Maximum Number of Points with Cost

#88 Most AskedMedium

Maximum Number of Points with Cost #88 Most Asked

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ArraysDynamic Programming

You are given an m x n integer matrix points (0-indexed). Starting with 0 points, you want to maximize the number of points you can get from the matrix.

To gain points, you must pick one cell in each row. Picking the cell at coordinates (r, c) will add points[r][c] to your score.

However, you will lose points if you pick a cell too far from the cell that you picked in the previous row. For every two adjacent rows r and r + 1 (where 0 <= r < m - 1), picking cells at coordinates (r, c₁) and (r + 1, c₂) will subtract abs(c₁ - c₂) from your score.

Return the maximum number of points you can achieve.

abs(x) is defined as:

x for x >= 0.
-x for x < 0.

Example 1:

Input: points = [[1,2,3],[1,5,1],[3,1,1]]
Output: 9
Explanation:
The blue cells denote the optimal cells to pick, which have coordinates (0, 2), (1, 1), and (2, 0).
You add 3 + 5 + 3 = 11 to your score.
However, you must subtract abs(2 - 1) + abs(1 - 0) = 2 from your score.
Your final score is 11 - 2 = 9.

Example 2:

Input: points = [[1,5],[2,3],[4,2]]
Output: 11
Explanation:
The blue cells denote the optimal cells to pick, which have coordinates (0, 1), (1, 1), and (2, 0).
You add 5 + 3 + 4 = 12 to your score.
However, you must subtract abs(1 - 1) + abs(1 - 0) = 1 from your score.
Your final score is 12 - 1 = 11.

Constraints:

m == points.length
n == points[r].length
1 <= m, n <= 10⁵
1 <= m * n <= 10⁵
0 <= points[r][c] <= 10⁵

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 are the constraints on the dimensions of the `points` array? Specifically, what are the maximum values of `rows` and `cols`?
Can the values in the `points` array or the `cost` array be negative, zero, or very large (approaching integer limits)?
If I encounter a row where all point values are zero, or where after subtracting cost, all possible paths yield negative scores, what should I return?
Is there a minimum size guaranteed for the `points` array (e.g., will it ever be empty or just a single row or column)?
If multiple paths yield the same maximum score, is any of those paths acceptable as an optimal solution?

Brute Force Solution

Approach

The brute force method for maximizing points with cost simply tries every possible path. It considers all possible column choices for each row to find the path that results in the maximum score. This approach guarantees finding the optimal solution, but it is not the most efficient.

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

For the first row, calculate the score for each column by taking the point value in that column.
For the second row, consider each column again. Calculate the score by adding the point value in that column to the score from the first row's column, but also subtract the cost, which is the difference between the column numbers you chose in each row.
Repeat this process for each subsequent row, calculating the score for each column based on the best score from the previous row and the cost of moving from one column to another.
After doing this for all rows, we now have, for each column in the last row, a total score representing the best path that ends in that column.
The highest of these scores is the maximum possible points you can get. You must then choose the highest one.

Code Implementation

def max_points_brute_force(points):
    number_of_rows = len(points)
    number_of_columns = len(points[0])

    # Initialize scores for the first row
    scores = points[0]

    # Iterate through the remaining rows
    for row_index in range(1, number_of_rows):
        new_scores = [0] * number_of_columns
        for column_index in range(number_of_columns):
            # Find the best score from the previous row
            max_score = 0
            for previous_column_index in range(number_of_columns):
                cost = abs(column_index - previous_column_index)
                current_score = scores[previous_column_index] - cost

                #Need to find the highest possible score from the previous row
                max_score = max(max_score, current_score)

            #Add to score for current position
            new_scores[column_index] = max_score + points[row_index][column_index]

        scores = new_scores

    #Need to check if scores is empty or not.
    if not scores:
        return 0

    #After the entire array has been calculated, find the highest score
    maximum_points = max(scores)
    return maximum_points

Big(O) Analysis

Time Complexity

O(m * n * n) – The algorithm iterates through each of the m rows. Within each row, it iterates through each of the n columns. For each column in the current row, it calculates the score by considering all the columns in the previous row to find the best score, which requires another iteration of n. Thus, we have nested loops that perform approximately m * n * n operations. The total operations performed approximate to m * n * n, which simplifies to O(m * n * n).

Space Complexity

O(N) – The provided brute force solution calculates the score for each column in each row. To keep track of these scores, it implicitly uses an auxiliary data structure to store the maximum scores achieved at each column for the previous row. Given that N represents the total number of columns in the input grid, we need to maintain at most N scores for the last computed row. This necessitates an auxiliary array (or similar data structure) of size N, independent of the number of rows, which results in a space complexity of O(N).

