maximum-number-of-eaten-apples

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Algorithms

There is a special kind of apple tree that grows apples every day for <code>n</code> days. On the <code>ith</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>.

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

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

&nbsp;
Example 1:

<pre>
Input: apples = [1,2,3,5,2], days = [3,2,1,4,2]
Output: 7
Explanation: 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>

Example 2:

<pre>
Input: apples = [3,0,0,0,0,2], days = [3,0,0,0,0,2]
Output: 5
Explanation: 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>

&nbsp;
Constraints:

<ul>
	<li><code>n == apples.length == days.length</code></li>
	<li><code>1 &lt;= n &lt;= 2 * 104</code></li>
	<li><code>0 &lt;= apples[i], days[i] &lt;= 2 * 104</code></li>
	<li><code>days[i] = 0</code> if and only if <code>apples[i] = 0</code>.</li>
</ul>

eatenApples

Maximum Number of Eaten Apples

Greedy

Array

Heap (Priority Queue)

有一棵特殊的苹果树，一连 <code>n</code> 天，每天都可以长出若干个苹果。在第 <code>i</code> 天，树上会长出 <code>apples[i]</code> 个苹果，这些苹果将会在 <code>days[i]</code> 天后（也就是说，第 <code>i + days[i]</code> 天时）腐烂，变得无法食用。也可能有那么几天，树上不会长出新的苹果，此时用 <code>apples[i] == 0</code> 且 <code>days[i] == 0</code> 表示。

你打算每天 最多 吃一个苹果来保证营养均衡。注意，你可以在这 <code>n</code> 天之后继续吃苹果。

给你两个长度为 <code>n</code> 的整数数组 <code>days</code> 和 <code>apples</code> ，返回你可以吃掉的苹果的最大数目。

 

示例 1：

<pre>输入：apples = [1,2,3,5,2], days = [3,2,1,4,2]
输出：7
解释：你可以吃掉 7 个苹果：
- 第一天，你吃掉第一天长出来的苹果。
- 第二天，你吃掉一个第二天长出来的苹果。
- 第三天，你吃掉一个第二天长出来的苹果。过了这一天，第三天长出来的苹果就已经腐烂了。
- 第四天到第七天，你吃的都是第四天长出来的苹果。
</pre>

示例 2：

<pre>输入：apples = [3,0,0,0,0,2], days = [3,0,0,0,0,2]
输出：5
解释：你可以吃掉 5 个苹果：
- 第一天到第三天，你吃的都是第一天长出来的苹果。
- 第四天和第五天不吃苹果。
- 第六天和第七天，你吃的都是第六天长出来的苹果。
</pre>

 

提示：

<ul>
	<li><code>apples.length == n</code></li>
	<li><code>days.length == n</code></li>
	<li><code>1 &lt;= n &lt;= 2 * 104</code></li>
	<li><code>0 &lt;= apples[i], days[i] &lt;= 2 * 104</code></li>
	<li>只有在 <code>apples[i] = 0</code> 时，<code>days[i] = 0</code> 才成立</li>
</ul>