LeetCode 546: Remove Boxes Solution in Python – A Step-by-Step Guide

What Is LeetCode 546: Remove Boxes?

In LeetCode 546: Remove Boxes, you’re given an array boxes representing a sequence of colored boxes (integers), and your task is to remove all boxes by selecting consecutive segments of the same color in any order, earning points equal to the square of the segment’s length each time, with the goal of maximizing the total score. For example, with boxes = [1, 3, 2, 2, 2, 3, 1], you can score 23 points by removing strategically (e.g., five 2s for 25 points isn’t optimal). This problem builds on LeetCode 312: Burst Balloons for dynamic programming with sequence partitioning.

Problem Statement

• Input: boxes (List[int])—array of box colors (integers).
• Output: Integer—maximum points achievable by removing all boxes.
• Rules: Remove consecutive same-color boxes, score = (number removed)² per move; maximize total.

Constraints

• 1 <= boxes.length <= 100
• 1 <= boxes[i] <= 100

Examples

• Input: boxes = [1,3,2,2,2,3,1]
• Output: 23
• Possible: Remove [2,2,2] (9), [3] (1), [3] (1), [1] (1), [1] (1) = 13; Optimal: 23 (explained later).
• Input: boxes = [1,1,1]
• Output: 9
• Remove all three 1s at once: 3² = 9.
• Input: boxes = [1]
• Output: 1
• Single box: 1² = 1.

Understanding the Problem: Maximizing Points

To solve LeetCode 546: Remove Boxes in Python, we need to maximize points by removing all boxes, where each removal of k consecutive same-color boxes scores k² points. The catch? Order matters: removing a segment can connect previously separate same-color boxes, increasing future scores (e.g., removing 3s in [1, 3, 2, 2, 2, 3, 1] connects 1s). A naive approach might try all permutations (exponential), but with up to 100 boxes, dynamic programming or optimized recursion is key. We’ll explore:

• Best Solution (DP with Memoization): O(n³) time, O(n³) space—optimal and efficient.
• Alternative Solution (Recursive DFS with Pruning): O(n * 2ⁿ) time, O(n)—intuitive but slower.

Let’s dive into the DP solution with a friendly breakdown!

Best Solution: Dynamic Programming with Memoization

Why DP Wins

The DP with memoization solution is the best for LeetCode 546 because it efficiently computes the maximum score in O(n³) time and O(n³) space by breaking the problem into subproblems—considering segments and their interactions—using a 3D memoization table to avoid recomputation. It’s like solving a puzzle by testing all ways to group boxes, caching each piece for speed!

How It Works

Think of this as a box-scoring strategist:

• Step 1: Define DP State:
• dp[i][j][k]: Max score for boxes[i:j+1] with k boxes of value boxes[i] appended before i.
• Step 2: Base Case:
• Empty range (i > j): 0 points.
• Single box (i = j): (k+1)² points (k boxes + current).
• Step 3: Recurrence:
• Remove boxes[i] with k boxes: (k+1)² + dp[i+1][j][0].
• Find next same value m, remove intermediate, connect: dp[i+1][m-1][0] + dp[m][j][k+1].
• Take max of all options.
• Step 4: Memoize:
• Cache results in 3D table.
• Why It Works:
• Captures all grouping possibilities.
• Memoization eliminates redundant subproblems.

It’s like a box-scoring optimizer!

Step-by-Step Example

Example: boxes = [1, 3, 2, 2, 2, 3, 1]

• Init: dp = {}.
• Goal: dp[0][6][0] (full array, no prepend).
• Step 1: dp[0][6][0]:
• Remove 1: 1² + dp[1][6][0].
• Find next 1 (m=6): dp[1][5][0] + dp[6][6][1].
• Step 2: dp[1][6][0]:
• Remove 3: 1² + dp[2][6][0].
• Next 3 (m=5): dp[2][4][0] + dp[5][6][1].
• Step 3: dp[2][6][0]:
• Remove [2,2,2]: 3² + dp[5][6][0].
• Split at 3 (m=5): dp[2][4][0] + dp[5][6][1].
• Compute: Bottom-up:
• dp[6][6][1]: (1+1)² = 4.
• dp[5][6][1]: Max(1² + 4, 2²) = 5.
• dp[2][4][0]: 3² = 9.
• dp[2][6][0]: Max(9, 9 + 5) = 14.
• Continue up: Final dp[0][6][0] = 23.
• Result: 23.

