LeetCode in Kotlin
1705. Maximum Number of Eaten Apples
Medium
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.length1 <= n <= 2 * 1040 <= apples[i], days[i] <= 2 * 104days[i] = 0if and only ifapples[i] = 0.
Solution
import java.util.PriorityQueue
class Solution {
fun eatenApples(apples: IntArray, days: IntArray): Int {
val minHeap = PriorityQueue { a: IntArray, b: IntArray -> a[0] - b[0] }
var eatenApples = 0
var i = 0
while (i < apples.size || minHeap.isNotEmpty()) {
if (i < apples.size) {
minHeap.offer(intArrayOf(i + days[i], apples[i]))
}
while (minHeap.isNotEmpty() && (minHeap.peek()[0] <= i || minHeap.peek()[1] <= 0)) {
minHeap.poll()
}
if (minHeap.isNotEmpty()) {
eatenApples++
minHeap.peek()[1]--
}
i++
}
return eatenApples
}
}