Maximum Number of Eaten Apples
<p>There is a special kind of apple tree that grows apples every day for <code>n</code> days. On the <code>i<sup>th</sup></code> day, the tree grows <code>apples[i]</code> apples that will rot after <code>days[i]</code> days, that is on day <code>i + days[i]</code> the apples will be rotten and cannot be eaten. On some days, the apple tree does not grow any apples, which are denoted by <code>apples[i] == 0</code> and <code>days[i] == 0</code>.</p>

<p>You decided to eat <strong>at most</strong> one apple a day (to keep the doctors away). Note that you can keep eating after the first <code>n</code> days.</p>

<p>Given two integer arrays <code>days</code> and <code>apples</code> of length <code>n</code>, return <em>the maximum number of apples you can eat.</em></p>

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

<pre>
<strong>Input:</strong> apples = [1,2,3,5,2], days = [3,2,1,4,2]
<strong>Output:</strong> 7
<strong>Explanation:</strong> You can eat 7 apples:
- On the first day, you eat an apple that grew on the first day.
- On the second day, you eat an apple that grew on the second day.
- On the third day, you eat an apple that grew on the second day. After this day, the apples that grew on the third day rot.
- On the fourth to the seventh days, you eat apples that grew on the fourth day.
</pre>

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

<pre>
<strong>Input:</strong> apples = [3,0,0,0,0,2], days = [3,0,0,0,0,2]
<strong>Output:</strong> 5
<strong>Explanation:</strong> You can eat 5 apples:
- On the first to the third day you eat apples that grew on the first day.
- Do nothing on the fouth and fifth days.
- On the sixth and seventh days you eat apples that grew on the sixth day.
</pre>

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

<ul>
	<li><code>n == apples.length == days.length</code></li>
	<li><code>1 &lt;= n &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>0 &lt;= apples[i], days[i] &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>days[i] = 0</code> if and only if <code>apples[i] = 0</code>.</li>
</ul>
Greedy; Array; Heap (Priority Queue)
class Solution {
public:
    int eatenApples(vector<int>& apples, vector<int>& days) {
        using pii = pair<int, int>;
        priority_queue<pii, vector<pii>, greater<pii>> q;
        int n = days.size();
        int ans = 0, i = 0;
        while (i < n || !q.empty()) {
            if (i < n && apples[i]) {
                q.emplace(i + days[i] - 1, apples[i]);
            }
            while (!q.empty() && q.top().first < i) {
                q.pop();
            }
            if (!q.empty()) {
                auto [t, v] = q.top();
                q.pop();
                --v;
                ++ans;
                if (v && t > i) {
                    q.emplace(t, v);
                }
            }
            ++i;
        }
        return ans;
    }
};
