Filling Bookcase Shelves

class Solution {
public:
    int minShelfHeight(vector<vector<int>>& bookList, int maxWidth) {
        vector<int> minHeightDP(bookList.size() + 1);
        minHeightDP[0] = 0;
        minHeightDP[1] = bookList[0][1];

        for (int idx = 1; idx <= bookList.size(); ++idx) {
            int widthLeft = maxWidth - bookList[idx - 1][0];
            int tallest = bookList[idx - 1][1];
            minHeightDP[idx] = minHeightDP[idx - 1] + tallest;

            int curIdx = idx - 1;
            while (curIdx > 0 && widthLeft - bookList[curIdx - 1][0] >= 0) {
                widthLeft -= bookList[curIdx - 1][0];
                tallest = max(tallest, bookList[curIdx - 1][1]);
                minHeightDP[idx] = min(minHeightDP[idx], minHeightDP[curIdx - 1] + tallest);
                curIdx--;
            }
        }

        return minHeightDP[bookList.size()];
    }
};

This C++ class named Solution provides a method minShelfHeight to solve the task of efficiently organizing book dimensions on a shelf that has a specified maximum width. Here’s the breakdown of the solution:

• The method accepts bookList, a vector of vectors containing two integers representing the width and height of each book, and maxWidth, the maximum allowed total width on a single shelf.
• The solution utilizes dynamic programming. The state minHeightDP[i] is defined as the minimum height required to place the first i books.
• Initialization:
  • minHeightDP[0] is set to 0 because no height is needed when no books are placed.
  • minHeightDP[1] is set to the height of the first book.
• Transitioning through each book:
  • Start by considering placing the current book on a new shelf. Update the minimum height considering the book might extend the previous configuration’s height.
  • Examine previous books to find an optimal arrangement for books on the current shelf. This step checks if adding previous books to the same shelf with the current book exceeds the maxWidth.
  • The height for the current configuration is adjusted to the tallest book on the current shelf.
  • Update minHeightDP[i] with the minimal calculated height for placing up to the i-th book.
• The method finally returns the minimal height needed to store all books from minHeightDP[bookList.size()], which represents the optimal height for all books given the restrictions.

This function provides an efficient way to determine the minimum shelf height needed for storing a list of books within a given shelf width constraint, ensuring all books are organized without exceeding the maximum allowable width.

class Solution {

    public int minShelvesHeight(int[][] bookDimensions, int maxWidth) {
        int[] heightDP = new int[bookDimensions.length + 1];
        heightDP[0] = 0;
        heightDP[1] = bookDimensions[0][1];

        for (int i = 2; i <= bookDimensions.length; i++) {
            int availableWidth = maxWidth - bookDimensions[i - 1][0];
            int currentMaxHeight = bookDimensions[i - 1][1];
            heightDP[i] = bookDimensions[i - 1][1] + heightDP[i - 1];

            int k = i - 1;
            while (k > 0 && availableWidth - bookDimensions[k - 1][0] >= 0) {
                currentMaxHeight = Math.max(currentMaxHeight, bookDimensions[k - 1][1]);
                availableWidth -= bookDimensions[k - 1][0];
                heightDP[i] = Math.min(heightDP[i], currentMaxHeight + heightDP[k - 1]);
                k--;
            }
        }

        return heightDP[bookDimensions.length];
    }
}

In the given Java solution to the "Filling Bookcase Shelves" problem, the goal is to find the minimal vertical height required to store all the books on shelves that have a maximum allowable width. The primary strategy involves dynamic programming to optimize the arrangement based on the dimensions of the books.

• The function accepts two parameters:
  • bookDimensions: a 2D array where each entry contains two values [width, height] representing a book's dimensions.
  • maxWidth: the maximum total width that each shelf can accommodate.

Here's a concise breakdown of how the solution operates:

• An array heightDP is initialized to store the minimum cumulative height at each step. The array size is the number of books plus one for easier index management, and starts with the first book inherently taking its own height as the shelf height considering it's the only book on the shelf.

• For each book from the second onward, the approach calculates:

  • The cumulative width of books that can be placed on the current shelf.
  • Updates the heightDP for each possible number of books on this shelf based on the height trade-offs of adding another book or starting a new shelf.

• This involves iterating over the books and using a nested loop (while loop) to check backwards from the current book to see if adding previous books to the current shelf:

  • Exceeds the maxWidth, and if not,
  • To ensure the least possible increase in total height when more books are included in the current shelf.

• During each iteration, the currentMaxHeight keeps track of the tallest book on the current shelf while the availableWidth ensures the total width doesn't exceed maxWidth. By comparing these dynamically, and adjusting shelves and their respective heights as new books are considered, the final value at heightDP[bookDimensions.length] provides the minimum shelf height required to store all books within the given width constraints.

The efficiency of the solution comes from its ability to dynamically decide, for each book, whether to place it on a new shelf or integrate it into the existing shelf setup, while always aiming to minimize the overall vertical dimensional requirement. This method avoids the need for redundant or exhaustive checks, leveraging dynamic programming to solve the problem efficiently.

class BookShelf:
    def calculateMinHeight(self, books: List[List[int]], maxShelfWidth: int) -> int:
        num_books = len(books)
        # minHeight[i] holds the minimum height of a bookshelf up till book i
        minHeight = [0] * (num_books + 1)
        
        # Initializing base cases
        minHeight[0] = 0
        minHeight[1] = books[0][1]
        
        for index in range(2, num_books + 1):
            width_left = maxShelfWidth - books[index - 1][0]
            current_height = books[index - 1][1]
            minHeight[index] = current_height + minHeight[index - 1]
            
            j = index - 1
            # Try placing earlier books on the same shelf
            while j > 0 and width_left - books[j - 1][0] >= 0:
                current_height = max(current_height, books[j - 1][1])
                width_left -= books[j - 1][0]
                minHeight[index] = min(minHeight[index], current_height + minHeight[j - 1])
                j -= 1
        
        return minHeight[num_books]

This Python solution tackles the problem of arranging a set of books on shelves in a manner that minimizes the total height of the bookcase. Each book is defined by its thickness and height, and the objective is to place these books on multiple shelves without exceeding a given shelf's maximum width.

• Start by initializing an array minHeight where each element at index i represents the minimal total height of shelves up to the i-th book.
• minHeight[0] is set to 0, representing no books, and minHeight[1] is initialized to the height of the first book.
• Iterate through each book and attempt to place it on a new shelf or with the previous books on the same shelf.
• Keep track of the remaining shelf width width_left as books are added.
• Calculate the shelf height dynamically by considering whether the current book should start a new shelf or be placed with the earlier books, ensuring to adjust height calculations based on the tallest book on the current shelf.
• The approach ensures optimal placement by examining all possible shelf configurations for each book, and updates the minHeight array to reflect the minimum height achieved at each step.

Ensure the shelves' width constraint is respected while determining the minimum height possible for the bookshelf. The result is found in minHeight[num_books], which holds the total minimal height configuration for all books.