What is the time complexity of Remove Boxes?

The optimal solution has a time complexity of O(n^4).

What data structures are used to solve Remove Boxes?

This problem uses Array, Dynamic Programming, Memoization. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Remove Boxes?

Insight 1: Dynamic Programming is crucial to avoid recomputing overlapping subproblems. Memoization (top-down DP) or tabulation (bottom-up DP) will significantly improve performance. Insight 2: The key state to track is the range (l, r) of boxes being considered and the number (k) of boxes with the same color as boxes[l] that are already adjacent to the left of the current range.

Which companies ask LeetCode 546. Remove Boxes in interviews?

This problem is frequently asked at Capital One, Cisco, Tencent.

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Remove Boxes - LeetCode 546 Solution

Q: What is the space complexity of Remove Boxes?

The space complexity is O(n^3).

Remove Boxes - Complete Solution Guide

Remove Boxes is LeetCode problem 546, 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 are given several boxes with different colors represented by different positive numbers. You may experience several rounds to remove boxes until there is no box left. Each time you can choose some continuous boxes with the same color (i.e., composed of k boxes, k >= 1 ), remove them and get k * k points. Return the maximum points you can get . Example 1: Input: boxes = [1,3,2,2,2,3,4,3,1] Output: 23 Explanation: [1, 3, 2, 2, 2, 3, 4, 3, 1] ----> [1, 3, 3, 4, 3, 1] (3*3=9 points) ----> [1, 3,

Detailed Explanation

The "Remove Boxes" problem challenges us to maximize the points earned by removing continuous groups of boxes with the same color. We are given an array of integers, where each integer represents the color of a box. We can remove a contiguous segment of `k` boxes of the same color in a single operation, earning `k * k` points. The objective is to find the maximum total score achievable by strategically removing boxes until none remain. The input is an array of integers representing the box colors, and the output is the maximum possible score.

Solution Approach

The provided solutions utilize a top-down Dynamic Programming approach with memoization. The core idea is to define a recursive function `dp(l, r, k)` that calculates the maximum score obtainable by removing boxes from index `l` to `r`, given that there are already `k` boxes of the same color as `boxes[l]` adjacent to the left of index `l`. The function explores two main possibilities: (1) removing the current block of boxes immediately to get `(k + count) * (k + count)` points, where count represents consecutive same colored boxes starting from index `l`, and then solving the subproblem from `l + count` to `r`, or (2) merging the current block with another block of the same color within the range `[l+1, r]`. The optimal solution is then the maximum of all these possibilities.

Step-by-Step Algorithm

Step 1: Define the recursive function `dp(l, r, k)` to compute the maximum score for removing boxes within the range [l, r] with k boxes of the same color as boxes[l] already adjacent to the left.
Step 2: Handle the base case: If l > r (empty range), return 0 (no score).
Step 3: Optimize by memoization. Before any calculations, check if `dp(l, r, k)` has been computed previously. If so, return the stored value.
Step 4: Calculate the count of continuous boxes starting from l with the same color. Update l to point after these continuous boxes.
Step 5: Calculate the score by directly removing the 'count' boxes, which is (k + count) * (k + count) and solving the subproblem `dp(l + 1, r, 0)` (starting from the next block with no adjacent same-color boxes, k = 0).
Step 6: Iterate through the remaining range from l+1 to r to find any index 'm' such that boxes[m] has the same color as boxes[l_orig] (original starting index).
Step 7: If such an 'm' is found, consider merging the 'count' boxes at l_orig with the box at index 'm'. This requires removing all boxes between `l + 1` and `m - 1` which is `dp(l+1, m-1, 0)` and then solving the subproblem of removing from m to r, with the existing boxes to the left, `dp(m, r, k + count)`.
Step 8: Store the result in the memoization table `memo[l][r][k]` and return the maximum score found.

Key Insights

Insight 1: Dynamic Programming is crucial to avoid recomputing overlapping subproblems. Memoization (top-down DP) or tabulation (bottom-up DP) will significantly improve performance.
Insight 2: The key state to track is the range (l, r) of boxes being considered and the number (k) of boxes with the same color as boxes[l] that are already adjacent to the left of the current range.
Insight 3: Optimizing the state transition involves considering both immediate removal of a block of same-colored boxes and merging existing blocks to create a larger block for a higher score.

Complexity Analysis

Time Complexity: O(n^4)

Space Complexity: O(n^3)

Topics

This problem involves: Array, Dynamic Programming, Memoization.

Companies

Asked at: Capital One, Cisco, Tencent.

