LeetCode每日一题(2279. Maximum Bags With Full Capacity of Rocks)

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

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把所有袋子的容量与当前实际储量的差值计算出来然后排序， 用多出来的石头来填补这些差值， 直到剩余的石头不足以填补这些差值或者剩下了石头

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use std::cmp::Reverse;
use std::collections::BinaryHeap;

impl Solution {
    pub fn maximum_bags(capacity: Vec<i32>, rocks: Vec<i32>, mut additional_rocks: i32) -> i32 {
        let mut heap: BinaryHeap<Reverse<i32>> = capacity
            .into_iter()
            .zip(rocks)
            .map(|(c, r)| Reverse(c - r))
            .collect();
        let mut ans = 0;
        while let Some(Reverse(d)) = heap.pop() {
            if d > additional_rocks {
                return ans;
            }
            ans += 1;
            additional_rocks -= d;
        }
        ans
    }
}