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained for beginners:

class Solution:
    def removeBoxes(self, boxes: List[int]) -> int:
        # Step 1: Initialize memoization table
        dp = {}

        # Step 2: DFS with memoization
        def dfs(i, j, k):
            if i > j:  # Empty range
                return 0
            if (i, j, k) in dp:
                return dp[(i, j, k)]

            # Base case: single box
            if i == j:
                dp[(i, j, k)] = (k + 1) * (k + 1)
                return dp[(i, j, k)]

            # Step 3: Remove current segment
            max_points = (k + 1) * (k + 1) + dfs(i + 1, j, 0)

            # Step 4: Find next same value to connect
            for m in range(i + 1, j + 1):
                if boxes[m] == boxes[i]:
                    points = dfs(i + 1, m - 1, 0) + dfs(m, j, k + 1)
                    max_points = max(max_points, points)

            dp[(i, j, k)] = max_points
            return max_points

        # Step 5: Start with full array
        return dfs(0, len(boxes) - 1, 0)

• Line 4: Memoization dictionary.
• Lines 7-9: Base case: empty range = 0, cached result.
• Lines 12-15: Single box: (k+1)² points.
• Line 18: Remove first segment, recurse.
• Lines 21-24: Try connecting with next same value, take max.
• Line 27: Cache and return max points.
• Line 30: Start with full array, k=0.
• Time Complexity: O(n³)—n states for i, j, k up to n.
• Space Complexity: O(n³)—memoization table.

It’s like a box-scoring mastermind!

Alternative Solution: Recursive DFS with Pruning

Why an Alternative Approach?

The recursive DFS with pruning explores all removal sequences, pruning suboptimal paths with basic checks, running in O(n * 2ⁿ) time and O(n) space (recursion stack). It’s intuitive but inefficient due to exponential growth, making it a good baseline for understanding before DP.

How It Works

Picture this as a box-removal adventurer:

• Step 1: Recursively try all removal options.
• Step 2: For each position, remove consecutive same values.
• Step 3: Prune by tracking current score vs. max.
• Step 4: Return max score.

It’s like a box-removal explorer!

Step-by-Step Example

Example: boxes = [1, 1]

• DFS:
• Remove [1,1]: 2² = 4.
• Split: Remove 1, then 1: 1² + 1² = 2.
• Result: 4.

Code for DFS Approach

class Solution:
    def removeBoxes(self, boxes: List[int]) -> int:
        def dfs(arr, memo={}):
            if not arr:
                return 0
            key = tuple(arr)
            if key in memo:
                return memo[key]

            max_score = 0
            i = 0
            while i < len(arr):
                j = i
                while j < len(arr) and arr[j] == arr[i]:
                    j += 1
                count = j - i
                new_arr = arr[:i] + arr[j:]
                score = count * count + dfs(new_arr, memo)
                max_score = max(max_score, score)
                i = j

            memo[key] = max_score
            return max_score

        return dfs(boxes)

• Line 4: Base case: empty array = 0.
• Lines 7-8: Memoization check.
• Lines 11-19: Try removing each consecutive segment, recurse.
• Line 22: Cache and return max.
• Time Complexity: O(n * 2ⁿ)—exponential due to all subsequences.
• Space Complexity: O(n)—recursion stack.

It’s a recursive box scorer!

Comparing the Two Solutions

• DP (Best):
• Pros: O(n³), O(n³), optimal.
• Cons: 3D DP complexity.
• DFS (Alternative):
• Pros: O(n * 2ⁿ), O(n), intuitive.
• Cons: Exponential time.

DP wins for efficiency!

Additional Examples and Edge Cases

• [1]: 1.
• [1, 1, 2, 2]: 8.
• [1, 2, 1]: 5.

DP handles them all!

Complexity Recap

• DP: Time O(n³), Space O(n³).
• DFS: Time O(n * 2ⁿ), Space O(n).

DP’s the efficiency champ!

Key Takeaways

• DP: Memoized mastery—learn at Python Basics!
• DFS: Recursive exploration.
• Boxes: Scoring is fun.
• Python: DP or recursion, your pick!

Final Thoughts: Score Those Boxes!

LeetCode 546: Remove Boxes in Python is a challenging DP puzzle. DP with memoization is your fast track, while recursive DFS offers an intuitive start. Want more? Try LeetCode 312: Burst Balloons or LeetCode 1143: Longest Common Subsequence. Ready to remove? Head to Solve LeetCode 546 on LeetCode and maximize that score today!