LeetCode 3074 - Apple Redistribution into Boxes

LeetCode Problem 3074

Difficulty: 🟢 Easy
Topics: Array, Greedy, Sorting

Solution

Problem Understanding

The problem asks us to redistribute a collection of apple packs into boxes while using the minimum number of boxes. Each pack apple[i] contains a certain number of apples, and each box capacity[j] can hold up to a certain number of apples. Apples from a single pack can be split across multiple boxes, so we are not constrained to keep a pack intact. The input arrays apple and capacity represent these quantities. The expected output is a single integer: the minimum number of boxes required so that the total capacity of the selected boxes is at least the total number of apples.

Constraints indicate that both arrays have a maximum length of 50, and each value is between 1 and 50. Importantly, the input guarantees that a solution always exists. Edge cases include scenarios where the total apple count is exactly the sum of the largest box capacities, when a single box can hold all apples, or when apples are spread such that almost all boxes are required.

Approaches

The brute-force approach would enumerate all combinations of boxes, checking whether the sum of capacities meets the total apples. We would start with 1 box, then 2, and so on until we find a valid combination. While this guarantees the correct answer, the combinatorial explosion makes it infeasible because the number of subsets of boxes grows exponentially with m (2^m).

The optimal approach uses a greedy observation: since we want the minimum number of boxes, we should always pick the boxes with the largest capacities first. Sorting the capacity array in descending order and iteratively adding boxes until the sum meets or exceeds the total apples ensures that we use the fewest boxes possible. This works because any smaller box we skip could only be replaced by larger boxes later, which would not reduce the number of boxes used.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(2^m * m)	O(m)	Check all subsets of boxes to find minimum number satisfying total apples
Optimal	O(m log m)	O(1)	Sort boxes descending and sum until total apples are met

Algorithm Walkthrough

1. Compute the total number of apples by summing all elements in apple. This gives the target we need to meet with box capacities.
2. Sort the capacity array in descending order so the largest boxes come first.
3. Initialize a running sum of box capacities and a counter for the number of boxes selected.
4. Iterate through the sorted capacities, adding each capacity to the running sum and incrementing the box counter.
5. Stop once the running sum is greater than or equal to the total number of apples. Return the counter as the minimum number of boxes required.

Why it works: Sorting by descending capacity ensures that we always use the largest boxes first, minimizing the number of boxes. Since each box contributes as much as possible to reaching the target, any solution using smaller boxes would require at least as many boxes. The greedy choice is therefore optimal. The problem asks us to redistribute a set of apple packs into boxes in a way that minimizes the number of boxes used. We are given an array apple where each element represents the number of apples in a pack, and an array capacity where each element represents the maximum number of apples a box can hold. The output should be the minimum number of boxes needed such that all apples can be placed without exceeding any box's capacity. Apples from the same pack can be split across multiple boxes, so the only constraint is the total number of apples relative to the total box capacity.

The constraints indicate that the arrays are small (n, m <= 50) and the maximum value in any array is 50. This ensures that any solution with complexity roughly O(n log n + m log m) or linear scans will be efficient. The problem guarantees that a solution always exists, so we do not need to handle impossible distributions.

Important edge cases include situations where a single box can hold all apples, where each box is smaller than some packs requiring multiple boxes to accommodate a single pack, and cases where all boxes must be used.

Approaches

A brute-force approach would be to try all combinations of boxes to find the minimum number needed that can accommodate all apples. This could involve generating all subsets of capacity and checking if their sum is greater than or equal to the total number of apples. While this guarantees correctness, it is infeasible because the number of subsets is 2^m, which can be over a million for m = 50.

The key observation for an optimal approach is that we do not need to consider complex distributions of individual packs. Since apples can be split freely, the problem reduces to a simple greedy strategy: we need to cover the total number of apples with the fewest boxes, which is achieved by selecting the largest boxes first. Sorting capacity in descending order and summing from largest to smallest ensures that we reach the required total with the fewest boxes.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(2^m)	O(m)	Generate all subsets of boxes and check if total capacity suffices
Optimal	O(m log m)	O(1)	Sort capacities descending, sum until total apples are covered

Algorithm Walkthrough

1. Compute the total number of apples by summing all elements in the apple array. This represents the total capacity we need to cover.
2. Sort the capacity array in descending order. By using the largest boxes first, we minimize the number of boxes needed.
3. Initialize a counter for the number of boxes used and a running sum of capacity.
4. Iterate through the sorted capacity array, adding each box's capacity to the running sum and incrementing the box counter.
5. Stop the iteration once the running sum is greater than or equal to the total number of apples.
6. Return the number of boxes counted as the minimum number required.

