Problem Description

In this problem, we have an apple tree that grows a certain number of apples every day for n days. Each apple has an expiration period given by the corresponding index in the days array. This means that if the tree grows apples[i] apples on the i-th day, these apples will rot and become inedible after days[i] days. It is also mentioned that on some days, the tree might not grow any apples, which is represented by apples[i] == 0 and days[i] == 0.

You have a strategy where you eat at most one apple per day, and you're looking to maximize the number of apples you can consume. It's important to note that your apple-eating period can extend beyond the n days of apple growth, meaning you can continue eating the apples you've collected after the tree stops producing new ones.

The goal is to determine the maximum number of apples you can eat using the given apples and days arrays.

Intuition

The intuition behind the solution is to simulate the process of eating apples in a way that prioritizes eating the apples that will rot soonest. This helps in maximizing the number of apples eaten and minimizing waste. We can achieve this through the use of a min-heap data structure.

The min-heap will store pairs of values: the expiration date of the apples and the quantity of those apples. By ensuring that the heap is always sorted by the expiration date (the sooner-to-rot apples at the top), we can efficiently determine which apple to eat next.

Our approach follows these steps:

1. Process each day from the given arrays, starting from day 0 until the last day when apples can potentially be eaten (i < n or q is not empty).
2. For each day, if the tree grows apples on that day (apples[i]), add these apples along with their expiration date to the min-heap.
3. After that, clean the heap by removing any apples that have already rotted (q[0][0] < i).
4. If the heap is not empty, eat one apple (the one at the top of the heap since it's the next to rot). Decrease the quantity of that type of apple and increment the total count of eaten apples (ans).
5. If there are still apples of the same type left and they haven’t expired yet, put them back into the min-heap with the updated quantity.
6. Repeat the above steps for each day, moving forward one day at a time (i += 1).
7. Continue even after day n until all the apples that we have are either eaten or rotten.

Using a heap is key here because it allows us to efficiently access the apples that will rot first (heap's root), eat them to prevent waste, and keep the remaining collection of apples well-organized for the upcoming days.

Learn more about Greedy and Heap (Priority Queue) patterns.

Solution Approach

The solution approach for this problem involves the use of a min-heap data structure, which in Python can be implemented using the heapq library. A min-heap helps us keep track of the apples with the earliest expiration date on top. Let's walk through the key steps in the implementation:

1. We start by initializing a heap q and setting i = 0 and ans = 0. The variable i tracks the current day, and ans is used to count the number of apples eaten.

2. We enter a while loop that continues as long as i < n (we're within the tree's apple-producing period) or we have apples in our heap q.

3. Inside the loop, if it's a productive day (apples[i] > 0), we add the apples and their expiration day (i + days[i] - 1) as a tuple to the heap. We use heappush from the heapq library to maintain the heap invariant (smallest expiration date on top).

4. After potentially adding new apples to the heap, we remove any apples that have already rotted. We do this by looping while the heap is not empty and the expiration date of the apples at the top of the heap is less than the current day q[0][0] < i. If the condition is met, we use heappop to remove the top element from the heap.

5. If there are any apples left in the heap (i.e., there are still unrotten apples), we eat one. Since we can only eat one apple a day, we remove the top element from the heap, decrease the apple count by one, and increase the ans variable to reflect the eaten apple.

6. If there are remaining apples of the type we just ate (their count is now v - 1) and they have not expired (t > i), we push them back onto the heap with the updated count.

7. We increment i to move to the next day and repeat the process until the heap is empty or we've simulated all days within the apple growth period n.

This solution approach ensures we are always eating the apple that is closest to rotting, thus maximizing the number of apples we can consume before they go bad.

Remember, the use of a heap is due to its efficient ability to always give us the smallest element, which in this case is the soonest expiration date. The heap's property of log-complexity for insertion and removal operations enables us to manage our collection of apples as they grow and rot throughout the simulated days effectively.

Example Walkthrough

Let's illustrate the solution approach with a small example. Suppose we have the following arrays for apples and days:

apples = [1, 2, 3, 0, 0]
days = [3, 2, 1, 0, 0]

This tells us that on day 0 we have 1 apple that expires after 3 days, on day 1 we have 2 apples that expire after 2 days, and on day 2 we have 3 apples that expire after 1 day.

Initialize the heap q, the current day i = 0, and the total eaten apples ans = 0.

Starting at day 0, let's walk through each day:

• Day 0: We have 1 apple that will expire after 3 days (i.e., it will rot on day 3). We add (3, 1) to the heap (the first value is the expiration day, and the second is the number of apples). Since there's only one apple and it's the only one in our heap, we eat it. Now, ans = 1.

• Day 1: We have 2 new apples expiring by day 3. Add (3, 2) to the heap. Remove the rotten apples from the top of the heap if any (but today, there are none). Now we eat one of the apples that expire by day 3. Thus, we update the heap to remain with (3, 1) for this type of apple, and ans becomes 2.

• Day 2: We have 3 new apples that will expire by the end of today, day 2. So we add (2, 3) to the heap. Before eating, we check for expired apples. Since it's day 2, we remove the (2, 3) set as they have expired, and then we eat one apple from the remaining set (3, 1), leaving none of that type. Now ans = 3.

• Day 3: We do not have any new apples. We check for expired apples, and the heaps top is (3, 1) which gets removed as it has expired today. Since there are no unexpired apples left, we can't eat any apples today.

• Day 4: No new apples, and the heap is empty. We don't have any apples to eat.

Therefore, using this strategy, the maximum number of apples we can eat is 3 (eaten on days 0, 1, and 2).

Solution Implementation

Time and Space Complexity

The time complexity of the code is O(n + k log k), where n is the length of the days array, and k is the total number of different apples received over the days. Each day up to day n, the operation either pushes (heappush) or pops (heappop) an item from the heap q, which takes O(log k) time. After day n, all elements in the heap q will be popped without any further pushes, so the complexity in that phase depends on the number of elements in the heap, which can be at most k. Since each element in the heap can only be pushed and popped once, the number of heap operations is limited to the number of elements k. Therefore, the total time spent on heap operations is O(k log k). Additionally, there is a constant amount of work done for each day i up to day n, contributing to the O(n) factor.

The space complexity of the code is O(k), as it requires storing each different apple's expiry date and count in the heap q. In the worst case, k could be as large as the number of days if we receive at least one apple on each day with different expiry dates. The heap size will never exceed the total number of apples because once an apple's expiry date has passed, it is removed from the heap.

Learn more about how to find time and space complexity quickly using problem constraints.