LeetCode #1105 — MEDIUM

Filling Bookcase Shelves

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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 of the bookcase 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.

For example, if we have an ordered list of 5 books, we might place the first and second book onto the first shelf, the third book on the second shelf, and the fourth and fifth book on the last shelf.

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

Example 1:

Input: books = [[1,1],[2,3],[2,3],[1,1],[1,1],[1,1],[1,2]], shelfWidth = 4
Output: 6
Explanation:
The sum of the heights of the 3 shelves is 1 + 3 + 2 = 6.
Notice that book number 2 does not have to be on the first shelf.

Example 2:

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

Constraints:

1 <= books.length <= 1000
1 <= thickness_i <= shelfWidth <= 1000
1 <= height_i <= 1000

Step 01

Brute Force Baseline

Problem summary: You are given an array books where books[i] = [thicknessi, heighti] indicates the thickness and height of the ith 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 of the bookcase 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. For example, if we have an ordered list of 5 books, we might place the first and second book onto the first shelf, the third book on the second shelf, and the

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming

Example 1

[[1,1],[2,3],[2,3],[1,1],[1,1],[1,1],[1,2]]
4

Example 2

[[1,3],[2,4],[3,2]]
6

Step 02

Core Insight

What unlocks the optimal approach

Use dynamic programming: dp(i) will be the answer to the problem for books[i:].

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1105: Filling Bookcase Shelves
class Solution {
    public int minHeightShelves(int[][] books, int shelfWidth) {
        int n = books.length;
        int[] f = new int[n + 1];
        for (int i = 1; i <= n; ++i) {
            int w = books[i - 1][0], h = books[i - 1][1];
            f[i] = f[i - 1] + h;
            for (int j = i - 1; j > 0; --j) {
                w += books[j - 1][0];
                if (w > shelfWidth) {
                    break;
                }
                h = Math.max(h, books[j - 1][1]);
                f[i] = Math.min(f[i], f[j - 1] + h);
            }
        }
        return f[n];
    }
}

// Accepted solution for LeetCode #1105: Filling Bookcase Shelves
func minHeightShelves(books [][]int, shelfWidth int) int {
	n := len(books)
	f := make([]int, n+1)
	for i := 1; i <= n; i++ {
		w, h := books[i-1][0], books[i-1][1]
		f[i] = f[i-1] + h
		for j := i - 1; j > 0; j-- {
			w += books[j-1][0]
			if w > shelfWidth {
				break
			}
			h = max(h, books[j-1][1])
			f[i] = min(f[i], f[j-1]+h)
		}
	}
	return f[n]
}

# Accepted solution for LeetCode #1105: Filling Bookcase Shelves
class Solution:
    def minHeightShelves(self, books: List[List[int]], shelfWidth: int) -> int:
        n = len(books)
        f = [0] * (n + 1)
        for i, (w, h) in enumerate(books, 1):
            f[i] = f[i - 1] + h
            for j in range(i - 1, 0, -1):
                w += books[j - 1][0]
                if w > shelfWidth:
                    break
                h = max(h, books[j - 1][1])
                f[i] = min(f[i], f[j - 1] + h)
        return f[n]

// Accepted solution for LeetCode #1105: Filling Bookcase Shelves
struct Solution;

impl Solution {
    fn min_height_shelves(books: Vec<Vec<i32>>, shelf_width: i32) -> i32 {
        let n = books.len();
        let mut dp = vec![0; n + 1];
        for i in 0..n {
            let mut width = books[i][0];
            let mut height = books[i][1];
            dp[i + 1] = height + dp[i];
            for j in (0..i).rev() {
                width += books[j][0];
                if width <= shelf_width {
                    height = height.max(books[j][1]);
                    dp[i + 1] = dp[i + 1].min(height + dp[j]);
                }
            }
        }
        dp[n]
    }
}

#[test]
fn test() {
    let books = vec_vec_i32![[1, 1], [2, 3], [2, 3], [1, 1], [1, 1], [1, 1], [1, 2]];
    let shelf_width = 4;
    let res = 6;
    assert_eq!(Solution::min_height_shelves(books, shelf_width), res);
}

// Accepted solution for LeetCode #1105: Filling Bookcase Shelves
function minHeightShelves(books: number[][], shelfWidth: number): number {
    const n = books.length;
    const f = new Array(n + 1).fill(0);
    for (let i = 1; i <= n; ++i) {
        let [w, h] = books[i - 1];
        f[i] = f[i - 1] + h;
        for (let j = i - 1; j > 0; --j) {
            w += books[j - 1][0];
            if (w > shelfWidth) {
                break;
            }
            h = Math.max(h, books[j - 1][1]);
            f[i] = Math.min(f[i], f[j - 1] + h);
        }
    }
    return f[n];
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n^2)

Space

O(n)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.