<p>You are given a <strong>0-indexed</strong> 2D array <code>grid</code> of size <code>2 x n</code>, where <code>grid[r][c]</code> represents the number of points at position <code>(r, c)</code> on the matrix. Two robots are playing a game on this matrix.</p>

<p>Both robots initially start at <code>(0, 0)</code> and want to reach <code>(1, n-1)</code>. Each robot may only move to the <strong>right</strong> (<code>(r, c)</code> to <code>(r, c + 1)</code>) or <strong>down </strong>(<code>(r, c)</code> to <code>(r + 1, c)</code>).</p>

<p>At the start of the game, the <strong>first</strong> robot moves from <code>(0, 0)</code> to <code>(1, n-1)</code>, collecting all the points from the cells on its path. For all cells <code>(r, c)</code> traversed on the path, <code>grid[r][c]</code> is set to <code>0</code>. Then, the <strong>second</strong> robot moves from <code>(0, 0)</code> to <code>(1, n-1)</code>, collecting the points on its path. Note that their paths may intersect with one another.</p>

<p>The <strong>first</strong> robot wants to <strong>minimize</strong> the number of points collected by the <strong>second</strong> robot. In contrast, the <strong>second </strong>robot wants to <strong>maximize</strong> the number of points it collects. If both robots play <strong>optimally</strong>, return <em>the <b>number of points</b> collected by the <strong>second</strong> robot.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/09/08/a1.png" style="width: 388px; height: 103px;" />
<pre>
<strong>Input:</strong> grid = [[2,5,4],[1,5,1]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> The optimal path taken by the first robot is shown in red, and the optimal path taken by the second robot is shown in blue.
The cells visited by the first robot are set to 0.
The second robot will collect 0 + 0 + 4 + 0 = 4 points.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/09/08/a2.png" style="width: 384px; height: 105px;" />
<pre>
<strong>Input:</strong> grid = [[3,3,1],[8,5,2]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> The optimal path taken by the first robot is shown in red, and the optimal path taken by the second robot is shown in blue.
The cells visited by the first robot are set to 0.
The second robot will collect 0 + 3 + 1 + 0 = 4 points.
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/09/08/a3.png" style="width: 493px; height: 103px;" />
<pre>
<strong>Input:</strong> grid = [[1,3,1,15],[1,3,3,1]]
<strong>Output:</strong> 7
<strong>Explanation: </strong>The optimal path taken by the first robot is shown in red, and the optimal path taken by the second robot is shown in blue.
The cells visited by the first robot are set to 0.
The second robot will collect 0 + 1 + 3 + 3 + 0 = 7 points.
</pre>

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

<ul>
	<li><code>grid.length == 2</code></li>
	<li><code>n == grid[r].length</code></li>
	<li><code>1 &lt;= n &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= grid[r][c] &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution:
    def gridGame(self, grid: List[List[int]]) -> int:
        length = len(grid[0])
        # set first and last cell null
        grid[0][0] = 0
        grid[1][length-1] = 0
        # first row sum
        sum_top = sum(grid[0])
        # accumulating second row sum
        sum_down = 0
        # if the first cell of the second row is bigger then all 
        # the cells in first row return first row sum
        if sum_top < grid[1][0]:
            return sum_top
        # set result the maximum possible to search lesser
        res = sum(grid[1]) + sum(grid[0]) # max
        
        for i in range(length-1):
            # moving right set upper row cell to zero and left bottom sell 
            sum_top -= grid[0][i+1]
            sum_down += grid[1][i]
            
            # setting passed upper row sell to null
            grid[0][i+1]=0
            
            # if last res is bigger set new res
            if res > max(sum_top, sum_down):
                res = max(sum_top, sum_down)
            # if not we already found the res
            else:
                return res
        return res

            



        
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