Problem Description

You have two arrays: boxes containing heights of boxes (all unit width) and warehouse containing heights of n rooms in a warehouse. The warehouse rooms are numbered from 0 to n-1 from left to right, where warehouse[i] represents the height of room i.

The rules for placing boxes are:

• Boxes cannot be stacked on top of each other
• You can rearrange the boxes in any order you want
• You can push boxes into the warehouse from either the left side or the right side
• When pushing a box through the warehouse, if it encounters a room with height less than the box's height, the box gets stuck there and cannot go further. All subsequent boxes pushed from the same side will also be blocked

Your goal is to find the maximum number of boxes that can be placed in the warehouse.

For example, if you have a warehouse with room heights [4, 3, 5, 2, 6] and you're pushing a box of height 4 from the left, it will stop at room index 1 (height 3). If you push the same box from the right, it will stop at room index 3 (height 2).

The solution involves preprocessing the warehouse to determine the effective height constraint for each room (considering boxes can come from either direction), then using a greedy approach by sorting both arrays and matching smaller boxes to available rooms.

Intuition

The key insight is understanding what happens when we push boxes from either side. When a box is pushed from the left and gets stuck at position i, it means the box couldn't pass through some room before i that was too short. Similarly, when pushing from the right.

For any room at position i, we need to figure out: what's the tallest box that could actually reach this room? This depends on the minimum height of all rooms the box must pass through to get there.

If we push from the left to reach room i, the box must pass through rooms 0, 1, ..., i-1. So the constraint is min(warehouse[0], warehouse[1], ..., warehouse[i-1]).

If we push from the right to reach room i, the box must pass through rooms n-1, n-2, ..., i+1. So the constraint is min(warehouse[i+1], warehouse[i+2], ..., warehouse[n-1]).

Since we can choose which side to push from, the effective height limit for room i is the better of these two options: max(left_constraint, right_constraint). But we also can't exceed the room's own height, so the final constraint is min(warehouse[i], max(left_constraint, right_constraint)).

Once we know the effective height constraint for each room, the problem becomes simpler: we have a list of room capacities and a list of box heights, and we want to match as many boxes as possible to rooms.

The greedy approach works here: sort both arrays and try to place the smallest boxes first. For each box, find the smallest room that can accommodate it. This maximizes our chances of fitting more boxes because we're using the "cheapest" resources (smallest acceptable rooms) for each box, leaving larger rooms available for larger boxes.

Learn more about Greedy and Sorting patterns.

Solution Approach

The implementation consists of three main phases: preprocessing, sorting, and greedy matching.

Phase 1: Preprocessing the Warehouse

We create two auxiliary arrays left and right to track the height constraints:

• left[i] stores the minimum height among all rooms to the left of position i (rooms that a box must pass through when pushed from the left)
• right[i] stores the minimum height among all rooms to the right of position i (rooms that a box must pass through when pushed from the right)

left[0] = right[-1] = inf  # No constraints at the boundaries
for i in range(1, n):
    left[i] = min(left[i - 1], warehouse[i - 1])
for i in range(n - 2, -1, -1):
    right[i] = min(right[i + 1], warehouse[i + 1])

Then we update each room's effective height based on the better access path:

for i in range(n):
    warehouse[i] = min(warehouse[i], max(left[i], right[i]))

This gives us the maximum box height that can actually be placed in each room, considering access from either direction.

Phase 2: Sorting

Both arrays are sorted in ascending order:

boxes.sort()
warehouse.sort()

Phase 3: Greedy Matching

We use two pointers to match boxes to rooms:

ans = i = 0
for x in boxes:
    while i < n and warehouse[i] < x:
        i += 1
    if i == n:
        break
    ans, i = ans + 1, i + 1

For each box (starting from smallest), we find the first available room that can accommodate it. If found, we place the box and move to the next room. If we run out of rooms, we stop.

The time complexity is O(n log n + m log m) for sorting, where n is the warehouse size and m is the number of boxes. The space complexity is O(n) for the auxiliary arrays.

Example Walkthrough

Let's walk through a concrete example with warehouse = [4, 3, 5, 2, 6] and boxes = [3, 5, 5, 2].

