2279.Maximum Bags With Full Capacity of Rocks
===

###### tags: `Medium`,`Array`,`Greedy`,`Sorting`

[2279. Maximum Bags With Full Capacity of Rocks](https://leetcode.com/problems/maximum-bags-with-full-capacity-of-rocks/)

### 題目描述

You have `n` bags numbered from `0` to `n - 1`. You are given two **0-indexed** integer arrays `capacity` and `rocks`. The i^th^ 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*.

### 範例

**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 * 10^4^
* 1 <= `capacity[i]` <= 10^9^
* 0 <= `rocks[i]` <= `capacity[i]`
* 1 <= `additionalRocks` <= 10^9^

### 解答
#### Python

def maximumBags(self, capacity: List[int], rocks: List[int], additionalRocks: int) -> int:

    sorted_remain_space = sorted( [x-y for x,y in zip(capacity, rocks)] )

    rock_cnt = 0
    for i in range(len(sorted_remain_space)):
        if rock_cnt + sorted_remain_space[i] > additionalRocks:
            return i
        rock_cnt += sorted_remain_space[i]

    return len(sorted_remain_space)

> [author=m4g0 ]

class Solution:
    def maximumBags(self, capacity: List[int], rocks: List[int], additionalRocks: int) -> int:
        remaining_capacity = [c - r for c, r in zip(capacity, rocks)]
        remaining_capacity = sorted(remaining_capacity)
        bags = 0

        for rc in remaining_capacity:
            if additionalRocks >= rc:
                additionalRocks -= rc
                bags += 1

        return bags

> [name=Kobe][time=Tue, Dec 27, 2022]

### Reference

[回到題目列表](https://hackmd.io/@Marsgoat/leetcode_every_day)