LeetCode #1710 — EASY

Maximum Units on a Truck

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

You are assigned to put some amount of boxes onto one truck. You are given a 2D array boxTypes, where boxTypes[i] = [numberOfBoxes_i, numberOfUnitsPerBox_i]:

numberOfBoxes_i is the number of boxes of type i.
numberOfUnitsPerBox_iis the number of units in each box of the type i.

You are also given an integer truckSize, which is the maximum number of boxes that can be put on the truck. You can choose any boxes to put on the truck as long as the number of boxes does not exceed truckSize.

Return the maximum total number of units that can be put on the truck.

Example 1:

Input: boxTypes = [[1,3],[2,2],[3,1]], truckSize = 4
Output: 8
Explanation: There are:
- 1 box of the first type that contains 3 units.
- 2 boxes of the second type that contain 2 units each.
- 3 boxes of the third type that contain 1 unit each.
You can take all the boxes of the first and second types, and one box of the third type.
The total number of units will be = (1 * 3) + (2 * 2) + (1 * 1) = 8.

Example 2:

Input: boxTypes = [[5,10],[2,5],[4,7],[3,9]], truckSize = 10
Output: 91

Constraints:

1 <= boxTypes.length <= 1000
1 <= numberOfBoxes_i, numberOfUnitsPerBox_i <= 1000
1 <= truckSize <= 10⁶

Step 01

Brute Force Baseline

Problem summary: You are assigned to put some amount of boxes onto one truck. You are given a 2D array boxTypes, where boxTypes[i] = [numberOfBoxesi, numberOfUnitsPerBoxi]: numberOfBoxesi is the number of boxes of type i. numberOfUnitsPerBoxi is the number of units in each box of the type i. You are also given an integer truckSize, which is the maximum number of boxes that can be put on the truck. You can choose any boxes to put on the truck as long as the number of boxes does not exceed truckSize. Return the maximum total number of units that can be put on the truck.

Baseline thinking

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

Pattern signal: Array · Greedy

Example 1

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

Example 2

[[5,10],[2,5],[4,7],[3,9]]
10

Core Insight

What unlocks the optimal approach

If we have space for at least one box, it's always optimal to put the box with the most units.
Sort the box types with the number of units per box non-increasingly.
Iterate on the box types and take from each type as many as you can.

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 #1710: Maximum Units on a Truck
class Solution {
    public int maximumUnits(int[][] boxTypes, int truckSize) {
        Arrays.sort(boxTypes, (a, b) -> b[1] - a[1]);
        int ans = 0;
        for (var e : boxTypes) {
            int a = e[0], b = e[1];
            ans += b * Math.min(truckSize, a);
            truckSize -= a;
            if (truckSize <= 0) {
                break;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1710: Maximum Units on a Truck
func maximumUnits(boxTypes [][]int, truckSize int) (ans int) {
	sort.Slice(boxTypes, func(i, j int) bool { return boxTypes[i][1] > boxTypes[j][1] })
	for _, e := range boxTypes {
		a, b := e[0], e[1]
		ans += b * min(truckSize, a)
		truckSize -= a
		if truckSize <= 0 {
			break
		}
	}
	return
}

# Accepted solution for LeetCode #1710: Maximum Units on a Truck
class Solution:
    def maximumUnits(self, boxTypes: List[List[int]], truckSize: int) -> int:
        ans = 0
        for a, b in sorted(boxTypes, key=lambda x: -x[1]):
            ans += b * min(truckSize, a)
            truckSize -= a
            if truckSize <= 0:
                break
        return ans

// Accepted solution for LeetCode #1710: Maximum Units on a Truck
impl Solution {
    pub fn maximum_units(mut box_types: Vec<Vec<i32>>, truck_size: i32) -> i32 {
        box_types.sort_by(|a, b| b[1].cmp(&a[1]));
        let mut sum = 0;
        let mut ans = 0;
        for box_type in box_types.iter() {
            if sum + box_type[0] < truck_size {
                sum += box_type[0];
                ans += box_type[0] * box_type[1];
            } else {
                ans += (truck_size - sum) * box_type[1];
                break;
            }
        }
        ans
    }
}

// Accepted solution for LeetCode #1710: Maximum Units on a Truck
function maximumUnits(boxTypes: number[][], truckSize: number): number {
    boxTypes.sort(([_, a], [__, b]) => b - a);
    let ans = 0;
    for (const [count, size] of boxTypes) {
        ans += Math.min(truckSize, count) * size;
        truckSize -= count;
        if (truckSize < 0) {
            break;
        }
    }
    return ans;
}

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 log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

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.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.