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

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

<p>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 <code>r</code> and <code>r + 1</code> (where <code>0 &lt;= r &lt; m - 1</code>), picking cells at coordinates <code>(r, c<sub>1</sub>)</code> and <code>(r + 1, c<sub>2</sub>)</code> will <strong>subtract</strong> <code>abs(c<sub>1</sub> - c<sub>2</sub>)</code> from your score.</p>

<p>Return <em>the <strong>maximum</strong> number of points you can achieve</em>.</p>

<p><code>abs(x)</code> is defined as:</p>

<ul>
	<li><code>x</code> for <code>x &gt;= 0</code>.</li>
	<li><code>-x</code> for <code>x &lt; 0</code>.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong><strong> </strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/07/12/screenshot-2021-07-12-at-13-40-26-diagram-drawio-diagrams-net.png" style="width: 300px; height: 300px;" />
<pre>
<strong>Input:</strong> points = [[1,2,3],[1,5,1],[3,1,1]]
<strong>Output:</strong> 9
<strong>Explanation:</strong>
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.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/07/12/screenshot-2021-07-12-at-13-42-14-diagram-drawio-diagrams-net.png" style="width: 200px; height: 299px;" />
<pre>
<strong>Input:</strong> points = [[1,5],[2,3],[4,2]]
<strong>Output:</strong> 11
<strong>Explanation:</strong>
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.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == points.length</code></li>
	<li><code>n == points[r].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= m * n &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= points[r][c] &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution {
public:
    long long maxPoints(vector<vector<int>>& points) {
        const auto& m = points.size();
        const auto& n = points[0].size();
        
        std::vector<std::vector<long long>> dp(m, std::vector<long long>(n, 0));

        long long ans = 0;

        for (int jdx = 0; jdx < n; ++jdx) {
            dp[0][jdx] = points[0][jdx];
            ans = std::max(dp[0][jdx], ans);
        }

        std::vector<long long> left_best(n, 0);
        std::vector<long long> right_best(n, 0);

        for (int row = 1; row < m; ++row) {
            left_best[0]      = dp[row - 1][0];
            right_best[n - 1] = dp[row - 1][n - 1];

            for (int col = 1; col < n; ++col) {
                left_best[col] = std::max(dp[row - 1][col], left_best[col - 1] - 1);
            }

            for (int col = n - 2; col >= 0; --col) {
                right_best[col] = std::max(dp[row - 1][col], right_best[col + 1] - 1);
            }

            for (int col = 0; col < n; ++col) {
                long long point = points[row][col];
                //std::cout << left_best[col] << "\t" << right_best[col] << std::endl;
                dp[row][col] = std::max(left_best[col] + point, right_best[col] + point);
                //std:cout << dp[row][col] << std::endl;;
                ans = std::max(ans, dp[row][col]);
            }

           // std::cout << std::endl;

            left_best.clear();
            right_best.clear();
        }


        

        return ans;
    }
};


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