<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(object):
    def maxPoints(self, points):
        
        :type points: List[List[int]]
        :rtype: int

        n=len(points)
        s=max(points[0])
        pc=points[0].index(s)
        m=0
        for i in points:
            print(i)
        if(n>=2):
            for i in range(len(points[0])):
                ts=0
                for j in range(len(points[1])):
                    ts=points[0][i]+points[1][j]-abs(i-j)
                    if(m<ts):
                        m=ts
                        pc=i
                        s=points[0][i]
        print(s,"&&&",pc)
        for i in range(1,n):
            m=len(points[i])
            cm,l=0,0
            sl=[]
            for j in range(m):
                ts=points[i][j]+s-abs(pc-j)
                sl.append(ts)
            s=max(sl)
            pc=sl.index(s)
            tm=points[i][pc]
            print(s,"**",pc,"**",tm)
        return s
"""
class Solution(object):
    def maxPoints(self, points):
        """
        :type points: List[List[int]]
        :rtype: int
        """
        m, n = len(points), len(points[0])

        # Initialize the dp array with the first row
        dp = points[0]

        # Iterate over each row starting from the second row
        for i in range(1, m):
            # Create a new dp array for the current row
            new_dp = [0] * n
            
            # Left sweep: Calculate maximum values considering left-to-right traversal
            left_max = [0] * n
            left_max[0] = dp[0]
            for j in range(1, n):
                left_max[j] = max(left_max[j - 1] - 1, dp[j])

            # Right sweep: Calculate maximum values considering right-to-left traversal
            right_max = [0] * n
            right_max[-1] = dp[-1]
            for j in range(n - 2, -1, -1):
                right_max[j] = max(right_max[j + 1] - 1, dp[j])

            # Combine the left_max and right_max to get the max score for each column
            for j in range(n):
                new_dp[j] = points[i][j] + max(left_max[j], right_max[j])

            # Move to the next row
            dp = new_dp

        # The result is the maximum value in the last dp array
        return max(dp)


                



        
                



        
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