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3074. Apple Redistribution into Boxes

Description

You are given an array apple of size n and an array capacity of size m.

There are n packs where the ith pack contains apple[i] apples. There are m boxes as well, and the ith box has a capacity of capacity[i] apples.

Return the minimum number of boxes you need to select to redistribute these n packs of apples into boxes.

Note that, apples from the same pack can be distributed into different boxes.

 

Example 1:

Input: apple = [1,3,2], capacity = [4,3,1,5,2]
Output: 2
Explanation: We will use boxes with capacities 4 and 5.
It is possible to distribute the apples as the total capacity is greater than or equal to the total number of apples.

Example 2:

Input: apple = [5,5,5], capacity = [2,4,2,7]
Output: 4
Explanation: We will need to use all the boxes.

 

Constraints:

  • 1 <= n == apple.length <= 50
  • 1 <= m == capacity.length <= 50
  • 1 <= apple[i], capacity[i] <= 50
  • The input is generated such that it's possible to redistribute packs of apples into boxes.

Solutions

Solution 1: Greedy + Sorting

To minimize the number of boxes needed, we should prioritize using boxes with larger capacities. Therefore, we can sort the boxes in descending order of capacity, and then use the boxes one by one until all the apples are packed. We return the number of boxes used at this point.

The time complexity is $O(m \times \log m + n)$ and the space complexity is $O(\log m)$, where $m$ and $n$ are the lengths of the arrays capacity and apple respectively.

  • class Solution {
    public int minimumBoxes(int[] apple, int[] capacity) {
        Arrays.sort(capacity);
        int s = 0;
        for (int x : apple) {
            s += x;
        }
        for (int i = 1, n = capacity.length;; ++i) {
            s -= capacity[n - i];
            if (s <= 0) {
                return i;
            }
        }
    }
}

  • class Solution {
public:
    int minimumBoxes(vector<int>& apple, vector<int>& capacity) {
        sort(capacity.rbegin(), capacity.rend());
        int s = accumulate(apple.begin(), apple.end(), 0);
        for (int i = 1;; ++i) {
            s -= capacity[i - 1];
            if (s <= 0) {
                return i;
            }
        }
    }
};

  • class Solution:
    def minimumBoxes(self, apple: List[int], capacity: List[int]) -> int:
        capacity.sort(reverse=True)
        s = sum(apple)
        for i, c in enumerate(capacity, 1):
            s -= c
            if s <= 0:
                return i


  • func minimumBoxes(apple []int, capacity []int) int {
	sort.Ints(capacity)
	s := 0
	for _, x := range apple {
		s += x
	}
	for i := 1; ; i++ {
		s -= capacity[len(capacity)-i]
		if s <= 0 {
			return i
		}
	}
}

  • function minimumBoxes(apple: number[], capacity: number[]): number {
    capacity.sort((a, b) => b - a);
    let s = apple.reduce((acc, cur) => acc + cur, 0);
    for (let i = 1; ; ++i) {
        s -= capacity[i - 1];
        if (s <= 0) {
            return i;
        }
    }
}

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