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Remove Boxes

Hard

Asked by:

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Topics:

ArraysDynamic ProgrammingRecursion

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, 3, 3, 1] (1*1=1 points) 
----> [1, 1] (3*3=9 points) 
----> [] (2*2=4 points)

Example 2:

Input: boxes = [1,1,1]
Output: 9

Example 3:

Input: boxes = [1]
Output: 1

Constraints:

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

Try coding it in LeetCode

Remove Boxes

Hard

Asked by:

23 views

Topics:

ArraysDynamic ProgrammingRecursion

You are given several boxes with different colors represented by different positive numbers.

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, 3, 3, 1] (1*1=1 points) 
----> [1, 1] (3*3=9 points) 
----> [] (2*2=4 points)

Example 2:

Input: boxes = [1,1,1]
Output: 9

Example 3:

Input: boxes = [1]
Output: 1

Constraints:

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

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values for the numbers within the boxes array?
Can the input array `boxes` be empty or null?
If multiple box removal sequences achieve the same maximum score, can I return any one of them?
Could you provide a small example input and the corresponding expected output to illustrate the scoring rules more concretely?
Are the values in the input array always integers, or can they be floating-point numbers?

Brute Force Solution

Approach

The problem asks us to maximize points by removing boxes. Brute force means trying every possible removal order. We explore all options until we find the one that gives us the highest score.

Here's how the algorithm would work step-by-step:

Start by considering all the boxes and figuring out what happens if we remove the first box.
Then, pretend we're back at the beginning and consider what happens if we remove the second box instead.
Keep doing this for every box in the beginning.
For each of those choices, we're left with a smaller set of boxes. Again, try removing each of the remaining boxes one by one.
Keep repeating this process: each time, explore the consequences of removing each of the boxes that are left.
Eventually, we'll reach a point where there are no boxes left. At that point, we can calculate the total score we got for that particular removal order.
Keep track of the scores for all the different removal orders we tried.
Finally, compare all the scores we calculated, and choose the highest score as the best possible result.

Code Implementation

def remove_boxes_brute_force(boxes):
    def calculate_max_score(remaining_boxes):
        # If no boxes are left, score is 0
        if not remaining_boxes:
            return 0

        max_score_found = 0
        
        # Try removing each box, one at a time
        for box_index in range(len(remaining_boxes)):

            # Simulate removing the current box
            current_box = remaining_boxes[box_index]
            boxes_after_removal = remaining_boxes[:box_index] + remaining_boxes[box_index+1:]

            # Calculate the score for removing this box
            score_for_removal = current_box * current_box

            # Recursively calculate the max score after removing box
            score_if_removed = score_for_removal + calculate_max_score(boxes_after_removal)

            # Keep track of the highest score
            max_score_found = max(max_score_found, score_if_removed)

        return max_score_found

    # Initiate the brute force search with the original set of boxes
    return calculate_max_score(boxes)

Big(O) Analysis

Time Complexity

O(n!) – The provided explanation describes a brute-force approach where we explore every possible removal order of the boxes. For 'n' boxes, there are n! (n factorial) possible orderings. Each removal order requires calculating a score. Therefore, the algorithm explores n! different paths to find the maximum score, leading to O(n!) time complexity. This is because the number of possible permutations grows factorially with the number of boxes.

Space Complexity

O(N^3) – The described solution uses recursion to explore all possible removal orders. To avoid redundant computations, it implicitly memoizes the results of subproblems. The number of possible subproblems that need to be memoized depends on the remaining sequence of boxes after some removals and potentially the number of identical boxes at the end of the remaining sequence. Since the remaining sequence can be identified by the start and end indices, and the count of the same colored boxes appended, a 3-dimensional array of size N x N x N is needed to store the memoized results, where N is the number of boxes. Thus, the auxiliary space used by the memoization table is O(N^3).

Optimal Solution

Approach

The goal is to maximize points by removing groups of boxes of the same color. Instead of trying all possible removal orders, this solution uses a clever approach that remembers the best scores for smaller sections of the boxes, building up to the full solution.

Here's how the algorithm would work step-by-step:

Imagine you're looking at a section of the boxes and there's a group of boxes of the same color at the very end.
Instead of removing those ending boxes right away, think about if you could find more boxes of the same color earlier in the sequence and bring them together to form an even bigger group with the boxes at the end.
If you can do that, you'll get a much higher score because you're removing a larger group all at once.
The clever part is to remember the best score you can get for each possible section of boxes, including the number of matching boxes at the end.
When you consider removing a section, check if it's better to remove the ending group right away, or to first try to bring more matching boxes to the end from somewhere else in the section.
You use the previously remembered best scores for the smaller sections that remain after these removals to figure out the best move.
By remembering the best scores for smaller sections and building upon them, you avoid trying every single removal order and find the highest possible score more efficiently.

