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)
func eatenApples(apples []int, days []int) int {
	var h hp
	ans, n := 0, len(apples)
	for i := 0; i < n || len(h) > 0; i++ {
		if i < n && apples[i] > 0 {
			heap.Push(&h, pair{i + days[i] - 1, apples[i]})
		}
		for len(h) > 0 && h[0].first < i {
			heap.Pop(&h)
		}
		if len(h) > 0 {
			h[0].second--
			if h[0].first == i || h[0].second == 0 {
				heap.Pop(&h)
			}
			ans++
		}
	}
	return ans
}

type pair struct {
	first  int
	second int
}

type hp []pair

func (a hp) Len() int           { return len(a) }
func (a hp) Swap(i, j int)      { a[i], a[j] = a[j], a[i] }
func (a hp) Less(i, j int) bool { return a[i].first < a[j].first }
func (a *hp) Push(x any)        { *a = append(*a, x.(pair)) }
func (a *hp) Pop() any          { l := len(*a); t := (*a)[l-1]; *a = (*a)[:l-1]; return t }