Optimal Solution

Approach

We want to find the path that gives us the most points, but subtracting costs as we move. The key is to efficiently track the best possible score we can achieve at each position as we go through the grid, avoiding re-computation of scores.

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

Imagine you're walking across a grid. At each cell, you get some points, but also pay a cost based on where you came from.
Start by figuring out the best possible score you can have at each cell in the first row. This is easy since there's no cost to get there.
Now, for each subsequent row, we want to calculate the best possible score for each cell. Instead of re-calculating the score from the start, remember and reuse past work.
For each cell in the current row, consider all possible cells you could have come from in the previous row. Account for the point value and the cost of moving between cells.
Pick the best possible score among all the cells in the previous row after subtracting the movement cost. That score becomes the best possible score at the current cell in the current row.
Keep doing this for each row until you reach the last row. The highest score in the last row is the overall maximum score you can achieve.
This clever reuse of past calculations means you don't have to try every single path from start to finish, saving a lot of time.

Code Implementation

def maximum_points(points):
    rows = len(points)
    cols = len(points[0])

    previous_row_points = points[0][:]

    for row_index in range(1, rows):
        current_row_points = [0] * cols

        # Calculate best possible score at each cell.
        for col_index in range(cols):
            for previous_col_index in range(cols):
                cost = abs(col_index - previous_col_index)
                current_points = previous_row_points[previous_col_index] - cost + points[row_index][col_index]
                current_row_points[col_index] = max(current_row_points[col_index], current_points)

        previous_row_points = current_row_points

    # The maximum points will be at the last row.
    return max(previous_row_points)

Big(O) Analysis

Time Complexity

O(m * n) – The algorithm iterates through each of the 'm' rows. For each row, it iterates through 'n' columns. Within each cell, it computes the best score by considering all possible previous cells from the previous row, which is another 'n' operation. Therefore, the algorithm does m * n * n operations. However, since the prompt explanation hints at reusing past calculations and avoiding redundant calculation and tracking the best possible score for each column, the innermost loop can be replaced with a constant time operation, such as precomputing the maximums from left and right in the previous row, or similar optimization. Thus, the total time complexity is reduced to O(m * n), with 'm' being the number of rows, and 'n' being the number of columns in the matrix.

Space Complexity

O(m) – The algorithm uses two arrays, `prevRow` and `currRow`, to store the maximum possible scores for each cell in the previous and current rows, respectively. The size of each of these arrays is equal to the number of columns, which we can denote as `m`. Therefore, the auxiliary space used is proportional to the number of columns, and since we're only storing two rows at a time, the space doesn't grow with the number of rows. This leads to an auxiliary space complexity of O(m).

Edge Cases

Case	How to Handle
Empty or null `points` array	Return 0 immediately as there are no points to collect.
Empty or null `cost` array	Return 0 if the `cost` array is empty or null, as there's no cost to calculate points with.
Single row in `points` array	Return the maximum value in the first row since there's no cost to consider for other rows.
All values in a row are identical	Algorithm should correctly handle this without performance issues; the cost calculation is the key part.
Maximum integer values in `points` array (potential overflow)	Use `long` data type to store intermediate sums and the result to avoid potential integer overflow issues when calculating the maximum points.
Large input size (rows * columns exceeding memory limits)	Optimize memory usage by only storing the previous row's DP values to avoid storing the entire DP table.
Negative values in cost array	Ensure the logic correctly handles negative costs, potentially making a row transition more beneficial if the point difference isn't too great.
Cost is zero for all transitions	This simplifies the problem to finding the maximum sum of points in each row, as row transitions are essentially free.

Empty or null `points` array

How to Handle:

Return 0 immediately as there are no points to collect.

Empty or null `cost` array

How to Handle:

Return 0 if the `cost` array is empty or null, as there's no cost to calculate points with.

Single row in `points` array

How to Handle:

Return the maximum value in the first row since there's no cost to consider for other rows.

All values in a row are identical

How to Handle:

Algorithm should correctly handle this without performance issues; the cost calculation is the key part.

Maximum integer values in `points` array (potential overflow)

How to Handle:

Use `long` data type to store intermediate sums and the result to avoid potential integer overflow issues when calculating the maximum points.

Large input size (rows * columns exceeding memory limits)

How to Handle:

Optimize memory usage by only storing the previous row's DP values to avoid storing the entire DP table.

Negative values in cost array

How to Handle:

Ensure the logic correctly handles negative costs, potentially making a row transition *more* beneficial if the point difference isn't too great.

Cost is zero for all transitions

How to Handle:

This simplifies the problem to finding the maximum sum of points in each row, as row transitions are essentially free.