Code Implementation

def remove_boxes(boxes):    number_of_boxes = len(boxes)
    memo = {}

    def calculate_max_points(left, right, consecutive_matches):
        if left > right:
            return 0

        if (left, right, consecutive_matches) in memo:
            return memo[(left, right, consecutive_matches)]

        # Optimize: Extend existing group
        while left < right and boxes[right] == boxes[right - 1]:
            right -= 1
            consecutive_matches += 1

        # Option 1: Remove the current consecutive group
        max_points = (consecutive_matches + 1) ** 2 + calculate_max_points(left, right - 1, 0)

        # Option 2: Merge with an earlier group
        for index in range(left, right):

            # Look for a same-color box to merge to the right
            if boxes[index] == boxes[right] and boxes[index] != boxes[right -1]:
                # Attempt to find best possible removal from merging
                max_points = max(max_points,
                                  calculate_max_points(left, index - 1, 0) +
                                  calculate_max_points(index + 1, right - 1, consecutive_matches + 1))

        memo[(left, right, consecutive_matches)] = max_points
        return max_points

    # Initiate recursive function with memoization
    return calculate_max_points(0, number_of_boxes - 1, 0)

Big(O) Analysis

Time Complexity

O(n^4) – The algorithm uses dynamic programming to consider all subproblems of removing boxes. The core idea is to consider sections of the boxes defined by a starting index i, an ending index j, and a count k representing the number of boxes of the same color as boxes[j] that are attached to the end. Therefore, the dp table is a 3D array of size n*n*n. The algorithm iterates through all possible subproblems, and for each subproblem, it potentially iterates through a range of boxes to find matching colors, which takes O(n) time. Therefore, the overall time complexity is O(n*n*n * n) which simplifies to O(n^4).

Space Complexity

O(N^3) – The solution utilizes a 3D array, typically named `dp`, to store the best scores for subproblems. The dimensions of this array are determined by the range of possible start indices, end indices, and the count of matching boxes at the end of a section. All three dimensions are directly influenced by the input size, N, the number of boxes. Thus, the `dp` array requires space proportional to N * N * N, or N^3. The recursive calls, while present, are optimized via memoization using the dp array, so they do not contribute significantly to space complexity beyond what is taken by the dp array. Therefore, the auxiliary space complexity is O(N^3).

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 immediately since there are no boxes to remove.
Array with only one box	Return 1 * 1 since removing the single box yields a score of 1.
Array with all identical boxes	The optimal solution involves removing all boxes at once, so calculate n*n.
Maximum array size (e.g., n=100)	Memoization is crucial to avoid exceeding time limit due to exponential recursive calls.
Boxes with values close to integer limits	The scoring calculations should be checked for potential integer overflow and handled appropriately by using larger data types or modular arithmetic if needed.
Recursion depth exceeding stack limit	Memoization is used within recursion to avoid recomputing the same subproblems, which implicitly prevents exceeding stack limit.
Input array with duplicate values intermixed.	Memoization will handle duplicate values correctly in determining the optimal solution through previously calculated values.
Highly skewed distribution of box values.	Memoization ensures all possible combinations are explored regardless of the input distribution.

Null or empty input array

How to Handle:

Return 0 immediately since there are no boxes to remove.

Array with only one box

How to Handle:

Return 1 * 1 since removing the single box yields a score of 1.

Array with all identical boxes

How to Handle:

The optimal solution involves removing all boxes at once, so calculate n*n.

Maximum array size (e.g., n=100)

How to Handle:

Memoization is crucial to avoid exceeding time limit due to exponential recursive calls.

Boxes with values close to integer limits

How to Handle:

The scoring calculations should be checked for potential integer overflow and handled appropriately by using larger data types or modular arithmetic if needed.

Recursion depth exceeding stack limit

How to Handle:

Memoization is used within recursion to avoid recomputing the same subproblems, which implicitly prevents exceeding stack limit.

Input array with duplicate values intermixed.

How to Handle:

Memoization will handle duplicate values correctly in determining the optimal solution through previously calculated values.

Highly skewed distribution of box values.

How to Handle:

Memoization ensures all possible combinations are explored regardless of the input distribution.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values for the numbers within the boxes array?
Can the input array `boxes` be empty or null?
If multiple box removal sequences achieve the same maximum score, can I return any one of them?
Could you provide a small example input and the corresponding expected output to illustrate the scoring rules more concretely?
Are the values in the input array always integers, or can they be floating-point numbers?

Brute Force Solution

Approach

The problem asks us to maximize points by removing boxes. Brute force means trying every possible removal order. We explore all options until we find the one that gives us the highest score.

Here's how the algorithm would work step-by-step:

Start by considering all the boxes and figuring out what happens if we remove the first box.
Then, pretend we're back at the beginning and consider what happens if we remove the second box instead.
Keep doing this for every box in the beginning.
For each of those choices, we're left with a smaller set of boxes. Again, try removing each of the remaining boxes one by one.
Keep repeating this process: each time, explore the consequences of removing each of the boxes that are left.
Eventually, we'll reach a point where there are no boxes left. At that point, we can calculate the total score we got for that particular removal order.
Keep track of the scores for all the different removal orders we tried.
Finally, compare all the scores we calculated, and choose the highest score as the best possible result.

