What is the time complexity of Maximum Candies You Can Get from Boxes?

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

What data structures are used to solve Maximum Candies You Can Get from Boxes?

This problem uses Array, Breadth-First Search, Graph. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Maximum Candies You Can Get from Boxes?

Insight 1: This is a graph traversal problem where boxes are nodes, and keys unlock access to other nodes. We can use Breadth-First Search (BFS) to explore the boxes. Insight 2: We need to keep track of which boxes we *have* (contained) and which boxes we can *access* (open). A boolean array is suitable for this.

Which companies ask LeetCode 1298. Maximum Candies You Can Get from Boxes in interviews?

This problem is frequently asked at Airbnb, Lyft.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Maximum Candies You Can Get from Boxes - LeetCode 1298 Solution

Maximum Candies You Can Get from Boxes - Complete Solution Guide

Maximum Candies You Can Get from Boxes is LeetCode problem 1298, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You have n boxes labeled from 0 to n - 1 . You are given four arrays: status , candies , keys , and containedBoxes where: status[i] is 1 if the i th box is open and 0 if the i th box is closed, candies[i] is the number of candies in the i th box, keys[i] is a list of the labels of the boxes you can open after opening the i th box. containedBoxes[i] is a list of the boxes you found inside the i th box. You are given an integer array initialBoxes that contains the labels of the boxes you initially

Detailed Explanation

The problem asks us to find the maximum number of candies we can collect from a set of boxes. We are given information about each box: whether it's initially open or closed, the number of candies it contains, the keys it provides (allowing us to open other boxes), and the boxes it contains. We start with a set of initial boxes. We can only collect candies from open boxes. Opening a box also gives us access to its keys and contained boxes. The goal is to maximize the total candies collected by strategically opening boxes and using keys.

Solution Approach

The solution uses a Breadth-First Search (BFS) approach. We maintain a queue of open boxes whose candies we can collect. We also use boolean arrays to track which boxes we *have* (contained or initially owned) and which boxes have been *processed* (candies collected and keys/contained boxes utilized). We iterate through the queue, collecting candies, unlocking new boxes using keys, and adding new boxes to our inventory. A key aspect is that if a key is obtained that opens a box we already have, we add that box to the queue if it hasn't already been processed.

Step-by-Step Algorithm

Step 1: Initialize data structures: a queue for BFS, a boolean array `have_box` to track which boxes we possess, and a boolean array `processed` to track which boxes have been processed.
Step 2: Add the initial boxes to the `have_box` array. If an initial box is open, add it to the queue and mark it as processed.
Step 3: While the queue is not empty, dequeue a box.
Step 4: Add the candies from the current box to the total candies collected.
Step 5: Iterate through the keys found in the current box. Open the corresponding boxes by setting their status to 1. If we have that box and it hasn't been processed yet, enqueue it and mark it as processed.
Step 6: Iterate through the boxes contained in the current box. Add these boxes to our inventory by setting `have_box[new_box] = true`. If the box is open and hasn't been processed yet, enqueue it and mark it as processed.
Step 7: Repeat steps 3-6 until the queue is empty.
Step 8: Return the total number of candies collected.

Key Insights

Insight 1: This is a graph traversal problem where boxes are nodes, and keys unlock access to other nodes. We can use Breadth-First Search (BFS) to explore the boxes.
Insight 2: We need to keep track of which boxes we *have* (contained) and which boxes we can *access* (open). A boolean array is suitable for this.
Insight 3: We need to handle cases where we have a box but it's closed, and we may get the key later. We need to revisit these boxes when new keys are acquired.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Breadth-First Search, Graph.

Companies

Asked at: Airbnb, Lyft.

Maximum Candies You Can Get from Boxes | LeetCode 1298

Hard

Array Breadth-First Search Graph

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

Solution Code

Available in:

import java.util.ArrayDeque;
import java.util.Queue;
import java.util.List;
import java.util.ArrayList;

class Solution {
    public int maxCandies(int[] status, int[] candies, int[][] keys, int[][] containedBoxes, int[] initialBoxes) {
        int total_candies = 0;
        
        Queue<Integer> q = new ArrayDeque<>();
        boolean[] have_box = new boolean[status.length];
        boolean[] processed = new boolean[status.length];

        for (int box : initialBoxes) {
            have_box[box] = true;
            if (status[box] == 1) {
                if (!processed[box]) {
                    q.offer(box);
                    processed[box] = true;
                }
            }
        }

        while (!q.isEmpty()) {
            int current_box = q.poll();
            total_candies += candies[current_box];

            for (int key : keys[current_box]) {
                status[key] = 1;
                if (have_box[key] && !processed[key]) {
                    q.offer(key);
                    processed[key] = true;
                }
            }

            for (int new_box : containedBoxes[current_box]) {
                have_box[new_box] = true;
                if (status[new_box] == 1 && !processed[new_box]) {
                    q.offer(new_box);
                    processed[new_box] = true;
                }
            }

        }

        return total_candies;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 1298: Maximum Candies You Can Get from Boxes - Detailed explanation and solution approaches.

