Problem Description

In this problem, we are given two arrays: 'boxes', which holds the heights of boxes, and 'warehouse', which holds the heights of the rooms in a warehouse. The boxes can only be pushed into the warehouse from left to right, and boxes cannot be stacked on top of one another. The goal is to find the maximum number of boxes that can fit into the warehouse according to the given rules.

One important rule to note is that if a box is too tall to fit into a room, it blocks all the boxes behind it from entering the warehouse as well. However, we can rearrange the boxes before we start placing them into the warehouse. Given these constraints, we need to find the best order of boxes and the strategy to place as many boxes as possible into the warehouse rooms.

Intuition

To solve this problem, we can start with a greedy approach. A greedy approach means that we make the locally optimal choice at each step with the hope that these choices lead us to a globally optimal solution.

Since we can rearrange the boxes, it will be beneficial to try to fit smaller boxes first as they have a higher chance of fitting into the rooms of the warehouse, leaving the larger boxes for the bigger rooms towards the end. We start by sorting the 'boxes' array to organize the boxes from smallest to largest.

Furthermore, as we can only push boxes from left to right, we need to pre-process the 'warehouse' array to accommodate the fact that if we encounter a smaller room, it will affect all subsequent rooms. To do that, we can create a new array, 'left', which represents the maximum height of a box that can be placed in the subsequent rooms after considering previous smaller rooms. We initialize the 'left' array with the first room's height and then, for each subsequent room, we take the minimum height between the current room and the previous room. This way, we account for the potential blocking effect caused by smaller rooms.

Finally, we iterate over both the sorted 'boxes' array and the pre-processed 'left' array from the end towards the front. We put a box into the warehouse if the box's height is lower than or equal to the current room's height. If the box is too tall, we move to the next room. We continue this process until we have placed all the possible boxes or we have checked all the rooms.

The variable 'i' will keep track of how many boxes have been placed, and when we either run out of boxes or rooms, the value of 'i' will give us the number of boxes that can be put into the warehouse.

Learn more about Greedy and Sorting patterns.

Solution Approach

The solution approach for this problem uses a greedy method, sorted arrays, and pre-processing of the warehouse data. The implementation in Python can be broken down into several clear steps:

1. Pre-Process the Warehouse Heights: The warehouse array is traversed to create a 'left' array. This array stores the maximum height of a box that can be inserted into the warehouse up to that point, where the value for each room is the minimum between the current room and the previous one:

   left = [warehouse[0]] * n
   for i in range(1, n):
       left[i] = min(left[i - 1], warehouse[i])

   This step ensures that any height restrictions imposed by a smaller room are carried over to all subsequent rooms.

2. Sort the Boxes: The boxes array is sorted in increasing order of their heights. This allows us to try fitting the smallest boxes first:

   boxes.sort()

3. Place Boxes into the Warehouse: Two pointers, i and j are used to iterate through the 'boxes' and 'left' arrays, respectively. The i pointer starts at 0 to point to the smallest box, whereas j starts at n - 1 to point to the last room of the warehouse. The algorithm proceeds to check if each box can fit into the current room of the warehouse. If a box fits (box height ≤ room height), both pointers are moved one step (to the next box and the next room), and the process is repeated. If a box does not fit, the pointer j is decremented to move to the next room that could potentially accommodate the current box:

   i, j = 0, n - 1
   while i < len(boxes):
       while j >= 0 and left[j] < boxes[i]:
           j -= 1
       if j < 0:
           break
       i, j = i + 1, j - 1

   The loop breaks when either we run out of boxes (i is equal to the length of boxes array) or when there are no more rooms to check (j is less than 0).

4. Return the Result: At the end of the iteration process, i represents the number of boxes that have been placed in the warehouse. Since i starts at 0, it also conveniently matches the count of boxes placed:

   return i

In essence, the algorithm sorts the boxes so that we attempt to place the smallest ones first, and then iteratively checks each box against the limitations of the warehouse. Using two pointers facilitates an efficient traversal over both arrays without the need to check all possible combinations, focusing only on feasible fits. The approach ensures that we maximize the number of boxes placed in the warehouse.

Example Walkthrough

Let's illustrate the solution approach with a small example:

Imagine we have a boxes array with heights [3, 8, 1] and a warehouse array with heights [5, 4, 3, 2, 1].

Step 1: Pre-Process the Warehouse Heights

First, we'll create the left array from the warehouse heights to make sure we account for the height constraints.

left = [5] (initialize with the first room's height)

Then we process the other rooms and get:

left = [5, 4, 3, 2, 1] (each entry is the minimum of the current room height and the previous entry in the left array)

Step 2: Sort the Boxes

We sort the boxes array to prioritize fitting smaller boxes first:

boxes = [1, 3, 8] (sorted in increasing order)

Step 3: Place Boxes into the Warehouse

Iterate with two pointers, i points to boxes, starting with the smallest, and j points to the left array, starting with the last position.

i = 0, j = 4

The smallest box (1) can fit in the last room (1), so we move to the next box and the next room.

i = 1, j = 3

The next box (3) can fit into the fourth room (2), but only just, so we skip this room and move to the one before it.

i = 1, j = 2

Now, the same box (3) fits into the third room (3), so we move to the next box and next room.

i = 2, j = 1

Our final box (8) cannot fit in the second room (4), nor in the first room (5), so we cannot place this box.

i = 2, j = -1

Step 4: Return the Result

We were able to place 2 boxes into the warehouse. Since i now equals 2, we return this value as the result.

The output for the example is 2, indicating the maximum number of boxes we can place in the warehouse with the given constraints.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the code provided can be analyzed in the following steps:

1. Constructing the left array by walking through the warehouse array and picking the minimum of the previous left entry and the current warehouse height.

   • This process takes O(n) time where n is the length of warehouse.

2. Sorting the boxes array.

   • Assuming the sorting algorithm is Timsort (the default in Python), this will take O(b log b) time where b is the number of boxes.

3. The two-pointer approach to count how many boxes can fit into the warehouse.

   • This loop runs at most min(b, n) times in the worst case, where b is the length of sorted boxes and n is the length of the warehouse array.
   • Therefore, the worst-case time complexity for this loop is O(min(b, n)).

Combining all the steps, the overall time complexity of the code is O(n) + O(b log b) + O(min(b, n)). Since the sorting step is likely to dominate the time complexity, we can simplify this to O(b log b) assuming b > n.

Space Complexity

The space complexity of the code provided is as follows:

1. The left array of size n is created to store the minimum height of the warehouse at every point.

   • This consumes O(n) space.

2. The sorting of the boxes is done in-place, which does not consume additional space (ignoring the space used by the sorting algorithm itself).

   • Python's Timsort requires O(log b) space.

Therefore, the total space complexity of the code is the larger of O(n) and O(log b), which simplifies to O(n) assuming n >= log b.

Learn more about how to find time and space complexity quickly using problem constraints.