How do you solve Maximum Number of Eaten Apples on LeetCode?

Maximum Number of Eaten Apples (LeetCode #1705) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: It's optimal to eat apples that will rot first, ensuring maximum consumption.

What is the brute force solution for Maximum Number of Eaten Apples?

The brute force approach for Maximum Number of Eaten Apples is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves simulating each day and checking which apples can be eaten. We can iterate through each day, and for each day, we check all the apples that are still edible to see if we can eat one. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Maximum Number of Eaten Apples?

Maximum Number of Eaten Apples uses the following concepts: Array, Greedy, Heap (Priority Queue). The recommended patterns to study are: Heap (Priority Queue), Greedy Algorithms.

What are the interview tips for Maximum Number of Eaten Apples?

Clearly explain your thought process and how you plan to manage the apples and their rot days. Be ready to discuss edge cases, such as when there are no apples or when all apples rot before they can be eaten. Practice implementing both brute-force and optimal solutions to solidify your understanding.

What are common mistakes when solving Maximum Number of Eaten Apples?

Not considering the days when apples are available vs. when they rot. Failing to manage the heap correctly, leading to incorrect counts of eaten apples.

#1705

Maximum Number of Eaten Apples

Medium

ArrayGreedyHeap (Priority Queue)Heap (Priority Queue)Greedy Algorithms

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Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

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Intuition

Time O(n log n)Space O(n)

The optimal solution uses a priority queue (or max-heap) to efficiently track the apples that can be eaten based on their rot days. This allows us to always eat the apple that will rot the soonest, maximizing the total number of apples eaten.

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Algorithm

4 steps

1Step 1: Initialize a max-heap to keep track of available apples and their expiration days.
2Step 2: Iterate through each day, adding new apples to the heap.
3Step 3: On each day, check the heap for the apple that will rot the soonest and eat it if available.
4Step 4: Remove any rotten apples from the heap before eating.

solution.py18 lines

1# Full working Python code
2import heapq
3
4def maxEatenApples(apples, days):
5    max_heap = []
6    total_eaten = 0
7    for day in range(len(apples) + max(days)):
8        if day < len(apples) and apples[day] > 0:
9            heapq.heappush(max_heap, (day + days[day], apples[day]))
10        while max_heap and max_heap[0][0] <= day:
11            heapq.heappop(max_heap)  # Remove rotten apples
12        if max_heap:
13            total_eaten += 1
14            max_heap[0] = (max_heap[0][0], max_heap[0][1] - 1)
15            if max_heap[0][1] == 0:
16                heapq.heappop(max_heap)
17    return total_eaten
18

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Complexity note: The complexity is O(n log n) due to the operations on the priority queue, where n is the number of days. Each insertion and deletion operation takes O(log n) time.

1It's optimal to eat apples that will rot first, ensuring maximum consumption.
2Using a priority queue helps efficiently manage which apples to eat based on their expiration.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.