Why it works: The invariant is that at each step, we are using the largest remaining box to cover as many apples as possible. Since we only need the total capacity and can split apples arbitrarily, this greedy choice is optimal. Any smaller box would require at least one additional box to achieve the same total capacity.

Python Solution

from typing import List

class Solution:
    def minimumBoxes(self, apple: List[int], capacity: List[int]) -> int:
        total_apples = sum(apple)
        capacity.sort(reverse=True)
        used_boxes = 0
        current_sum = 0
        
        for c in capacity:
            current_sum += c
            used_boxes += 1
            if current_sum >= total_apples:
                return used_boxes
        current_capacity = 0
        
        for c in capacity:
            current_capacity += c
            used_boxes += 1
            if current_capacity >= total_apples:
                break
        
        return used_boxes

The Python implementation first calculates the total number of apples. It then sorts the capacities in descending order. Using a loop, it accumulates capacities and counts the boxes used. As soon as the accumulated capacity meets or exceeds the total apples, it returns the count. This follows the algorithm steps exactly and handles edge cases naturally. This Python implementation first calculates the total number of apples. It then sorts the capacities in descending order and iteratively sums them until the total capacity meets or exceeds the total apples. The used_boxes counter tracks how many boxes have been needed so far, and this is returned as the final answer.

Go Solution

import "sort"

func minimumBoxes(apple []int, capacity []int) int {
    totalApples := 0
    for _, a := range apple {
        totalApples += a
    }
    
    sort.Slice(capacity, func(i, j int) bool {
        return capacity[i] > capacity[j]
    })
    
    usedBoxes := 0
    currentSum := 0
    for _, c := range capacity {
        currentSum += c
        usedBoxes++
        if currentSum >= totalApples {
            return usedBoxes
    sort.Sort(sort.Reverse(sort.IntSlice(capacity)))
    
    usedBoxes := 0
    currentCapacity := 0
    
    for _, c := range capacity {
        currentCapacity += c
        usedBoxes++
        if currentCapacity >= totalApples {
            break
        }
    }
    
    return usedBoxes
}

The Go implementation mirrors Python. Sorting uses sort.Slice with a descending comparator. The main loop accumulates box capacities and counts used boxes. Nil slices are not an issue because the constraints guarantee at least one box.

Worked Examples

Example 1: apple = [1,3,2], capacity = [4,3,1,5,2]

Total apples = 1+3+2 = 6

Sort capacities descending: [5,4,3,2,1]

Iteration:

1. Add 5 → current sum = 5, boxes = 1 (not enough)
2. Add 4 → current sum = 9, boxes = 2 (≥ 6, stop)

Output: 2

Example 2: apple = [5,5,5], capacity = [2,4,2,7]

Total apples = 15

Sort capacities descending: [7,4,2,2]

Iteration:

1. Add 7 → sum = 7, boxes = 1
2. Add 4 → sum = 11, boxes = 2
3. Add 2 → sum = 13, boxes = 3
4. Add 2 → sum = 15, boxes = 4 (≥ 15, stop)

Output: 4 In Go, we calculate the total number of apples similarly. We sort the capacity slice in descending order using sort.Reverse. The main loop sums capacities and counts boxes until the total apples are covered. Go requires explicit type handling and slice operations, but the algorithm logic mirrors the Python version.

Worked Examples

Example 1: apple = [1,3,2], capacity = [4,3,1,5,2]

Total apples = 1 + 3 + 2 = 6

Sort capacities descending: [5,4,3,2,1]

Step	Box Capacity Added	Running Sum	Boxes Used
1	5	5	1
2	4	9	2

Running sum 9 >= 6 → stop. Return 2.

Example 2: apple = [5,5,5], capacity = [2,4,2,7]

Total apples = 15

Sort capacities descending: [7,4,2,2]

Step	Box Capacity Added	Running Sum	Boxes Used
1	7	7	1
2	4	11	2
3	2	13	3
4	2	15	4

Running sum 15 >= 15 → stop. Return 4.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(m log m)	Sorting the capacity array dominates, the loop is O(m)
Space	O(1)	Only a few integers are stored, sorting is in-place

The reasoning: total apples sum is O(n), sorting is O(m log m), and the loop is O(m). Since n, m ≤ 50, the algorithm is efficient.