Remove Boxes | LeetCode 546

Hard

Array Dynamic Programming Memoization

Time: O(n^4)Space: O(n^3)

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    private int[][][] dp;
    private int[] boxes;
    private int n;

    public int removeBoxes(int[] boxes) {
        this.boxes = boxes;
        this.n = boxes.length;
        this.dp = new int[n][n][n];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                Arrays.fill(dp[i][j], -1);
            }
        }
        return solve(0, n - 1, 0);
    }

    private int solve(int l, int r, int k) {
        if (l > r) {
            return 0;
        }

        if (dp[l][r][k] != -1) {
            return dp[l][r][k];
        }

        int l_orig = l;
        while (l + 1 <= r && boxes[l + 1] == boxes[l_orig]) {
            l += 1;
        }

        int count = l - l_orig + 1;

        int res = (k + count) * (k + count) + solve(l + 1, r, 0);

        for (int m = l + 1; m <= r; m++) {
            if (boxes[m] == boxes[l_orig]) {
                res = Math.max(res, solve(l + 1, m - 1, 0) + solve(m, r, k + count));
            }
        }

        dp[l][r][k] = res;
        return res;
    }
}

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Problem Info

Description

Problem 546: Remove Boxes - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Memoization

Companies

Capital OneCiscoTencent

Stats

Difficulty:Hard

Languages:4

Companies:3

Problem #:546

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: Define the recursive function `dp(l, r, k)` to compute the maximum score for removing boxes within the range [l, r] with k boxes of the same color as boxes[l] already adjacent to the left.
2
Step 2: Handle the base case: If l > r (empty range), return 0 (no score).
3
Step 3: Optimize by memoization. Before any calculations, check if `dp(l, r, k)` has been computed previously. If so, return the stored value.
4
Step 4: Calculate the count of continuous boxes starting from l with the same color. Update l to point after these continuous boxes.
5
Step 5: Calculate the score by directly removing the 'count' boxes, which is (k + count) * (k + count) and solving the subproblem `dp(l + 1, r, 0)` (starting from the next block with no adjacent same-color boxes, k = 0).
6
Step 6: Iterate through the remaining range from l+1 to r to find any index 'm' such that boxes[m] has the same color as boxes[l_orig] (original starting index).
7
Step 7: If such an 'm' is found, consider merging the 'count' boxes at l_orig with the box at index 'm'. This requires removing all boxes between `l + 1` and `m - 1` which is `dp(l+1, m-1, 0)` and then solving the subproblem of removing from m to r, with the existing boxes to the left, `dp(m, r, k + count)`.
8
Step 8: Store the result in the memoization table `memo[l][r][k]` and return the maximum score found.

💡

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^4)

💾

Space Complexity

O(n^3)

📝

Why?

The time complexity is O(n^4) because the `dp` function has three state variables (l, r, k), each ranging up to n. Within the `dp` function, there is a loop that iterates up to n times, contributing an additional factor of n, hence O(n^4). The space complexity is O(n^3) because the memoization table stores results for all possible combinations of (l, r, k), where each ranges from 0 to n.

📖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

Bottom-Up Dynamic Programming

Instead of recursion with memoization, a bottom-up (tabulation) DP approach can be used. This approach iteratively builds the solution from smaller subproblems to larger ones. The `dp` table is filled in a specific order to ensure that all dependencies are calculated before they are needed.

Time: O(n^4)Space: O(n^3)

Trade-offs: Bottom-up DP can avoid recursion overhead and might be slightly faster in some cases. However, it might be harder to understand and implement than the top-down approach for this specific problem. The space complexity remains the same.

✨

Key Insights

1Insight 1: Dynamic Programming is crucial to avoid recomputing overlapping subproblems. Memoization (top-down DP) or tabulation (bottom-up DP) will significantly improve performance.
2Insight 2: The key state to track is the range (l, r) of boxes being considered and the number (k) of boxes with the same color as boxes[l] that are already adjacent to the left of the current range.
3Insight 3: Optimizing the state transition involves considering both immediate removal of a block of same-colored boxes and merging existing blocks to create a larger block for a higher score.

📚

Key Terms Explained

For Students

Array

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

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: Empty input array. The code handles this implicitly because if the input array is empty, n = 0, and the dp function immediately returns 0.
Edge case 2: Input array with a single element. The code correctly calculates the score as 1*1 = 1. The base case l > r is immediately hit, and the outer cases are solved.
Edge case 3: Input array with all elements of the same color. In this case, the initial 'while' loop will consume the entire array, and the final removal score is simply (k + n) * (k + n), which is correct.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the 'k' parameter (number of same colored boxes already adjacent). This leads to incorrect calculations of the score, as merging existing blocks is not properly accounted for.

✗Common mistake 2: Incorrect base case handling. Forgetting the case when l > r leads to errors and potentially infinite recursion without memoization.

✗Common mistake 3: Not using memoization. This results in exponential time complexity due to redundant calculations and causes Time Limit Exceeded (TLE) errors on larger test cases.

✅

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: Resource optimization: Imagine optimizing the removal of redundant data blocks from a storage system to maximize efficiency. The 'Remove Boxes' algorithm can be adapted to find the optimal way to remove contiguous blocks of identical data.

🏦Real-world scenario 2: Game development: In certain puzzle games, a player earns points by removing groups of same-colored objects. This algorithm can be used to calculate the maximum score achievable in such games.

💡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

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ 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: June 7, 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

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

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 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 identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

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 "Remove 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 dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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