README_EN.md

1705. Maximum Number of Eaten Apples

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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.

Solutions

Python3

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

Java

class Solution {
    public int eatenApples(int[] apples, int[] days) {
        PriorityQueue<int[]> q = new PriorityQueue<>(Comparator.comparingInt(a -> a[0]));
        int n = days.length;
        int ans = 0, i = 0;
        while (i < n || !q.isEmpty()) {
            if (i < n && apples[i] > 0) {
                q.offer(new int[] {i + days[i] - 1, apples[i]});
            }
            while (!q.isEmpty() && q.peek()[0] < i) {
                q.poll();
            }
            if (!q.isEmpty()) {
                var p = q.poll();
                ++ans;
                if (--p[1] > 0 && p[0] > i) {
                    q.offer(p);
                }
            }
            ++i;
        }
        return ans;
    }
}

C++

using pii = pair<int, int>;

class Solution {
public:
    int eatenApples(vector<int>& apples, vector<int>& days) {
        priority_queue<pii, vector<pii>, greater<pii>> q;
        int n = days.size();
        int ans = 0, i = 0;
        while (i < n || !q.empty()) {
            if (i < n && apples[i]) q.emplace(i + days[i] - 1, apples[i]);
            while (!q.empty() && q.top().first < i) q.pop();
            if (!q.empty()) {
                auto [t, v] = q.top();
                q.pop();
                --v;
                ++ans;
                if (v && t > i) q.emplace(t, v);
            }
            ++i;
        }
        return ans;
    }
};

Go

func eatenApples(apples []int, days []int) int {
	var h hp
	ans, n := 0, len(apples)
	for i := 0; i < n || len(h) > 0; i++ {
		if i < n && apples[i] > 0 {
			heap.Push(&h, pair{i + days[i] - 1, apples[i]})
		}
		for len(h) > 0 && h[0].first < i {
			heap.Pop(&h)
		}
		if len(h) > 0 {
			h[0].second--
			if h[0].first == i || h[0].second == 0 {
				heap.Pop(&h)
			}
			ans++
		}
	}
	return ans
}

type pair struct {
	first  int
	second int
}

type hp []pair

func (a hp) Len() int            { return len(a) }
func (a hp) Swap(i, j int)       { a[i], a[j] = a[j], a[i] }
func (a hp) Less(i, j int) bool  { return a[i].first < a[j].first }
func (a *hp) Push(x interface{}) { *a = append(*a, x.(pair)) }
func (a *hp) Pop() interface{}   { l := len(*a); t := (*a)[l-1]; *a = (*a)[:l-1]; return t }

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