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Maximum Number of Eaten Apples

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.

Solution Explanation: Maximum Number of Eaten Apples

This problem asks to find the maximum number of apples that can be eaten, given that you can eat at most one apple per day and apples rot after a certain number of days. The solution uses a greedy approach combined with a priority queue (min-heap) for efficient management of apples and their rotting times.

Approach:

The core idea is to prioritize eating apples that are closest to rotting. This ensures you maximize the number of apples consumed before they spoil. A min-heap data structure perfectly suits this task because it allows us to efficiently retrieve the apple with the soonest expiry date.

  1. Data Structure: A priority queue (min-heap) stores tuples of (expiry_day, apple_count). expiry_day represents the last day the apple can be eaten, and apple_count is the number of apples available on that expiry day.

  2. Iteration: The algorithm iterates through the days. For each day:

    • Add new apples: If apples are available on that day, a tuple representing the expiry date and quantity is added to the heap.
    • Remove rotten apples: Apples whose expiry date is before the current day are removed from the heap.
    • Eat an apple: If the heap isn't empty, the apple with the earliest expiry date is popped. The apple_count is decremented, and if there are still apples remaining (apple_count > 0), the updated tuple is pushed back onto the heap.
    • Increment day counter: The day counter is incremented.

Time Complexity Analysis:

  • Adding and removing elements from the heap takes O(log n) time, where n is the number of apples.
  • The loop iterates at most through all days and the maximum number of apples (in worst case), which is bounded by the total number of apples across all days.

Therefore, the overall time complexity is O(N log N + M), where N is the number of days, and M is the total number of apples (which could be greater than N). In many cases, M will be roughly proportional to N, so a simpler approximation of O(N log N) is often used.

Space Complexity Analysis:

The space complexity is dominated by the priority queue, which in the worst case, could store up to N tuples (one for each day if all days have apples). Therefore, the space complexity is O(N).

Code Examples:

The code examples provided in Python, Java, C++, and Go all follow the same algorithmic approach, using a priority queue (heap) to maintain the apples based on their expiry dates. The differences are primarily in the syntax and specific library functions used for heap implementation. Each language's specific heap implementation details are well-suited for optimal performance in their respective environments. The choice of language would depend on project requirements and personal preferences.