Put Boxes Into the Warehouse II

class Solution {
public:
    int maxBoxesInWarehouse(vector<int>& boxes, vector<int>& warehouse) {
        int n = warehouse.size();
    
        // Sorting boxes by their sizes
        sort(boxes.begin(), boxes.end());
    
        int start = 0;
        int end = n - 1;
        int totalBoxes = 0;
        int boxIter = boxes.size() - 1;
    
        // Going through each box from largest to smallest
        while (start <= end && boxIter >= 0) {
            // Put the current largest box if possible in the leftmost part
            if (boxes[boxIter] <= warehouse[start]) {
                totalBoxes++;
                start++;
            // Otherwise try to place it in the rightmost part
            } else if (boxes[boxIter] <= warehouse[end]) {
                totalBoxes++;
                end--;
            }
            boxIter--;
        }
    
        return totalBoxes;
    }
};

The problem involves placing as many of the given boxes as possible into a warehouse with spatial constraints given by two entry points. Boxes and spaces in the warehouse have definite sizes and the goal is to maximize the number of boxes stored. The solution employs a two-pointer technique and involves sorting the array of box sizes for strategic placement from largest to smallest.

To understand the approach:

• Sort the list of box sizes in ascending order.
• Use two pointers to represent the leftmost and rightmost available positions in the warehouse.
• Iterate over the sorted list of boxes from the largest to the smallest.
• Check if the largest box can fit into the leftmost position. If it fits, move the left pointer to the right, indicating that this position is now taken, and increment the count of boxes placed.
• If the box does not fit on the left, check the rightmost position. If it fits, move the right pointer to the left, and increment the count of boxes placed.
• Continue this process until all boxes are either placed or no suitable positions are left.

Key concepts utilized:

• Sorting
• Two-pointer technique
• Iteration
• Conditional logic

This method ensures that the largest boxes are considered first for placement, optimizing the space they can occupy either from the leftmost or rightmost sides of the warehouse, which maximizes the total number of boxes stored.

class Solution {
    
    public int maximumBoxesInWarehouse(int[] boxes, int[] warehouse) {
        int sizeOfWarehouse = warehouse.length;
    
        // Sorting the boxes array
        Arrays.sort(boxes);
    
        int minIndex = 0;
        int maxIndex = sizeOfWarehouse - 1;
        int numOfBoxes = 0;
        int indexOfBox = boxes.length - 1;
    
        // Descending through the boxes array
        while (minIndex <= maxIndex && indexOfBox >= 0) {
            // Fitting the box in the leftmost space if possible
            if (boxes[indexOfBox] <= warehouse[minIndex]) {
                numOfBoxes++;
                minIndex++;
            // Fitting the box in the rightmost space if possible
            } else if (boxes[indexOfBox] <= warehouse[maxIndex]) {
                numOfBoxes++;
                maxIndex--;
            }
            indexOfBox--;
        }
    
        return numOfBoxes;
    }
}

This code defines a method in Java for determining the maximum number of boxes that can be placed inside a warehouse given certain constraints. The sizes of the boxes and the maximum heights available in parts of the warehouse are provided.

• Begin by sorting the boxes array to organize them by size.
• Initialize pointers minIndex and maxIndex to track the smallest and largest available positions in the warehouse array, respectively.
• numOfBoxes, which starts at zero, tracks the count of boxes successfully placed.
• Iterate from the largest box in the sorted boxes array towards the smallest, checking each box against the available heights at minIndex and maxIndex in the warehouse.
  • If the current box fits at the minIndex, increment numOfBoxes and move the minIndex pointer to the right.
  • If it doesn't fit at the minIndex but fits at the maxIndex, increment numOfBoxes and move the maxIndex pointer to the left.
• The loop continues until either all boxes are considered or no fitting spaces are available in the warehouse.
• Lastly, return the total number of numOfBoxes that fits inside the warehouse.

This approach closely utilizes a two-pointer technique to effectively and efficiently match boxes to the available space in the warehouse from both ends.

class Solution:
    def countBoxes(self, boxes: List[int], rooms: List[int]) -> int:
        # Sorting the list of boxes
        boxes.sort()
    
        total_rooms = len(rooms)
        min_idx = 0
        max_idx = total_rooms - 1
        entered_boxes = 0
        current_box = len(boxes) - 1
    
        # Assign the tallest boxes to either the smallest room they can fit into
        while min_idx <= max_idx and current_box >= 0:
            if boxes[current_box] <= rooms[min_idx]:
                entered_boxes += 1
                min_idx += 1
            elif boxes[current_box] <= rooms[max_idx]:
                entered_boxes += 1
                max_idx -= 1
            current_box -= 1
        return entered_boxes

The Python solution for the problem of placing boxes into warehouses with entrance limits on height from both ends employs a two-pointer strategy. The solution class is defined with a method countBoxes, which accepts a list of box heights and a list of available room or doorway heights. Follow these detailed steps for the provided approach:

1. Start by sorting the array of box heights. This arrangement aids in efficiently determining the fitability of the tallest boxes into the available spaces.

2. Initialize pointers for both ends of the warehouse room list (min_idx for the leftmost, and max_idx for the rightmost) and a pointer (current_box) for the boxes starting from the last index.

3. Use a while loop to traverse from the tallest box down to the smallest, checking against both the smallest and largest available room sizes from the warehouse list:

   • If the current box fits into the room at the min_idx position, increment the entered_boxes counter and shift the min_idx pointer to the right, indicating that this room is now occupied.
   • If it doesn't fit there but fits in the max_idx room, increment the entered_boxes counter as well and shift the max_idx pointer to the left.
   • If the box doesn't fit in either, simply move to the next smaller box by decrementing current_box.

4. The loop terminates when either all boxes have been tried (current_box becomes -1) or the warehouse room pointers overlap (min_idx exceeds max_idx), meaning no more rooms are available.

5. The function then returns the total count of entered_boxes, representing how many boxes were successfully placed into the warehouse.

This approach is efficient and ensures the maximum number of boxes are placed into the warehouse, given the constraints of varying room sizes available from both sides of the warehouse.