Introduction

In this blog post, we will explore how to solve the LeetCode problem "1705" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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.

Explanation

Here's the breakdown of the approach, complexities, and the Python code:

• High-Level Approach:

  • Use a min-heap to store the expiration dates of the available apples. The heap will store tuples: (expiration_date, number_of_apples).
  • Iterate through each day, adding new apples to the heap (if any).
  • Eat an apple from the heap with the earliest expiration date, ensuring to remove expired apples first.

• Complexity:

  • Runtime Complexity: O(n log n), where n is the length of the input arrays. The heap operations (push and pop) take logarithmic time, and these operations are performed at most n times.
  • Storage Complexity: O(n). In the worst case, the heap could store information about apples from each of the n days.

• Python Code:

Code

    import heapq
def eatenApples(apples, days):
    n = len(apples)
    heap = []  # (expiration_date, number_of_apples)
    eaten = 0
    day = 0

    while day < n or heap:
        # Add new apples to the heap
        if day < n and apples[day] > 0:
            heapq.heappush(heap, (day + days[day], apples[day]))

        # Remove expired apples from the heap
        while heap and heap[0][0] <= day:
            heapq.heappop(heap)

        # Eat an apple if available
        if heap:
            expiration, count = heapq.heappop(heap)
            eaten += 1
            if count > 1:
                heapq.heappush(heap, (expiration, count - 1))

        day += 1

    return eaten