Filling Bookcase Shelves - Practice Coding Problems

Filling Bookcase Shelves - Problem

You are given an array books where books[i] = [thickness_i, height_i] indicates the thickness and height of the i^th book. You are also given an integer shelfWidth.

We want to place these books in order onto bookcase shelves that have a total width shelfWidth. We choose some of the books to place on this shelf such that the sum of their thickness is less than or equal to shelfWidth, then build another level of the shelf so that the total height of the bookcase has increased by the maximum height of the books we just put down.

We repeat this process until there are no more books to place. Note that at each step of the above process, the order of the books we place is the same order as the given sequence of books.

Return the minimum possible height that the total bookshelf can be after placing shelves in this manner.

Input & Output

Example 1 — Basic Case

$ Input: books = [[1,1],[2,3],[2,1],[1,3]], shelfWidth = 4

› Output: 6

💡 Note: Optimal arrangement: Shelf 1: [1,1],[2,3] (width=3, height=3), Shelf 2: [2,1] (width=2, height=1), Shelf 3: [1,3] (width=1, height=3). Total height = 3+1+3 = 6.

Example 2 — All Books on One Shelf

$ Input: books = [[1,3],[2,4],[3,2]], shelfWidth = 6

› Output: 4

💡 Note: All books fit on one shelf: total width = 1+2+3 = 6 ≤ 6, height = max(3,4,2) = 4.

Example 3 — Each Book Separate

$ Input: books = [[3,5],[3,4],[3,3]], shelfWidth = 3

› Output: 12

💡 Note: Each book must be on its own shelf since each has width 3. Total height = 5+4+3 = 12.

Constraints

1 ≤ books.length ≤ 1000
1 ≤ thickness_i ≤ shelfWidth ≤ 1000
1 ≤ height_i ≤ 1000

Visualization

Tap to expand

Understanding the Visualization

Input

Books with [thickness, height] and shelf width limit

Process

Find optimal way to group consecutive books on shelves

Output

Minimum possible total height of bookcase

Key Takeaway

🎯 Key Insight: Use dynamic programming to try all valid shelf partitions and choose the one with minimum total height

Asked in

a Amazon 8 G Google 6 M Microsoft 4 f Facebook 3

The key insight is that this is a dynamic programming problem where we need to decide optimal shelf partitions. For each book position, we try all valid ways to end the current shelf and recursively solve for remaining books. The optimal approach uses bottom-up DP with Time: O(n²), Space: O(n).

Common Approaches

Approach	Time	Space	Notes
✓ Recursive Brute Force	O(2ⁿ)	O(n)	Try all possible ways to partition consecutive books into shelves
Memoized Recursion	O(n²)	O(n)	Add memoization to avoid recomputing same subproblems
Bottom-Up Dynamic Programming	O(n²)	O(n)	Build solution iteratively from last book to first

Recursive Brute Force — Algorithm Steps

For each book position, try all valid shelf endings
Recursively solve for remaining books
Return minimum height across all partitions

Visualization

Tap to expand

Step-by-Step Walkthrough

Start

Begin with first book, try all possible shelf endings

Recurse

For each shelf choice, recursively solve remaining books

Minimum

Return minimum height across all possibilities

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>

int solve(int books[][2], int n, int shelfWidth, int i) {
    if (i >= n) return 0;
    
    int minHeight = INT_MAX;
    int currentWidth = 0;
    int maxHeight = 0;
    
    for (int j = i; j < n; j++) {
        currentWidth += books[j][0];
        if (currentWidth > shelfWidth) break;
        if (books[j][1] > maxHeight) maxHeight = books[j][1];
        int totalHeight = maxHeight + solve(books, n, shelfWidth, j + 1);
        if (totalHeight < minHeight) minHeight = totalHeight;
    }
    
    return minHeight;
}

int solution(int books[][2], int n, int shelfWidth) {
    return solve(books, n, shelfWidth, 0);
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    int books[1000][2];
    int n = 0;
    
    // Parse books array
    char* ptr = line;
    while (*ptr) {
        if (*ptr == '[' && *(ptr+1) != '[') {
            ptr++;
            books[n][0] = atoi(ptr);
            while (*ptr && *ptr != ',') ptr++;
            if (*ptr == ',') ptr++;
            books[n][1] = atoi(ptr);
            n++;
        }
        ptr++;
    }
    
    int shelfWidth;
    scanf("%d", &shelfWidth);
    
    printf("%d\n", solution(books, n, shelfWidth));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(2ⁿ)

For each book, we have 2 choices: start new shelf or continue current shelf, leading to exponential combinations

✓ Linear Growth

Space Complexity

O(n)

Recursion stack depth up to n books

⚡ Linearithmic Space

95.0K Views

Medium Frequency

~25 min Avg. Time

2.8K Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Recursive Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler