<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>

#pragma GCC optimize("Ofast","inline","fast-math","unroll-loops","no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native","f16c")
static const auto fast = []() {ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0); return 0; } ();
class Solution {
public:
    int maxProfit(std::vector<int>& inventory, int orders) {
        const int MOD = 1e9 + 7;
        long long ans = 0;

        // Sort inventory in descending order
        std::sort(inventory.begin(), inventory.end(), std::greater<int>());

        int n = inventory.size();
        int currentIndex = 0;

        while (orders > 0) {
            int current = inventory[currentIndex];
            int next = (currentIndex + 1 < n) ? inventory[currentIndex + 1] : 0;  // The next distinct value in the inventory

            int height = current - next;  // The number of levels between current and next
            int width = currentIndex + 1;  // How many stacks of balls have the same current max value

            // How many balls we can sell in this step
            long long sell = (long long)height * width;

            if (orders >= sell) {
                // We can sell all the balls from current to next level
                ans = (ans + (long long)(current + next + 1) * height / 2 * width) % MOD;
                orders -= sell;
                currentIndex++;
            } else {
                // We can't sell all the way down to the next level, so sell the remainder at this level
                int fullRows = orders / width;
                int remainder = orders % width;

                ans = (ans + (long long)(current + current - fullRows + 1) * fullRows / 2 * width) % MOD;
                ans = (ans + (long long)(current - fullRows) * remainder) % MOD;

                orders = 0;
            }
        }

        return ans;
    }
};
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