1705. Maximum Number of Eaten Apples

https://leetcode.com/problems/maximum-number-of-eaten-apples/

There is a special kind of apple tree that grows apples every day for n days. On the ith day, the tree grows apples[i] apples that will rot after days[i] days, that is on day i + days[i] the apples will be rotten and cannot be eaten. On some days, the apple tree does not grow any apples, which are denoted by apples[i] == 0 and days[i] == 0.

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

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

Example 1:

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.

Example 2:

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.

Constraints:

• apples.length == n

• days.length == n

• 1 <= n <= 2 * 104

• 0 <= apples[i], days[i] <= 2 * 104

• days[i] = 0 if and only if apples[i] = 0.

apples[i]表示在第i天能得到多少苹果，该天得到的苹果会在days[i]后腐烂，问每天最多吃一个苹果，最后最多能吃到多少苹果。遍历每一天，贪心的每次选最快过期的苹果吃一个，选最早/小用heap。

class Solution:
    def eatenApples(self, apples: List[int], days: List[int]) -> int:
        n, q, res, i = len(days), [], 0, 0
        while i < n or q:
            if i < n:
                heappush(q, [i + days[i], apples[i]])
            while q and (q[0][0] <= i or q[0][1] <= 0):
                heappop(q)
            if q:
                res += 1
                q[0][1] -= 1
            i += 1
        return res