<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:
    #define mod (long long)(pow(10, 9) + 7)
    int maxProfit(vector<int>& inventory, int orders) {
        ios::sync_with_stdio(false);
        cin.tie(nullptr);
        cout.tie(nullptr);

        long long mini = 0, low = 1, high = (long long)pow(10, 9), n = inventory.size();
        while(low <= high)
        {
            long long mid = (low + high) >> 1;
            long long curr = 0;
            for(int i = 0; i < n; i++)
                curr += max(0LL, inventory[i] - mid + 1);
            if(curr >= orders)
                mini = mid, low = mid + 1;
            else
                high = mid - 1;
        }

        long long ans = 0, count = 0;
        for(int i = 0; i < n; i++)
        {
            if(inventory[i] < mini)
                continue;
            long long curr_val = inventory[i];
            ans += curr_val * (curr_val + 1) / 2 - mini * (mini - 1) / 2;
            count += curr_val - mini + 1;
        }
        if(count > orders)
            ans -= (count - orders) * mini;
        ans %= mod;
        return ans;
    }
};
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