Teaching Kids Programming – Greedy Algorithm to Redistribute Items to Boxes (Knapsack Problem)

Teaching Kids Programming: Videos on Data Structures and Algorithms

Apple Redistribution into Boxes

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.

Greedy Algorithm to Redistribute Items to Boxes (Knapsack Problem)

This is a type of the Knapsack Problems. To formulize the given problem as a Knapsack problem, we need to adapt it to the classic knapsack model. The standard knapsack problem typically involves selecting items with given weights to maximize the total weight without exceeding the capacity of the knapsack. However, the given problem involves distributing apple packs into boxes with certain capacities, and we need to minimize the number of boxes used.

We can solve this by greedily selecting a larger box to fill the items/apples. As we can unpack the packs of apples, so the only thing matters is the total number of the items:

Then, we start by sorting the capacity array in descending order. This helps to try and fill the boxes with the largest capacity first, potentially minimizing the number of boxes needed.

Then, we use a greedy approach to select the minimum number of boxes. Iterate through the sorted capacities and keep adding boxes until the sum of the capacities is at least total_apples. Alternatively, we can iterate over the boxes from larger to smaller capacity, and then deduct the apples until we have filled the boxes with the items.

class Solution:
    def minimumBoxes(self, apple: List[int], capacity: List[int]) -> int:
        ## O(N)
        apple = sum(apple)
        if apple > sum(capacity):
            return -1
        ## O(MLogM)
        capacity.sort(reverse=True)
        i = 0
        n = len(capacity)
        ## O(M)
        while apple:
            if apple > capacity[i]:
                apple -= capacity[i]
                i += 1
            else:
                break
        return i + 1

The time complexity of this solution is O(N+MLogM). We can improve this a little bit by applying the Binary Search: Teaching Kids Programming – Redistribute Items to Boxes (Knapsack Problem, Binary Search Algorithm)

–EOF (The Ultimate Computing & Technology Blog) —