Introduction

In this blog post, we will explore how to solve the LeetCode problem "2279" 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

You have n bags numbered from 0 to n - 1. You are given two 0-indexed integer arrays capacity and rocks. The ith bag can hold a maximum of capacity[i] rocks and currently contains rocks[i] rocks. You are also given an integer additionalRocks, the number of additional rocks you can place in any of the bags. Return the maximum number of bags that could have full capacity after placing the additional rocks in some bags. Example 1: Input: capacity = [2,3,4,5], rocks = [1,2,4,4], additionalRocks = 2 Output: 3 Explanation: Place 1 rock in bag 0 and 1 rock in bag 1. The number of rocks in each bag are now [2,3,4,4]. Bags 0, 1, and 2 have full capacity. There are 3 bags at full capacity, so we return 3. It can be shown that it is not possible to have more than 3 bags at full capacity. Note that there may be other ways of placing the rocks that result in an answer of 3. Example 2: Input: capacity = [10,2,2], rocks = [2,2,0], additionalRocks = 100 Output: 3 Explanation: Place 8 rocks in bag 0 and 2 rocks in bag 2. The number of rocks in each bag are now [10,2,2]. Bags 0, 1, and 2 have full capacity. There are 3 bags at full capacity, so we return 3. It can be shown that it is not possible to have more than 3 bags at full capacity. Note that we did not use all of the additional rocks. Constraints: n == capacity.length == rocks.length 1 <= n <= 5 * 104 1 <= capacity[i] <= 109 0 <= rocks[i] <= capacity[i] 1 <= additionalRocks <= 109

Explanation

• Calculate Remaining Capacity: Determine the remaining capacity for each bag by subtracting the number of rocks it currently contains from its capacity.
  • Prioritize Bags: Sort the bags based on their remaining capacity in ascending order. This allows us to fill the bags requiring the least amount of additional rocks first.
  • Greedy Filling: Iterate through the sorted bags and greedily fill them with additional rocks until either all bags are full or we run out of additional rocks.

• Runtime Complexity: O(n log n), dominated by the sorting step. Storage Complexity: O(n), to store the remaining capacities.

Code

    def maximumBags(capacity, rocks, additionalRocks):
    n = len(capacity)
    remaining_capacity = []
    for i in range(n):
        remaining_capacity.append(capacity[i] - rocks[i])

    remaining_capacity.sort()

    full_bags = 0
    for i in range(n):
        if additionalRocks >= remaining_capacity[i]:
            additionalRocks -= remaining_capacity[i]
            full_bags += 1
        else:
            break

    return full_bags