Step 1: Preprocessing the Warehouse

First, we build the left and right arrays to track minimum heights:

• left[i] = minimum height of rooms to the left of position i
• right[i] = minimum height of rooms to the right of position i

Building left array (rooms that must be passed when pushing from left):

• left[0] = inf (no rooms to the left)
• left[1] = min(inf, warehouse[0]) = min(inf, 4) = 4
• left[2] = min(4, warehouse[1]) = min(4, 3) = 3
• left[3] = min(3, warehouse[2]) = min(3, 5) = 3
• left[4] = min(3, warehouse[3]) = min(3, 2) = 2

Building right array (rooms that must be passed when pushing from right):

• right[4] = inf (no rooms to the right)
• right[3] = min(inf, warehouse[4]) = min(inf, 6) = 6
• right[2] = min(6, warehouse[3]) = min(6, 2) = 2
• right[1] = min(2, warehouse[2]) = min(2, 5) = 2
• right[0] = min(2, warehouse[1]) = min(2, 3) = 2

Now we update each room's effective height:

• Room 0: min(4, max(left[0], right[0])) = min(4, max(inf, 2)) = min(4, inf) = 4
• Room 1: min(3, max(left[1], right[1])) = min(3, max(4, 2)) = min(3, 4) = 3
• Room 2: min(5, max(left[2], right[2])) = min(5, max(3, 2)) = min(5, 3) = 3
• Room 3: min(2, max(left[3], right[3])) = min(2, max(3, 6)) = min(2, 6) = 2
• Room 4: min(6, max(left[4], right[4])) = min(6, max(2, inf)) = min(6, inf) = 6

Updated warehouse = [4, 3, 3, 2, 6]

Step 2: Sorting

• Sort boxes: [2, 3, 5, 5]
• Sort warehouse: [2, 3, 3, 4, 6]

Step 3: Greedy Matching

Using two pointers, match boxes to rooms:

1. Box height 2, room pointer at index 0:

   • Room height 2 ≥ box height 2 ✓
   • Place box, increment both pointers
   • Count = 1

2. Box height 3, room pointer at index 1:

   • Room height 3 ≥ box height 3 ✓
   • Place box, increment both pointers
   • Count = 2

3. Box height 5, room pointer at index 2:

   • Room height 3 < box height 5, skip room
   • Room height 4 < box height 5, skip room
   • Room height 6 ≥ box height 5 ✓
   • Place box, increment both pointers
   • Count = 3

4. Box height 5, room pointer at index 5:

   • No more rooms available
   • Stop

Result: Maximum 3 boxes can be placed in the warehouse.

Solution Implementation

Time and Space Complexity

The time complexity of this code is O(n + m log m), where n is the length of the warehouse array and m is the length of the boxes array.

Breaking down the time complexity:

• The first loop to build the left array: O(n)
• The second loop to build the right array: O(n)
• The third loop to update warehouse values: O(n)
• Sorting the boxes array: O(m log m)
• Sorting the warehouse array: O(n log n)
• The final loop to match boxes with warehouse positions: O(m + n) in the worst case

The dominant operations are the sorting steps, giving us O(n log n + m log m). However, if we consider that typically m might be different from n, the more precise complexity would be O(n log n + m log m). When the reference answer states O(n log n), it appears to be considering the case where the warehouse sorting dominates, or potentially treating both arrays as having similar sizes.

The space complexity is O(n), where n is the length of the warehouse array.

Breaking down the space complexity:

• The left array: O(n)
• The right array: O(n)
• The sorting operations use O(log n) and O(log m) for the recursive stack space (assuming Python's Timsort)

The total space complexity is O(n) as the auxiliary arrays dominate the space usage.

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

Common Pitfalls

Pitfall 1: Misunderstanding the Warehouse Update Logic

A common mistake is incorrectly calculating the effective height for each warehouse position. Developers often write:

Incorrect:

warehouse[i] = max(left[i], right[i])  # Wrong!

This would mean a box needs to be tall enough for BOTH paths, when actually we want the BETTER of the two paths.

Correct:

warehouse[i] = min(warehouse[i], max(left[i], right[i]))

The logic is: we take the maximum of the two path constraints (choosing the better path), then ensure it doesn't exceed the room's actual height.

Pitfall 2: Off-by-One Errors in Preprocessing

Another frequent error occurs when building the left and right arrays with incorrect index boundaries:

Incorrect:

# Wrong boundary conditions
for i in range(n):  # Should start from 1
    left[i] = min(left[i - 1], warehouse[i - 1])  # Would cause index error

for i in range(n - 1, -1, -1):  # Should start from n-2
    right[i] = min(right[i + 1], warehouse[i + 1])  # Would cause index error

Correct:

for i in range(1, n):  # Start from 1 since left[0] is already set
    left[i] = min(left[i - 1], warehouse[i - 1])

for i in range(n - 2, -1, -1):  # Start from n-2 since right[n-1] is already set
    right[i] = min(right[i + 1], warehouse[i + 1])

Pitfall 3: Modifying Original Input Unnecessarily

If the problem requires preserving the original warehouse array, directly modifying it would be incorrect:

Better Practice:

# Create a copy if original needs to be preserved
effective_heights = warehouse.copy()
# Then modify effective_heights instead of warehouse

Pitfall 4: Forgetting Edge Cases

Not handling edge cases like empty arrays:

Add validation:

if not boxes or not warehouse:
    return 0

These pitfalls can lead to incorrect results or runtime errors, so careful attention to the preprocessing logic and boundary conditions is essential.