Topics

Array Breadth-First Search Graph

Companies

AirbnbLyft

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:1298

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

8 steps

1
Step 1: Initialize data structures: a queue for BFS, a boolean array `have_box` to track which boxes we possess, and a boolean array `processed` to track which boxes have been processed.
2
Step 2: Add the initial boxes to the `have_box` array. If an initial box is open, add it to the queue and mark it as processed.
3
Step 3: While the queue is not empty, dequeue a box.
4
Step 4: Add the candies from the current box to the total candies collected.
5
Step 5: Iterate through the keys found in the current box. Open the corresponding boxes by setting their status to 1. If we have that box and it hasn't been processed yet, enqueue it and mark it as processed.
6
Step 6: Iterate through the boxes contained in the current box. Add these boxes to our inventory by setting `have_box[new_box] = true`. If the box is open and hasn't been processed yet, enqueue it and mark it as processed.
7
Step 7: Repeat steps 3-6 until the queue is empty.
8
Step 8: Return the total number of candies collected.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The algorithm iterates through the boxes using a queue. In the worst case, all boxes might be added to the queue, contributing O(n) to both time and space complexity. The loops iterating through keys and contained boxes contribute a maximum of O(n) in total across all boxes, because the total number of keys and contained boxes cannot exceed n. The boolean arrays `have_box` and `processed` contribute O(n) to space complexity.

📖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

Depth-First Search (DFS)

We could use DFS instead of BFS. The main difference would be the order in which boxes are processed. The outcome would still be the same.

Time: O(n)Space: O(n) in the worst case (recursive call stack).

Trade-offs: BFS is generally preferred for finding the shortest path (though that's not relevant here). DFS might be slightly simpler to implement recursively, but could potentially lead to stack overflow issues for very large inputs, while BFS uses iterative approach and is generally safer.

✨

Key Insights

1Insight 1: This is a graph traversal problem where boxes are nodes, and keys unlock access to other nodes. We can use Breadth-First Search (BFS) to explore the boxes.
2Insight 2: We need to keep track of which boxes we *have* (contained) and which boxes we can *access* (open). A boolean array is suitable for this.
3Insight 3: We need to handle cases where we have a box but it's closed, and we may get the key later. We need to revisit these boxes when new keys are acquired.

📚

Key Terms Explained

For Students

Array

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

Graph

Nodes connected by edges. Like a social network or city map!

Edge Cases

Edge case 1: An initial box is closed and never opened. The solution correctly handles this because the box is never added to the queue.
Edge case 2: A box contains itself. The problem statement guarantees uniqueness of contained boxes, so this isn't an issue. Even if it happened, `processed` boolean array ensures the box is not added to the queue infinitely.
Edge case 3: Empty initialBoxes array. The algorithm handles this case gracefully, as the loop for processing initial boxes simply doesn't execute, and the algorithm returns 0 if no other boxes are accessible.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to check if a box is already processed before adding it to the queue. This can lead to infinite loops.

✗Common mistake 2: Not updating the `have_box` array when a box is found inside another box. This means the algorithm won't realize it has a box even after finding it.

✗Common mistake 3: Failing to update the `status` when a new key is obtained. This will prevent you from opening boxes even if you have their keys.

✅

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: Dependency resolution in software package management. Boxes can represent software packages, keys can represent dependencies, and candies can represent the value of the package. The goal is to maximize the 'candies' (value) by satisfying dependencies.

🏦Real-world scenario 2: Network security access control. Boxes can represent resources, candies represent the value of accessing a resource, keys represent permissions, and containedBoxes represent resources that are automatically included when certain other resource is accessed. Maximize the acquired value by smartly utilizing the permission keys.

💡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

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

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

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: July 9, 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

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 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 approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

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 Candies You Can Get from Boxes"?

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

🤖 Need help with this problem?