LeetCode #1691 — HARD

Maximum Height by Stacking Cuboids

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

Given n cuboids where the dimensions of the i^th cuboid is cuboids[i] = [width_i, length_i, height_i] (0-indexed). Choose a subset of cuboids and place them on each other.

You can place cuboid i on cuboid j if width_i <= width_j and length_i <= length_j and height_i <= height_j. You can rearrange any cuboid's dimensions by rotating it to put it on another cuboid.

Return the maximum height of the stacked cuboids.

Example 1:

Input: cuboids = [[50,45,20],[95,37,53],[45,23,12]]
Output: 190
Explanation:
Cuboid 1 is placed on the bottom with the 53x37 side facing down with height 95.
Cuboid 0 is placed next with the 45x20 side facing down with height 50.
Cuboid 2 is placed next with the 23x12 side facing down with height 45.
The total height is 95 + 50 + 45 = 190.

Example 2:

Input: cuboids = [[38,25,45],[76,35,3]]
Output: 76
Explanation:
You can't place any of the cuboids on the other.
We choose cuboid 1 and rotate it so that the 35x3 side is facing down and its height is 76.

Example 3:

Input: cuboids = [[7,11,17],[7,17,11],[11,7,17],[11,17,7],[17,7,11],[17,11,7]]
Output: 102
Explanation:
After rearranging the cuboids, you can see that all cuboids have the same dimension.
You can place the 11x7 side down on all cuboids so their heights are 17.
The maximum height of stacked cuboids is 6 * 17 = 102.

Constraints:

n == cuboids.length
1 <= n <= 100
1 <= width_i, length_i, height_i <= 100

Step 01

Brute Force Baseline

Problem summary: Given n cuboids where the dimensions of the ith cuboid is cuboids[i] = [widthi, lengthi, heighti] (0-indexed). Choose a subset of cuboids and place them on each other. You can place cuboid i on cuboid j if widthi <= widthj and lengthi <= lengthj and heighti <= heightj. You can rearrange any cuboid's dimensions by rotating it to put it on another cuboid. Return the maximum height of the stacked cuboids.

Baseline thinking

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

Pattern signal: Array · Dynamic Programming

Example 1

[[50,45,20],[95,37,53],[45,23,12]]

Example 2

[[38,25,45],[76,35,3]]

Example 3

[[7,11,17],[7,17,11],[11,7,17],[11,17,7],[17,7,11],[17,11,7]]

Core Insight

What unlocks the optimal approach

Does the dynamic programming sound like the right algorithm after sorting?
Let's say box1 can be placed on top of box2. No matter what orientation box2 is in, we can rotate box1 so that it can be placed on top. Why don't we orient everything such that height is the biggest?

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

Largest constraint values

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 #1691: Maximum Height by Stacking Cuboids 
class Solution {
    public int maxHeight(int[][] cuboids) {
        for (var c : cuboids) {
            Arrays.sort(c);
        }
        Arrays.sort(cuboids,
            (a, b) -> a[0] == b[0] ? (a[1] == b[1] ? a[2] - b[2] : a[1] - b[1]) : a[0] - b[0]);
        int n = cuboids.length;
        int[] f = new int[n];
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < i; ++j) {
                if (cuboids[j][1] <= cuboids[i][1] && cuboids[j][2] <= cuboids[i][2]) {
                    f[i] = Math.max(f[i], f[j]);
                }
            }
            f[i] += cuboids[i][2];
        }
        return Arrays.stream(f).max().getAsInt();
    }
}

// Accepted solution for LeetCode #1691: Maximum Height by Stacking Cuboids 
func maxHeight(cuboids [][]int) int {
	for _, c := range cuboids {
		sort.Ints(c)
	}
	sort.Slice(cuboids, func(i, j int) bool {
		a, b := cuboids[i], cuboids[j]
		return a[0] < b[0] || a[0] == b[0] && (a[1] < b[1] || a[1] == b[1] && a[2] < b[2])
	})
	n := len(cuboids)
	f := make([]int, n)
	for i := range f {
		for j := 0; j < i; j++ {
			if cuboids[j][1] <= cuboids[i][1] && cuboids[j][2] <= cuboids[i][2] {
				f[i] = max(f[i], f[j])
			}
		}
		f[i] += cuboids[i][2]
	}
	return slices.Max(f)
}

# Accepted solution for LeetCode #1691: Maximum Height by Stacking Cuboids 
class Solution:
    def maxHeight(self, cuboids: List[List[int]]) -> int:
        for c in cuboids:
            c.sort()
        cuboids.sort()
        n = len(cuboids)
        f = [0] * n
        for i in range(n):
            for j in range(i):
                if cuboids[j][1] <= cuboids[i][1] and cuboids[j][2] <= cuboids[i][2]:
                    f[i] = max(f[i], f[j])
            f[i] += cuboids[i][2]
        return max(f)

// Accepted solution for LeetCode #1691: Maximum Height by Stacking Cuboids 
struct Solution;

impl Solution {
    fn max_height(mut cuboids: Vec<Vec<i32>>) -> i32 {
        let n = cuboids.len();
        for i in 0..n {
            cuboids[i].sort_unstable();
        }
        cuboids.sort_unstable();
        let mut dp: Vec<i32> = vec![0; n];
        for i in 0..n {
            dp[i] = cuboids[i][2];
            'outer: for j in 0..i {
                for k in 0..3 {
                    if cuboids[j][k] > cuboids[i][k] {
                        continue 'outer;
                    }
                }
                dp[i] = dp[i].max(dp[j] + cuboids[i][2]);
            }
        }
        dp.into_iter().max().unwrap()
    }
}

#[test]
fn test() {
    let cuboids = vec_vec_i32![[50, 45, 20], [95, 37, 53], [45, 23, 12]];
    let res = 190;
    assert_eq!(Solution::max_height(cuboids), res);
    let cuboids = vec_vec_i32![[38, 25, 45], [76, 35, 3]];
    let res = 76;
    assert_eq!(Solution::max_height(cuboids), res);
    let cuboids = vec_vec_i32![
        [7, 11, 17],
        [7, 17, 11],
        [11, 7, 17],
        [11, 17, 7],
        [17, 7, 11],
        [17, 11, 7]
    ];
    let res = 102;
    assert_eq!(Solution::max_height(cuboids), res);
}

// Accepted solution for LeetCode #1691: Maximum Height by Stacking Cuboids 
function maxHeight(cuboids: number[][]): number {
    for (const c of cuboids) {
        c.sort((a, b) => a - b);
    }
    cuboids.sort((a, b) => {
        if (a[0] !== b[0]) {
            return a[0] - b[0];
        }
        if (a[1] !== b[1]) {
            return a[1] - b[1];
        }
        return a[2] - b[2];
    });
    const n = cuboids.length;
    const f = Array(n).fill(0);
    for (let i = 0; i < n; ++i) {
        for (let j = 0; j < i; ++j) {
            const ok = cuboids[j][1] <= cuboids[i][1] && cuboids[j][2] <= cuboids[i][2];
            if (ok) f[i] = Math.max(f[i], f[j]);
        }
        f[i] += cuboids[i][2];
    }
    return Math.max(...f);
}

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.