3074. Apple Redistribution into Boxes
Problem Description
You have two lists: one represents the number of apples in n
different packs (apple
), and the other represents the capacity of m
boxes (capacity
). The goal here is to fit all the apples from the packs into the boxes. Now, you can distribute apples from any single pack into multiple boxes if necessary, but what you're trying to find out is the smallest number of boxes you can use to hold all the apples.
Imagine you're moving and have a collection of differently sized boxes and many items of varying amounts. You'll want to use as few boxes as possible, by filling up the largest boxes first. This problem demands a similar strategy. We want to use the biggest boxes to their full potential to minimize the number of boxes used overall.
Intuition
The underlying intuition of the solution is based on maximizing the utilization of larger boxes to minimize the total number used. Think about filling up a water tank using different-sized bucketsāyou'd use the largest buckets first to fill it more quickly.
Applying this mentality to the apples and boxes, you would sort the boxes from largest to smallest capacity. By using the bigger boxes first, you ensure that each box holds as many apples as possible, decreasing the total number of boxes needed.
Once sorted, it's simply a matter of going through the boxes in order, adding their capacity to a running total. You keep adding the capacities until you've accounted for all the apples. The number of boxes added at the point where the running total of capacity exceeds or meets the total number of apples is the minimum number of boxes needed.
This approach is known as a greedy algorithm because at each step, you're making the choice that seems best at the moment (using the largest box available).
Solution Approach
The given solution uses a simple yet effective greedy algorithm. The algorithm can be described by the following steps:
-
Sorting: First, the algorithm sorts the box capacities in descending order. This is done using the built-in
sort
method with thereverse=True
flag set, which reorders thecapacity
list from highest to lowest. This allows us to use the boxes with the most capacity first. -
Summation: Before we start allocating apples to boxes, we calculate the total number of apples we need to pack by summing all the elements of the
apple
array. This gives us the variables
, which represents the sum of all apples. -
Allocation: We then go through the sorted list of boxes and subtract the capacity of each box from the running total of apples
s
. We start with the largest box and work our way down to the smallest. -
Check and Return: After each box's capacity is subtracted from the total
s
, we check ifs
becomes less than or equal to zero. This check is done after each box is accounted for in the loopfor i, c in enumerate(capacity, 1)
. Ifs
is less than or equal to zero, it means all apples have been allocated into the boxes we've considered so far. We returni
, the index representing the count of boxes used, at that point.
This approach makes efficient use of the available space by prioritizing larger boxes, ensuring that the number of boxes we end up using is minimized. It's a classic example of a greedy algorithm, where making the locally optimal choice (using the biggest box next) also leads to a global optimum (using the smallest number of boxes).
class Solution:
def minimumBoxes(self, apple: List[int], capacity: List[int]) -> int:
capacity.sort(reverse=True) # Step 1: [Sorting](/problems/sorting_summary)
s = sum(apple) # Step 2: Summation
for i, c in enumerate(capacity, 1): # Step 3: Allocation
s -= c
if s <= 0: # Step 4: Check and Return
return i
We don't need complex data structures here; a simple list and basic operations like sorting and iteration are sufficient to implement this algorithm effectively.
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Start EvaluatorExample Walkthrough
Suppose we have the following small example:
apple
= [4, 5, 7]capacity
= [6, 4, 10, 3]
The question is: What is the smallest number of boxes we can use to fit all the apples?
Let's follow the solution steps to find out:
-
Sorting:
- We sort
capacity
in descending order: [10, 6, 4, 3]
- We sort
-
Summation:
- By summing [4, 5, 7], we find that
s
(total number of apples) is 16.
- By summing [4, 5, 7], we find that
-
Allocation:
- We take boxes in order from our sorted list, subtract their capacity from
s
, and count the number of boxes used. - 1st box:
s
= 16 - 10 = 6 (1 box used) - 2nd box:
s
= 6 - 6 = 0 (2 boxes used)
- We take boxes in order from our sorted list, subtract their capacity from
-
Check and Return:
- After using the second box,
s
is now 0, which means we have allocated all the apples into the boxes.
- After using the second box,
According to the walk through, to fit all our apples, we need a minimum of 2 boxes. This concludes our allocation, and the function would return i = 2
.
Solution Implementation
1class Solution:
2 def minimum_boxes(self, apples: List[int], capacities: List[int]) -> int:
3 # Sort the capacities in descending order
4 capacities.sort(reverse=True)
5
6 # Calculate the total number of apples to distribute
7 total_apples = sum(apples)
8
9 # Initialize the number of boxes used
10 boxes_used = 0
11
12 # Iterate over the sorted capacities to distribute the apples
13 for capacity in capacities:
14 # Subtract the current box capacity from the total apples
15 total_apples -= capacity
16
17 # Increment the number of boxes used
18 boxes_used += 1
19
20 # Check if all apples have been distributed
21 if total_apples <= 0:
22 # If all apples are distributed, return the number of boxes used
23 return boxes_used
24
25 # If the code reaches here, it implies more boxes are needed
26 # than are available in 'capacities' to store all 'apples'
27 raise ValueError("Insufficient number of boxes to store all apples")
28
29# The List type needs to be imported from typing module
30from typing import List
31
1import java.util.Arrays; // Required for using the Arrays.sort() method
2
3class Solution {
4
5 /**
6 * Finds the minimum number of containers required to store all apples.
7 *
8 * @param apples Array representing the number of apples in each box.
