#1705

Medium

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Maximum Number of Eaten Apples

Maximize apples eaten by choosing the best apples first, considering their rot days and available days to eat them.

Problem Statement

Given two arrays, apples and days, representing apples grown and the number of days until they rot, your task is to eat as many apples as possible. On each day, you can eat at most one apple, and you must decide which apple to eat based on how long it will last before it rots.

You are allowed to eat apples after the last day in the apples array, and on some days, no apples grow. The goal is to determine the maximum number of apples you can eat, considering the days on which apples rot.

Examples

Example 1

Input: apples = [1,2,3,5,2], days = [3,2,1,4,2]

Output: 7

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

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 Approach

Greedy Approach with Priority Queue

The best approach to solve this problem is by using a greedy algorithm combined with a priority queue. On each day, check the apples that are still fresh and select the apple that will rot soonest. This ensures you prioritize consuming apples that are about to decay.

Priority Queue to Track Apples

Use a max-heap (priority queue) to keep track of the apples available to eat, sorted by their expiration times. Every day, add new apples to the heap and remove the apples that have already rotted, ensuring efficient consumption of the available apples.

Invariant Validation

The greedy choice is validated by always eating the apple that is about to rot first. This invariant ensures that you are not wasting any available apples, thus maximizing the total number of apples you can eat.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of this approach is O(n log n) due to the heap operations for inserting and removing apples. The space complexity is O(n) as we store the apples in a priority queue.

What Interviewers Usually Probe

Look for a solution that handles the priority of rotting apples first.
Candidates should demonstrate the ability to use a priority queue or heap for tracking the apples.
Be cautious of inefficiencies when trying to handle the apples and days arrays.

Common Pitfalls or Variants

Common pitfalls

Not efficiently removing apples that have already rotted from the heap.
Failing to handle cases where no apples grow on a given day.
Not considering the scenario where you continue eating after the last day of apples.

Follow-up variants

What if the apples array has large gaps between days of growth?
How would this solution change if you had a maximum number of apples you could eat per day?
What if apples can be eaten multiple times before they rot?

FAQ

What is the optimal approach for the 'Maximum Number of Eaten Apples' problem?

The optimal approach involves using a greedy strategy combined with a priority queue to prioritize eating apples that will rot soonest.

How does the priority queue help in solving this problem?

A priority queue helps efficiently track the apples available to eat, allowing for quick selection of apples that will rot the soonest.

Can the 'Maximum Number of Eaten Apples' problem be solved with dynamic programming?

While dynamic programming is a potential option, the greedy strategy combined with a priority queue is more efficient for this specific problem.

What is the time complexity of the solution to the 'Maximum Number of Eaten Apples' problem?

The time complexity is O(n log n) due to the heap operations required for inserting and removing apples.

How does the solution handle cases where no apples grow on a particular day?

The solution accounts for days with no apples by simply skipping those days and continuing the process with the next available apples.

terminal

Solution

Solution 1: Greedy + Priority Queue

We can greedily choose the apples that are closest to rotting among the unrotten apples, so that we can eat as many apples as possible.

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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

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