Maximum Number of Points with Cost

Problem Statement

In this challenge, you are presented with a two-dimensional integer matrix named points, indexed from 0. The matrix dimension is m x n, where m is the number of rows and n is the number of columns. Starting with a score of zero, your goal is to maximize the score by selecting one cell from each row.

When you choose a cell (r, c) in the matrix, the value points[r][c] is added to your score. However, there's a catch: if the positions of the cells chosen in consecutive rows are too far apart horizontally, you incur a penalty. This penalty is calculated as the absolute difference between the column indices of the selected cells from two consecutive rows. Your task is to determine the maximum score possible after considering all additions and subtractions.

Examples

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

Approach and Intuition

Let's dissect the problem with the assistance of the provided examples and the stated constraints:

Understanding the Dynamics with Examples

1. In Example 1:

   • Input matrix: [[1,2,3],[1,5,1],[3,1,1]]
   • The optimal path starts at the cell (0, 2) then goes to (1, 1) and ends at (2, 0).
   • Total points collected: 3 (from points[0][2]) + 5 (from points[1][1]) + 3 (from points[2][0]) = 11
   • The penalty incurred: abs(2-1) + abs(1-0) = 2 from moving between selected cells.
   • Final score: 11 - 2 = 9

   This demonstrates the need to balance between choosing high-value cells and minimizing the movement penalty between consecutive rows.

2. In Example 2:

   • Input matrix: [[1,5],[2,3],[4,2]]
   • The optimal path starts at the cell (0, 1) then goes to (1, 1), and ends at (2, 0).
   • Total collected points: 5 + 3 + 4 = 12
   • The penalty incurred: abs(1-1) + abs(1-0) = 1
   • Final score: 12 - 1 = 11

From these examples, we observe the following:

• Structural Understanding: The problem is inherently dynamic, where the optimum choice at each step relies on previous decisions.
• Dynamic Programming Strategy: Modeling this problem with a dynamic programming approach seems crucial. For every row r, and every possible column selection c, maintain the maximum possible score up to that point. Transition between states would consider all possible previous column choices c' from row r-1, maximizing points[r][c] - abs(c - c') + dp[r-1][c'].
• Initialization and Iteration: The first row serves as the base case where no penalty applies, and subsequent rows build upon previous computations.
• Efficient Computation and Pruning: Given the large size of m and n (up to 105), optimizing the dynamic transitions and reducing redundant calculations is significant. Strategies like maintaining maximum and second maximum values from the previous row could be explored to reduce the time complexity per row transition.

In essence, understanding the penalty mechanism and designing an efficient state transition strategy are central to solving this problem optimally within given constraints.

Solutions

class Solution {
public:
    long long maxScore(vector<vector<int>>& matrix) {
        int nCols = matrix[0].size();
        vector<long long> prevRow(nCols);
    
        for (auto& row : matrix) {
            long long maxLeft = 0;
    
            // Traverse from left to right
            for (int j = 0; j < nCols; ++j) {
                maxLeft = max(maxLeft - 1, prevRow[j]);
                prevRow[j] = maxLeft;
            }
    
            long long maxRight = 0;
            // Traverse from right to left
            for (int j = nCols - 1; j >= 0; --j) {
                maxRight = max(maxRight - 1, prevRow[j]);
                prevRow[j] = max(prevRow[j], maxRight) + row[j];
            }
        }
    
        // Extract the maximum score
        long long highestScore = 0;
        for (int j = 0; j < nCols; ++j) {
            highestScore = max(highestScore, prevRow[j]);
        }
    
        return highestScore;
    }
};