Apple Redistribution into Boxes

Easy

#2730Time: O(n + m log m), where `n` is the length of `apple` and `m` is the length of `capacity`. Summing the apples takes O(n) time. Sorting the capacities takes O(m log m) time. The final loop takes at most O(m) time. The sorting step dominates the complexity.Space: O(log m) or O(m), depending on the implementation of the sorting algorithm. In Java, `Arrays.sort` for primitive types uses a dual-pivot quicksort, which requires O(log m) space on average for the recursion stack.

Patterns

Greedy

Algorithms

Sorting

Data structures

Array

Prompt

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

There are n packs where the i^th pack contains apple[i] apples. There are m boxes as well, and the i^th 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.

Approaches

2 approaches with complexity analysis and trade-offs.

The core idea is to first determine the total number of apples that need to be stored. Since apples from a single pack can be split among multiple boxes, we only need the sum of all apples. To use the minimum number of boxes, we should greedily pick the boxes with the largest capacities first. A straightforward way to achieve this is by sorting the capacity array in descending order and then picking boxes one by one until all apples are accommodated.

Algorithm

Calculate totalApples by summing the apple array.
If totalApples is 0, return 0.
Sort the capacity array in ascending order.
Initialize boxesCount = 0.
Iterate through the capacity array from the last element to the first.
- Subtract the current capacity from totalApples.
- Increment boxesCount.
- If totalApples is now less than or equal to 0, break the loop.
Return boxesCount.

Walkthrough

The algorithm proceeds in these steps:

Calculate the total number of apples by summing all elements in the apple array. Let's call this totalApples.
If totalApples is zero, no boxes are needed, so we return 0.
Sort the capacity array. To easily access the largest capacities, we can sort it in ascending order and iterate from the end, or sort it in descending order and iterate from the beginning.
Initialize a boxesCount to 0. Iterate through the sorted capacities (from largest to smallest).
In each iteration, subtract the current box's capacity from totalApples and increment boxesCount.
Continue this process until totalApples is less than or equal to zero.
The final boxesCount is the minimum number of boxes required.

1import java.util.Arrays;2 3class Solution {4    public int minimumBoxes(int[] apple, int[] capacity) {5        int totalApples = 0;6        for (int a : apple) {7            totalApples += a;8        }9 10        if (totalApples == 0) {11            return 0;12        }13 14        // Sort capacity in ascending order15        Arrays.sort(capacity);16 17        int boxesCount = 0;18        // Iterate from the end to use largest capacities first19        for (int i = capacity.length - 1; i >= 0; i--) {20            totalApples -= capacity[i];21            boxesCount++;22            if (totalApples <= 0) {23                break;24            }25        }26        return boxesCount;27    }28}

Complexity

Time

O(n + m log m), where `n` is the length of `apple` and `m` is the length of `capacity`. Summing the apples takes O(n) time. Sorting the capacities takes O(m log m) time. The final loop takes at most O(m) time. The sorting step dominates the complexity.

Space

O(log m) or O(m), depending on the implementation of the sorting algorithm. In Java, `Arrays.sort` for primitive types uses a dual-pivot quicksort, which requires O(log m) space on average for the recursion stack.

Trade-offs

Pros

Simple to understand and implement.
It's a general solution that works regardless of the range of values in the capacity array.

Cons

The O(m log m) time complexity from sorting is not the most optimal solution possible given the problem's constraints.

This approach enhances the greedy strategy by avoiding a comparison-based sort. Given that the capacity values are limited to a small range (1 to 50), we can use a frequency array (a form of counting sort) to organize the boxes by capacity. This allows us to access the largest capacity boxes in constant time and leads to a more efficient linear time solution.

Algorithm

Calculate totalApples by summing the apple array.
If totalApples is 0, return 0.
Create a frequency array counts of size 51.
Populate counts by iterating through the capacity array: counts[cap]++.
Initialize boxesCount = 0.
Iterate c from 50 down to 1:
- Get the number of available boxes numBoxes = counts[c].
- Loop numBoxes times:
  - Increment boxesCount.
  - Subtract c from totalApples.
  - If totalApples <= 0, return boxesCount.
Return boxesCount.

Walkthrough

The improved algorithm works as follows:

As before, calculate the totalApples by summing the apple array.
Create a frequency array, counts, of size 51 (to cover capacities 1-50), initialized to all zeros.
Iterate through the capacity array. For each capacity c, increment counts[c]. This step takes O(m) time and groups all boxes by their capacity.
Initialize boxesCount = 0.
Iterate from the maximum possible capacity, 50, down to 1.
For each capacity c, check counts[c] to see how many boxes of this size are available.
Use these boxes one by one, decrementing totalApples and incrementing boxesCount for each box used, until totalApples is covered or you run out of boxes of capacity c.
As soon as totalApples becomes less than or equal to 0, we have found our answer and can return boxesCount immediately.

1class Solution {2    public int minimumBoxes(int[] apple, int[] capacity) {3        int totalApples = 0;4        for (int a : apple) {5            totalApples += a;6        }7 8        if (totalApples == 0) {9            return 0;10        }11 12        // Frequency array for capacities 1-5013        int[] counts = new int[51];14        for (int cap : capacity) {15            counts[cap]++;16        }17 18        int boxesCount = 0;19        // Iterate from largest capacity down to smallest20        for (int c = 50; c >= 1; c--) {21            int numBoxesAvailable = counts[c];22            for (int i = 0; i < numBoxesAvailable; i++) {23                boxesCount++;24                totalApples -= c;25                if (totalApples <= 0) {26                    return boxesCount;27                }28            }29        }30        return boxesCount; // Should be unreachable given problem constraints31    }32}

Complexity

Time

O(n + m + C), where `n` is the length of `apple`, `m` is the length of `capacity`, and `C` is the maximum possible capacity (50). This simplifies to O(n + m) because C is a constant. This is a linear time complexity, which is more efficient than the sorting approach.

Space

O(C), where `C` is the maximum capacity. Since `C` is a constant (50), the space complexity is O(1).

Trade-offs

Pros

Most efficient solution with linear time complexity.
Uses constant extra space, making it very memory-efficient.

Cons

This approach is specialized for problems where the range of input values (capacities) is small and bounded. It would be less practical if capacities could be very large.

Solutions

Solution

1class Solution {2public3  int minimumBoxes(int[] apple, int[] capacity) {4    Arrays.sort(capacity);5    int s = 0;6    for (int x : apple) {7      s += x;8    }9    for (int i = 1, n = capacity.length;; ++i) {10      s -= capacity[n - i];11      if (s <= 0) {12        return i;13      }14    }15  }16}

1class Solution {2public:3  int minimumBoxes(vector<int> &apple, vector<int> &capacity) {4    sort(capacity.rbegin(), capacity.rend());5    int s = accumulate(apple.begin(), apple.end(), 0);6    for (int i = 1;; ++i) {7      s -= capacity[i - 1];8      if (s <= 0) {9        return i;10      }11    }12  }13};

1class Solution:2    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

Video walkthrough

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