How do you solve Remove Boxes on LeetCode?

Remove Boxes (LeetCode #546) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n³) time complexity and O(n³) space complexity. Key insight: The order of removing boxes can significantly affect the total score.

What is the brute force solution for Remove Boxes?

The brute force approach for Remove Boxes is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves trying every possible way to remove boxes and calculating the points for each scenario. This means we will recursively remove boxes and keep track of the maximum points we can achieve. The optimal Optimal Solution approach improves this to O(n³).

What algorithm or data structure is used to solve Remove Boxes?

Remove Boxes uses the following concepts: Array, Dynamic Programming, Memoization. The recommended patterns to study are: Dynamic Programming, Memoization.

What are the interview tips for Remove Boxes?

Clearly explain your thought process and how you plan to optimize the brute force solution. Be prepared to discuss the trade-offs between time and space complexity. Practice writing the DP table and how states relate to each other.

What are common mistakes when solving Remove Boxes?

Not considering all possible segments for removal. Failing to correctly handle the memoization state.

#546

Remove Boxes

Hard

ArrayDynamic ProgrammingMemoizationDynamic ProgrammingMemoization

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n³)
Space	O(1)	O(n³)

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Intuition

Time O(n³)Space O(n³)

The optimal approach uses dynamic programming to store results of subproblems, which avoids redundant calculations. By memoizing the results, we can efficiently compute the maximum points for any configuration of boxes.

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Algorithm

3 steps

1Step 1: Create a 3D DP array where dp[l][r][k] represents the maximum points obtainable from boxes[l] to boxes[r] with k additional boxes of the same color as boxes[r].
2Step 2: Iterate through all possible segments of boxes and fill the DP table based on previous results.
3Step 3: Return the value stored in dp[0][n-1][0] which gives the maximum points for the entire array.

solution.py14 lines

1def removeBoxes(boxes):
2    n = len(boxes)
3    dp = [[[0] * (n + 1) for _ in range(n)] for _ in range(n)]
4    for l in range(n - 1, -1, -1):
5        for r in range(l, n):
6            for k in range(n):
7                while r + 1 < n and boxes[r] == boxes[r + 1]:
8                    r += 1
9                    k += 1
10                dp[l][r][k] = max(dp[l][r][k], (k + 1) * (k + 1) + (dp[l][r - 1][0] if r > 0 else 0))
11                for m in range(l, r):
12                    if boxes[m] == boxes[r]:
13                        dp[l][r][k] = max(dp[l][r][k], dp[l][m][k] + dp[m + 1][r - 1][0])
14    return dp[0][n - 1][0]

ℹ

Complexity note: The time complexity is O(n³) due to the three nested loops for the DP table, where n is the length of the boxes array. Each state in the DP table is computed based on previous states.

1The order of removing boxes can significantly affect the total score.
2Using memoization helps in avoiding recalculating results for the same state.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.