9 * @param capacities Array representing the capacity of each container.
10 * @return The minimum number of containers required.
11 */
12 public int minimumBoxes(int[] apples, int[] capacities) {
13 // Sort the capacities array in ascending order so we can use the largest capacities last
14 Arrays.sort(capacities);
15
16 // Calculate the total number of apples that need to be stored.
17 int totalApples = 0;
18 for (int apple : apples) {
19 totalApples += apple;
20 }
21
22 // Start using containers from the largest to store the apples.
23 for (int i = 1, n = capacities.length;; ++i) {
24 // Subtract the capacity of the used container from the total apples count.
25 totalApples -= capacities[n - i];
26
27 // Check if all apples are stored. If so, return the number of containers used.
28 if (totalApples <= 0) {
29 return i;
30 }
31 }
32 // Note: The loop will always terminate with a return inside the loop,
33 // so there is no need for an additional return statement here.
34 }
35}
36
1#include <vector>
2#include <algorithm>
3#include <numeric>
4
5class Solution {
6public:
7 // Function to determine the minimum number of boxes required to hold
8 // a specific number of apples.
9 int minimumBoxes(vector<int>& apples, vector<int>& capacities) {
10 // Sort capacities in non-increasing order to use the largest boxes first.
11 sort(capacities.rbegin(), capacities.rend());
12
13 // Accumulate the total number of apples that need to be boxed.
14 int totalApples = accumulate(apples.begin(), apples.end(), 0);
15
16 // Iterate through the sorted capacities to find the minimum number of boxes required.
17 for (int boxCount = 1; ; ++boxCount) {
18 // Subtract the current box capacity from the total apples.
19 totalApples -= capacities[boxCount - 1];
20
21 // If all apples are accounted for with the current number of boxes, return it.
22 if (totalApples <= 0) {
23 return boxCount;
24 }
25 }
26 // Note: The loop has no exit condition besides the return within the loop,
27 // which assumes that the given 'capacities' vector is sufficient.
28 }
29};
30
1function minimumBoxes(apples: number[], capacities: number[]): number {
2 // Sort the capacities array in descending order
3 capacities.sort((a, b) => b - a);
4
5 // Calculate the total number of apples
6 let totalApples = apples.reduce((accumulator, current) => accumulator + current, 0);
7
8 // Initialize the index (which will represent the number of boxes used)
9 let boxIndex = 0;
10
11 // Iterate until all apples are placed in boxes
12 while (totalApples > 0 && boxIndex < capacities.length) {
13 // Deduct the capacity of the current largest box from total apples
14 totalApples -= capacities[boxIndex];
15
16 // Move to the next box
17 boxIndex++;
18 }
19
20 // If totalApples is less than or equal to 0, all apples are in boxes
21 // Return the count of boxes used (boxIndex)
22 // If totalApples is not less than or equal to 0, we've run out of boxes
23 // before accommodating all apples, hence return boxIndex
24 return boxIndex;
25}
26
Time and Space Complexity
The time complexity of the function minimumBoxes
is O(m * log(m) + n)
. This is because the sort function applied to the capacity
list has a complexity of O(m * log(m))
, where m
is the length of the capacity
list. Following the sort, there is a for loop which iterates over the sorted capacity
list, and this loop may run up to n
times, where n
is the length of the apple
list. Therefore, the iteration adds an O(n)
complexity to the total, making the combined time complexity O(m * log(m) + n)
.
The space complexity of the function is O(log(m))
. This is due to the space needed for the sorting algorithm for a list of length m
. Most sorting algorithms, such as Timsort (used in Python's sort function), have a logarithmic space footprint because they need additional space to temporarily store elements while sorting.
Learn more about how to find time and space complexity quickly using problem constraints.
What does the following code do?
1def f(arr1, arr2):
2 i, j = 0, 0
3 new_arr = []
4 while i < len(arr1) and j < len(arr2):
5 if arr1[i] < arr2[j]:
6 new_arr.append(arr1[i])
7 i += 1
8 else:
9 new_arr.append(arr2[j])
10 j += 1
11 new_arr.extend(arr1[i:])
12 new_arr.extend(arr2[j:])
13 return new_arr
14
1public static List<Integer> f(int[] arr1, int[] arr2) {
2 int i = 0, j = 0;
3 List<Integer> newArr = new ArrayList<>();
4
5 while (i < arr1.length && j < arr2.length) {
6 if (arr1[i] < arr2[j]) {
7 newArr.add(arr1[i]);
8 i++;
9 } else {
10 newArr.add(arr2[j]);
11 j++;
12 }
13 }
14
15 while (i < arr1.length) {
16 newArr.add(arr1[i]);
17 i++;
18 }
19
20 while (j < arr2.length) {
21 newArr.add(arr2[j]);
22 j++;
23 }
24
25 return newArr;
26}
27
1function f(arr1, arr2) {
2 let i = 0, j = 0;
3 let newArr = [];
4
5 while (i < arr1.length && j < arr2.length) {
6 if (arr1[i] < arr2[j]) {
7 newArr.push(arr1[i]);
8 i++;
9 } else {
10 newArr.push(arr2[j]);
11 j++;
12 }
13 }
14
15 while (i < arr1.length) {
16 newArr.push(arr1[i]);
17 i++;
18 }
19
20 while (j < arr2.length) {
21 newArr.push(arr2[j]);
22 j++;
23 }
24
25 return newArr;
26}
27
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