Code Implementation

def remove_boxes_brute_force(boxes):
    def calculate_max_score(remaining_boxes):
        # If no boxes are left, score is 0
        if not remaining_boxes:
            return 0

        max_score_found = 0
        
        # Try removing each box, one at a time
        for box_index in range(len(remaining_boxes)):

            # Simulate removing the current box
            current_box = remaining_boxes[box_index]
            boxes_after_removal = remaining_boxes[:box_index] + remaining_boxes[box_index+1:]

            # Calculate the score for removing this box
            score_for_removal = current_box * current_box

            # Recursively calculate the max score after removing box
            score_if_removed = score_for_removal + calculate_max_score(boxes_after_removal)

            # Keep track of the highest score
            max_score_found = max(max_score_found, score_if_removed)

        return max_score_found

    # Initiate the brute force search with the original set of boxes
    return calculate_max_score(boxes)

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Imagine you're looking at a section of the boxes and there's a group of boxes of the same color at the very end.
Instead of removing those ending boxes right away, think about if you could find more boxes of the same color earlier in the sequence and bring them together to form an even bigger group with the boxes at the end.
If you can do that, you'll get a much higher score because you're removing a larger group all at once.
The clever part is to remember the best score you can get for each possible section of boxes, including the number of matching boxes at the end.
When you consider removing a section, check if it's better to remove the ending group right away, or to first try to bring more matching boxes to the end from somewhere else in the section.
You use the previously remembered best scores for the smaller sections that remain after these removals to figure out the best move.
By remembering the best scores for smaller sections and building upon them, you avoid trying every single removal order and find the highest possible score more efficiently.

Code Implementation

def remove_boxes(boxes):    number_of_boxes = len(boxes)
    memo = {}

    def calculate_max_points(left, right, consecutive_matches):
        if left > right:
            return 0

        if (left, right, consecutive_matches) in memo:
            return memo[(left, right, consecutive_matches)]

        # Optimize: Extend existing group
        while left < right and boxes[right] == boxes[right - 1]:
            right -= 1
            consecutive_matches += 1

        # Option 1: Remove the current consecutive group
        max_points = (consecutive_matches + 1) ** 2 + calculate_max_points(left, right - 1, 0)

        # Option 2: Merge with an earlier group
        for index in range(left, right):

            # Look for a same-color box to merge to the right
            if boxes[index] == boxes[right] and boxes[index] != boxes[right -1]:
                # Attempt to find best possible removal from merging
                max_points = max(max_points,
                                  calculate_max_points(left, index - 1, 0) +
                                  calculate_max_points(index + 1, right - 1, consecutive_matches + 1))

        memo[(left, right, consecutive_matches)] = max_points
        return max_points

    # Initiate recursive function with memoization
    return calculate_max_points(0, number_of_boxes - 1, 0)

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 immediately since there are no boxes to remove.
Array with only one box	Return 1 * 1 since removing the single box yields a score of 1.
Array with all identical boxes	The optimal solution involves removing all boxes at once, so calculate n*n.
Maximum array size (e.g., n=100)	Memoization is crucial to avoid exceeding time limit due to exponential recursive calls.
Boxes with values close to integer limits	The scoring calculations should be checked for potential integer overflow and handled appropriately by using larger data types or modular arithmetic if needed.
Recursion depth exceeding stack limit	Memoization is used within recursion to avoid recomputing the same subproblems, which implicitly prevents exceeding stack limit.
Input array with duplicate values intermixed.	Memoization will handle duplicate values correctly in determining the optimal solution through previously calculated values.
Highly skewed distribution of box values.	Memoization ensures all possible combinations are explored regardless of the input distribution.

Null or empty input array

How to Handle:

Return 0 immediately since there are no boxes to remove.

Array with only one box

How to Handle:

Return 1 * 1 since removing the single box yields a score of 1.

Array with all identical boxes

How to Handle:

The optimal solution involves removing all boxes at once, so calculate n*n.

Maximum array size (e.g., n=100)

How to Handle:

Memoization is crucial to avoid exceeding time limit due to exponential recursive calls.

Boxes with values close to integer limits

How to Handle:

The scoring calculations should be checked for potential integer overflow and handled appropriately by using larger data types or modular arithmetic if needed.

Recursion depth exceeding stack limit

How to Handle:

Memoization is used within recursion to avoid recomputing the same subproblems, which implicitly prevents exceeding stack limit.

Input array with duplicate values intermixed.

How to Handle:

Memoization will handle duplicate values correctly in determining the optimal solution through previously calculated values.

Highly skewed distribution of box values.

How to Handle:

Memoization ensures all possible combinations are explored regardless of the input distribution.