1705. Maximum Number of Eaten Apples

Problem

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 * 10^4

• 0 <= apples[i], days[i] <= 2 * 10^4

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

Solution (Java)

class Solution {
    public int eatenApples(int[] apples, int[] days) {
        PriorityQueue<int[]> minHeap = new PriorityQueue<>((a, b) -> a[0] - b[0]);
        int eatenApples = 0;
        for (int i = 0; i < apples.length || !minHeap.isEmpty(); i++) {
            if (i < apples.length) {
                minHeap.offer(new int[] {i + days[i], apples[i]});
            }
            while (!minHeap.isEmpty() && (minHeap.peek()[0] <= i || minHeap.peek()[1] <= 0)) {
                minHeap.poll();
            }
            if (!minHeap.isEmpty()) {
                eatenApples++;
                minHeap.peek()[1]--;
            }
        }
        return eatenApples;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).