Maximum Total Beauty of the Gardens
<p>Alice is a caretaker of <code>n</code> gardens and she wants to plant flowers to maximize the total beauty of all her gardens.</p>

<p>You are given a <strong>0-indexed</strong> integer array <code>flowers</code> of size <code>n</code>, where <code>flowers[i]</code> is the number of flowers already planted in the <code>i<sup>th</sup></code> garden. Flowers that are already planted <strong>cannot</strong> be removed. You are then given another integer <code>newFlowers</code>, which is the <strong>maximum</strong> number of flowers that Alice can additionally plant. You are also given the integers <code>target</code>, <code>full</code>, and <code>partial</code>.</p>

<p>A garden is considered <strong>complete</strong> if it has <strong>at least</strong> <code>target</code> flowers. The <strong>total beauty</strong> of the gardens is then determined as the <strong>sum</strong> of the following:</p>

<ul>
	<li>The number of <strong>complete</strong> gardens multiplied by <code>full</code>.</li>
	<li>The <strong>minimum</strong> number of flowers in any of the <strong>incomplete</strong> gardens multiplied by <code>partial</code>. If there are no incomplete gardens, then this value will be <code>0</code>.</li>
</ul>

<p>Return <em>the <strong>maximum</strong> total beauty that Alice can obtain after planting at most </em><code>newFlowers</code><em> flowers.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> flowers = [1,3,1,1], newFlowers = 7, target = 6, full = 12, partial = 1
<strong>Output:</strong> 14
<strong>Explanation:</strong> Alice can plant
- 2 flowers in the 0<sup>th</sup> garden
- 3 flowers in the 1<sup>st</sup> garden
- 1 flower in the 2<sup>nd</sup> garden
- 1 flower in the 3<sup>rd</sup> garden
The gardens will then be [3,6,2,2]. She planted a total of 2 + 3 + 1 + 1 = 7 flowers.
There is 1 garden that is complete.
The minimum number of flowers in the incomplete gardens is 2.
Thus, the total beauty is 1 * 12 + 2 * 1 = 12 + 2 = 14.
No other way of planting flowers can obtain a total beauty higher than 14.
</pre>

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

<pre>
<strong>Input:</strong> flowers = [2,4,5,3], newFlowers = 10, target = 5, full = 2, partial = 6
<strong>Output:</strong> 30
<strong>Explanation:</strong> Alice can plant
- 3 flowers in the 0<sup>th</sup> garden
- 0 flowers in the 1<sup>st</sup> garden
- 0 flowers in the 2<sup>nd</sup> garden
- 2 flowers in the 3<sup>rd</sup> garden
The gardens will then be [5,4,5,5]. She planted a total of 3 + 0 + 0 + 2 = 5 flowers.
There are 3 gardens that are complete.
The minimum number of flowers in the incomplete gardens is 4.
Thus, the total beauty is 3 * 2 + 4 * 6 = 6 + 24 = 30.
No other way of planting flowers can obtain a total beauty higher than 30.
Note that Alice could make all the gardens complete but in this case, she would obtain a lower total beauty.
</pre>

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

<ul>
	<li><code>1 &lt;= flowers.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= flowers[i], target &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= newFlowers &lt;= 10<sup>10</sup></code></li>
	<li><code>1 &lt;= full, partial &lt;= 10<sup>5</sup></code></li>
</ul>
Greedy; Array; Two Pointers; Binary Search; Enumeration; Prefix Sum; Sorting
class Solution {
public:
    long long maximumBeauty(vector<int>& flowers, long long newFlowers, int target, int full, int partial) {
        sort(flowers.begin(), flowers.end());
        int n = flowers.size();
        long long s[n + 1];
        s[0] = 0;
        for (int i = 1; i <= n; ++i) {
            s[i] = s[i - 1] + flowers[i - 1];
        }
        long long ans = 0;
        int i = flowers.end() - lower_bound(flowers.begin(), flowers.end(), target);
        for (int x = i; x <= n; ++x) {
            newFlowers -= (x == 0 ? 0 : max(target - flowers[n - x], 0));
            if (newFlowers < 0) {
                break;
            }
            int l = 0, r = n - x - 1;
            while (l < r) {
                int mid = (l + r + 1) >> 1;
                if (1LL * flowers[mid] * (mid + 1) - s[mid + 1] <= newFlowers) {
                    l = mid;
                } else {
                    r = mid - 1;
                }
            }
            int y = 0;
            if (r != -1) {
                long long cost = 1LL * flowers[l] * (l + 1) - s[l + 1];
                long long mx = flowers[l] + (newFlowers - cost) / (l + 1);
                long long threshold = target - 1;
                y = min(mx, threshold);
            }
            ans = max(ans, 1LL * x * full + 1LL * y * partial);
        }
        return ans;
    }
};