What is the time complexity of Maximum Units on a Truck?

The optimal solution has a time complexity of O(n log n).

What data structures are used to solve Maximum Units on a Truck?

This problem uses Array, Greedy, Sorting. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Maximum Units on a Truck?

Insight 1: Greedy Approach: We can solve this problem greedily by prioritizing box types with the highest units per box. This is because loading more boxes with higher units per box will always yield a higher total number of units. Insight 2: Sorting: Sorting the box types based on units per box in descending order allows us to efficiently pick the best boxes first. This enables us to make optimal choices in a greedy manner.

Which companies ask LeetCode 1710. Maximum Units on a Truck in interviews?

This problem is frequently asked at Arista Networks, IBM, J.P. Morgan.

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Maximum Units on a Truck - LeetCode 1710 Solution

Maximum Units on a Truck - Complete Solution Guide

Maximum Units on a Truck is LeetCode problem 1710, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 i 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 .

Detailed Explanation

The problem asks you to maximize the number of units you can transport on a truck with a limited box capacity. You're given a list of box types, each with a number of boxes available and the number of units per box. The `truckSize` represents the maximum number of boxes the truck can hold. The goal is to strategically select boxes to maximize the total number of units transported, not exceeding the truck's capacity.

Solution Approach

The solution uses a greedy approach combined with sorting. It first sorts the `boxTypes` array based on the number of units per box in descending order. Then, it iterates through the sorted array. For each box type, it checks if there's enough space on the truck to load all available boxes of that type. If there is, it loads all of them and updates the remaining truck space. Otherwise, it loads only as many boxes as the remaining truck space allows and stops.

Step-by-Step Algorithm

Step 1: Sort `boxTypes` array in descending order based on the number of units per box (using a lambda function or comparator for sorting).
Step 2: Initialize `total_units` to 0 (to store the total number of units loaded).
Step 3: Iterate through the sorted `boxTypes` array. For each box type:
Step 4: Check if the remaining `truckSize` is greater than or equal to the number of boxes of the current type.
Step 5: If true, add the total units of that box type (`numberOfBoxes * numberOfUnitsPerBox`) to `total_units` and subtract the number of boxes used from `truckSize`.
Step 6: If false, add the units that can fit in the remaining truck space (`truckSize * numberOfUnitsPerBox`) to `total_units` and set `truckSize` to 0 (since the truck is full). Break the loop because no more boxes can be loaded.
Step 7: Return `total_units`.

Key Insights

Insight 1: Greedy Approach: We can solve this problem greedily by prioritizing box types with the highest units per box. This is because loading more boxes with higher units per box will always yield a higher total number of units.
Insight 2: Sorting: Sorting the box types based on units per box in descending order allows us to efficiently pick the best boxes first. This enables us to make optimal choices in a greedy manner.
Insight 3: Capacity Check: We need to carefully check if we can load all available boxes of a type without exceeding the truck's capacity (`truckSize`). If not, we load only as many boxes as remaining space allows.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(1)

Topics

This problem involves: Array, Greedy, Sorting.

Companies

Asked at: Arista Networks, IBM, J.P. Morgan.

Maximum Units on a Truck | LeetCode 1710

Easy

Array Greedy Sorting

Time: O(n log n)Space: O(1)

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public int maximumUnits(int[][] boxTypes, int truckSize) {
        Arrays.sort(boxTypes, (a, b) -> b[1] - a[1]);
        int totalUnits = 0;
        for (int[] boxType : boxTypes) {
            int numberOfBoxes = boxType[0];
            int numberOfUnitsPerBox = boxType[1];
            if (truckSize >= numberOfBoxes) {
                totalUnits += numberOfBoxes * numberOfUnitsPerBox;
                truckSize -= numberOfBoxes;
            } else {
                totalUnits += truckSize * numberOfUnitsPerBox;
                truckSize = 0;
                break;
            }
        }
        return totalUnits;
    }
}

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Problem Info

Description

Problem 1710: Maximum Units on a Truck - Detailed explanation and solution approaches.

Topics

Array Greedy Sorting

Companies

Arista NetworksIBMJ.P. Morgan

Stats

Difficulty:Easy

Languages:4

Companies:3

Problem #:1710

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Sort `boxTypes` array in descending order based on the number of units per box (using a lambda function or comparator for sorting).
2
Step 2: Initialize `total_units` to 0 (to store the total number of units loaded).
3
Step 3: Iterate through the sorted `boxTypes` array. For each box type:
4
Step 4: Check if the remaining `truckSize` is greater than or equal to the number of boxes of the current type.
5
Step 5: If true, add the total units of that box type (`numberOfBoxes * numberOfUnitsPerBox`) to `total_units` and subtract the number of boxes used from `truckSize`.
6
Step 6: If false, add the units that can fit in the remaining truck space (`truckSize * numberOfUnitsPerBox`) to `total_units` and set `truckSize` to 0 (since the truck is full). Break the loop because no more boxes can be loaded.
7
Step 7: Return `total_units`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The dominant operation is sorting the `boxTypes` array, which takes O(n log n) time using efficient sorting algorithms like merge sort or quicksort. The subsequent iteration through the sorted array takes O(n) time. The space complexity is O(1) because we are using a constant amount of extra space regardless of the input size.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Using a Priority Queue (Heap)

Instead of sorting, you can use a priority queue (min-heap) to store the box types. The priority queue is ordered by units per box. We can then repeatedly extract the maximum and update `truckSize` until it reaches 0.

Time: O(n log n)Space: O(n)

Trade-offs: This approach has the same time complexity but higher space complexity due to the priority queue. It might be slightly less efficient in practice due to heap operations. Sorting is generally preferred for this problem unless there are other constraints.

✨

Key Insights

1Insight 1: Greedy Approach: We can solve this problem greedily by prioritizing box types with the highest units per box. This is because loading more boxes with higher units per box will always yield a higher total number of units.
2Insight 2: Sorting: Sorting the box types based on units per box in descending order allows us to efficiently pick the best boxes first. This enables us to make optimal choices in a greedy manner.
3Insight 3: Capacity Check: We need to carefully check if we can load all available boxes of a type without exceeding the truck's capacity (`truckSize`). If not, we load only as many boxes as remaining space allows.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Greedy Algorithm

Making the best choice at each step without looking ahead.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty `boxTypes` array: The solution handles this correctly; the loop will not execute, and 0 will be returned.
Edge case 2: `truckSize` is 0: The loop will not execute, and 0 will be returned.
Edge case 3: `truckSize` is smaller than the number of boxes of the highest-unit box type: The algorithm correctly loads as many boxes as possible from the highest-unit box type.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the `boxTypes` array: This will lead to an incorrect solution as the boxes won't be considered in the optimal order.

✗Common mistake 2: Incorrectly handling the case where `truckSize` is less than the number of boxes for a given type: Failing to load only the available space in such cases will lead to exceeding the truck's capacity.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Optimizing cargo loading in a truck or airplane to maximize profit or value. This algorithm can determine the optimal selection of goods to load given weight and volume constraints.

🏦Real-world scenario 2: Resource allocation in manufacturing or logistics. Prioritizing the production or shipment of high-value items to maximize efficiency.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

This is one of the first problems I solved when learning Array. It's a great foundation for more complex problems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 30, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I verify if a greedy approach is correct?

Prove the greedy choice property: the locally optimal choice leads to globally optimal solution. Use exchange argument: show that swapping any element with the greedy choice doesn't improve the solution. If you can't prove it, test extensively or consider DP.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Maximum Units on a Truck"?

For this easy-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether greedy techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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