2279. Maximum Bags With Full Capacity of Rocks

Question

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

Solution

计算每个背包的剩余空间，然后进行排序。
按从小到大的顺序依次减少石头的总数。
设置i作为选取石头的计数。每次循环将i增加，直到剩余的石头数量小于0。
如果最后剩下的石头仍然大于等于0，则返回计数。
否则返回计数-1。

Code

class Solution {
    public int maximumBags(int[] capacity, int[] rocks, int additionalRocks) {
        int[] diff = new int[capacity.length];
        for(int i = 0 ; i < capacity.length; i++){
            diff[i] = capacity[i] - rocks[i];
        }
        
        Arrays.sort(diff);
        int i = 0;
        while(additionalRocks > 0 && i < diff.length){
            additionalRocks -= diff[i];
            i++;
        }
        return additionalRocks >= 0 ? i : i - 1;
    }
}