<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>

'''
thoughts:
fill each r,c 's points -> O(m*n)
how?
    p[r][c] = max of (last row's each point + points[r][c] - c_diff)
        for c1 in range(n):
            p[r][c] = p[r-1][c1] + points[r][c] - abs(c - c1)
fill the table, and get max of last row

like dp
base first row == points

improve: only need prev row to fill cur row -> dp: 2 rows --> O(m*n*n) -- ETL
need max(dp[1])

improve: 
p[1][c] = points[r][c] + p[0][j] - abs(c - j) --> only do once 
    -> pre cal left, right max for p[0][j] - abs(c - j)

'''

#     [[1,2,3],
#      [1,5,1],
#      [3,1,1]]
# p[0]: 2 7 4
# p[1]: 5 9 6

#     [[1,2,3],
#      [1,5,1],
#      [3,1,1]]
# left: [2,7,4]
# right:[1,2,3]
#     

class Solution:
    def maxPoints(self, points: List[List[int]]) -> int:
        '''
        pre cal left, right max for p[0][j] - abs(c - j)
        cur row's p: point + max(left, right)
        '''
        m, n = len(points), len(points[0])
        prev = points[0]
        for r in range(1, m):
            left_max = prev[:]
            right_max = prev[:]
            cur = points[r][:]
            for c in range(1, n):
                left_max[c] = max(left_max[c], left_max[c-1] - 1)
            for c in range(n-2, -1, -1):
                right_max[c] = max(right_max[c], right_max[c+1] - 1)
            for c in range(n):
                cur[c] += max(left_max[c], right_max[c])

            prev = cur[:]
        return max(prev)
            











        # # correct thougts, but exceed time limit
        # m, n = len(points), len(points[0])
        # p = [points[0][:], [0] * n]
        
        # for r in range(1, m):
        #     for c in range(n):
        #         for j in range(n):
        #             p[1][c] = max(p[1][c], p[0][j] + points[r][c] - abs(c - j))
        #     p[0] = p[1]
        #     p[1] = [0] * n
        # return max(p[0])
1542