Test Cases

# Provided examples
assert Solution().minimumBoxes([1,3,2], [4,3,1,5,2]) == 2  # minimal 2 boxes
assert Solution().minimumBoxes([5,5,5], [2,4,2,7]) == 4    # all boxes needed

# Edge cases
assert Solution().minimumBoxes([50], [50]) == 1             # single box fits exactly
assert Solution().minimumBoxes([1]*50, [1]*50) == 50       # each box holds one apple
assert Solution().minimumBoxes([1]*50, [50]) == 1          # one box enough for all small packs
assert Solution().minimumBoxes([10,20,30], [15,25,20,5,30]) == 2  # mix, greedy picks largest first
| Time | O(m log m) | Sorting the `capacity` array dominates; summing apples is O(n) but n ≤ m |
| Space | O(1) | Only a few integer variables are used; in-place sort can avoid extra memory |

The sorting step is the most expensive operation. The linear pass to accumulate capacity is at most O(m), which is negligible compared to the sort. Space is constant since no additional data structures are required.

## Test Cases

```sql
# Basic examples
assert Solution().minimumBoxes([1,3,2], [4,3,1,5,2]) == 2  # Example 1
assert Solution().minimumBoxes([5,5,5], [2,4,2,7]) == 4  # Example 2

# Edge cases
assert Solution().minimumBoxes([10], [10]) == 1  # Single pack fits exactly in one box
assert Solution().minimumBoxes([1,1,1], [1,1,1]) == 3  # Each apple needs a separate box
assert Solution().minimumBoxes([50,50], [100,1,1]) == 1  # Large box covers all apples

# Random distribution
assert Solution().minimumBoxes([2,3,5], [5,5,2]) == 2  # Greedy selects 5+5 to cover 10 apples
assert Solution().minimumBoxes([1,2,3,4], [2,2,2,5]) == 3  # Select boxes 5+2+2 to cover 10 apples

Test	Why
[1,3,2], [4,3,1,5,2]	Standard case with several options
[5,5,5], [2,4,2,7]	Requires all boxes
[50], [50]	Exact fit with one box
[1]*50, [1]*50	Worst-case scenario, each apple requires a separate box
[1]*50, [50]	All small packs fit in one large box
[10,20,30], [15,25,20,5,30]	Greedy choice of largest capacities

Edge Cases

The first edge case is when a single box can hold all apples. This tests that the algorithm correctly stops after the first iteration rather than unnecessarily using more boxes. The implementation naturally handles this because the loop checks after each addition.

The second edge case is when each box can hold only one apple, and there are many small packs. This ensures that the algorithm counts each box individually and does not assume packs can be merged; the greedy selection still works because all boxes are equal.

The third edge case is when the largest boxes are not enough individually but a combination suffices. This tests that the algorithm correctly accumulates capacities in order and does not prematurely return, ensuring correctness for non-trivial combinations. | [1,3,2], [4,3,1,5,2] | Standard example, checks greedy choice | | [5,5,5], [2,4,2,7] | Requires all boxes, ensures full accumulation logic | | [10], [10] | Single element edge case | | [1,1,1], [1,1,1] | Each box exactly matches pack, no extra space | | [50,50], [100,1,1] | Tests greedy handling of one very large box | | [2,3,5], [5,5,2] | Random moderate numbers, check sum accumulation | | [1,2,3,4], [2,2,2,5] | Mix of capacities, ensures minimal boxes chosen |

Edge Cases

One important edge case is when a single box can hold all apples. A naive algorithm that tries to fit packs individually might use multiple boxes unnecessarily. Our greedy approach immediately selects the largest box first, handling this optimally.

Another edge case occurs when the total number of apples exactly equals the sum of the smallest boxes needed. If the implementation does not sort capacities descending, it might select smaller boxes first, using more than necessary. Sorting ensures we minimize box count.

A third edge case is when all packs are of size 1 but capacities vary. Some naive implementations might fail if they try to assign each pack to a single box instead of splitting across boxes. Our approach handles splitting naturally because we only consider total apples versus total capacity.