1705. Maximum Number of Eaten Apples

Problem Link

Description

---

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:

• n == apples.length == days.length
• 1 <= n <= 2 * 104
• 0 <= apples[i], days[i] <= 2 * 104
• days[i] = 0 if and only if apples[i] = 0.

Solution

---

Python3

from heapq import heappop, heappush
class Solution:
    def eatenApples(self, apples: List[int], days: List[int]) -> int:
        n = len(apples)
        i = res = 0
        
        heap = []
        
        while True:
            if i < n:
                heappush(heap, (days[i] + i, apples[i]))
            
            while heap and i >= heap[0][0]:
                heappop(heap)
                
            if heap:
                rot, apple = heappop(heap)
                res += 1
                if apple > 1:
                    heappush(heap, (rot, apple - 1))
            
            
            if not heap and i >= n: break
            
            i += 1
        
        return res