code> color that you initially own. You are also given an integer <code>orders</code>, which represents the total number of balls that the customer wants. You can sell the balls <strong>in any order</strong>.</p>

<p>Return <em>the <strong>maximum</strong> total value that you can attain after selling </em><code>orders</code><em> colored balls</em>. As the answer may be too large, return it <strong>modulo </strong><code>10<sup>9 </sup>+ 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/05/jj.gif" style="width: 480px; height: 270px;" />
<pre>
<strong>Input:</strong> inventory = [2,5], orders = 4
<strong>Output:</strong> 14
<strong>Explanation:</strong> Sell the 1st color 1 time (2) and the 2nd color 3 times (5 + 4 + 3).
The maximum total value is 2 + 5 + 4 + 3 = 14.
</pre>

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

<pre>
<strong>Input:</strong> inventory = [3,5], orders = 6
<strong>Output:</strong> 19
<strong>Explanation: </strong>Sell the 1st color 2 times (3 + 2) and the 2nd color 4 times (5 + 4 + 3 + 2).
The maximum total value is 3 + 2 + 5 + 4 + 3 + 2 = 19.
</pre>

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

<ul>
	<li><code>1 &lt;= inventory.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= inventory[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= orders &lt;= min(sum(inventory[i]), 10<sup>9</sup>)</code></li>
</ul>

class Solution {
public:
    int maxProfit(vector<int>& inventory, int orders) {
        map<int, int> hm;
        for(int occur : inventory){
            if(not hm.count(occur)) hm.insert({occur, 0});
            hm[occur]++;
        }
        long long ans = 0;
        int MOD = 1000000007;
        
//         hm.rbegin()->second;
//         prev(hm.end())->first - (prev(prev(hm.end())))->first;
        
        ///*
        while(orders > 0){
            long long avail;
            int x = hm.rbegin()->first;
            int y = hm.rbegin()->second;
            int z = (prev(prev(hm.end())))->first;
            if(hm.size() == 1){
                avail = x;
                avail *= y;
            }else{
                avail = y;
                avail *= x - z;
            }
            if(orders <= avail){
                int depth = orders / y;
                int rem = orders % y;
                long long incr = x;
                incr *= (x + 1);
                incr /= 2;
                long long decr = x - depth;
                decr *= (decr + 1);
                decr /= 2;
                incr -= decr;
                incr *= y;
                if(rem > 0){
                    long long incr1 = rem;
                    incr1 *= (x - depth);
                    incr += incr1;
                }
                ans += incr;
                ans %= MOD;
                
                break;
            }
            long long incr = x;
            incr *= (x + 1);
            incr /= 2;
            long long decr = z;
            decr *= (z + 1);
            decr /= 2;
            incr -= decr;
            incr *= y;
            
            ans += incr;
            ans %= MOD;
            
            orders -= avail;
            hm.erase(x);
            hm[z] += y;
        }
        //*/
        return ans;
    }
};
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