How do you solve Maximum Number of Points with Cost on LeetCode?

Maximum Number of Points with Cost (LeetCode #1937) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n) time complexity and O(n) space complexity. Key insight: Dynamic programming allows us to build solutions incrementally, reducing the need to explore all combinations.

What is the brute force solution for Maximum Number of Points with Cost?

The brute force approach for Maximum Number of Points with Cost is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking all possible combinations of cell selections across rows. This means we would explore every possible way to pick cells, which is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(m * n).

What algorithm or data structure is used to solve Maximum Number of Points with Cost?

Maximum Number of Points with Cost uses the following concepts: Array, Dynamic Programming, Matrix. The recommended patterns to study are: Dynamic Programming, Matrix Traversal.

What are the interview tips for Maximum Number of Points with Cost?

Always explain your thought process clearly when transitioning from brute force to optimal solutions. Be prepared to discuss time and space complexities in detail. Practice explaining your code as you write it to simulate the interview environment.

What are common mistakes when solving Maximum Number of Points with Cost?

Failing to account for penalties when transitioning between rows. Not initializing the dp array correctly, leading to incorrect scores.

#1937

Maximum Number of Points with Cost

Medium

ArrayDynamic ProgrammingMatrixDynamic ProgrammingMatrix Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(m * n)
Space	O(1)	O(n)

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Intuition

Time O(m * n)Space O(n)

The optimal solution uses dynamic programming to build up the maximum score row by row, considering the penalties for column differences. This allows us to efficiently compute the maximum score without exploring all combinations.

⚙️

Algorithm

4 steps

1Step 1: Initialize a dp array to store the maximum points for each column in the current row.
2Step 2: For each row, compute the maximum points for each column by considering the previous row's points and applying the penalties for column differences.
3Step 3: Update the dp array for the current row and repeat until all rows are processed.
4Step 4: Return the maximum value from the last row of the dp array.

solution.py22 lines

1# Full working Python code
2
3def maxPoints(points):
4    m, n = len(points), len(points[0])
5    dp = points[0][:]
6    for i in range(1, m):
7        new_dp = [0] * n
8        max_left = [0] * n
9        max_right = [0] * n
10
11        max_left[0] = dp[0]
12        for j in range(1, n):
13            max_left[j] = max(max_left[j - 1], dp[j] + j)
14
15        max_right[n - 1] = dp[n - 1] - (n - 1)
16        for j in range(n - 2, -1, -1):
17            max_right[j] = max(max_right[j + 1], dp[j] - j)
18
19        for j in range(n):
20            new_dp[j] = points[i][j] + max(max_left[j] - j, max_right[j] + j)
21        dp = new_dp
22    return max(dp)

ℹ

Complexity note: The time complexity is O(m * n) because we process each cell in the matrix once, while the space complexity is O(n) due to the dp array storing results for the current row.

1Dynamic programming allows us to build solutions incrementally, reducing the need to explore all combinations.
2Understanding how to manage penalties effectively is crucial for optimizing the score.

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