<p>You are given a <code>rows x cols</code> matrix <code>grid</code> representing a field of cherries where <code>grid[i][j]</code> represents the number of cherries that you can collect from the <code>(i, j)</code> cell.</p>

<p>You have two robots that can collect cherries for you:</p>

<ul>
	<li><strong>Robot #1</strong> is located at the <strong>top-left corner</strong> <code>(0, 0)</code>, and</li>
	<li><strong>Robot #2</strong> is located at the <strong>top-right corner</strong> <code>(0, cols - 1)</code>.</li>
</ul>

<p>Return <em>the maximum number of cherries collection using both robots by following the rules below</em>:</p>

<ul>
	<li>From a cell <code>(i, j)</code>, robots can move to cell <code>(i + 1, j - 1)</code>, <code>(i + 1, j)</code>, or <code>(i + 1, j + 1)</code>.</li>
	<li>When any robot passes through a cell, It picks up all cherries, and the cell becomes an empty cell.</li>
	<li>When both robots stay in the same cell, only one takes the cherries.</li>
	<li>Both robots cannot move outside of the grid at any moment.</li>
	<li>Both robots should reach the bottom row in <code>grid</code>.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/04/29/sample_1_1802.png" style="width: 374px; height: 501px;" />
<pre>
<strong>Input:</strong> grid = [[3,1,1],[2,5,1],[1,5,5],[2,1,1]]
<strong>Output:</strong> 24
<strong>Explanation:</strong> Path of robot #1 and #2 are described in color green and blue respectively.
Cherries taken by Robot #1, (3 + 2 + 5 + 2) = 12.
Cherries taken by Robot #2, (1 + 5 + 5 + 1) = 12.
Total of cherries: 12 + 12 = 24.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/04/23/sample_2_1802.png" style="width: 500px; height: 452px;" />
<pre>
<strong>Input:</strong> grid = [[1,0,0,0,0,0,1],[2,0,0,0,0,3,0],[2,0,9,0,0,0,0],[0,3,0,5,4,0,0],[1,0,2,3,0,0,6]]
<strong>Output:</strong> 28
<strong>Explanation:</strong> Path of robot #1 and #2 are described in color green and blue respectively.
Cherries taken by Robot #1, (1 + 9 + 5 + 2) = 17.
Cherries taken by Robot #2, (1 + 3 + 4 + 3) = 11.
Total of cherries: 17 + 11 = 28.
</pre>

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

<ul>
	<li><code>rows == grid.length</code></li>
	<li><code>cols == grid[i].length</code></li>
	<li><code>2 &lt;= rows, cols &lt;= 70</code></li>
	<li><code>0 &lt;= grid[i][j] &lt;= 100</code></li>
</ul>

class Solution {
public:
    int cherryPickup(vector<vector<int>>& grid) {
        int rows = grid.size(); // Number of rows in the grid
        int cols = grid[0].size(); // Number of columns in the grid

        // Initialize a 3D DP table
        // dp[r][c1][c2] will store the maximum cherries collected if
        // robot1 is at column c1 and robot2 is at column c2 in row r.
        vector<vector<vector<int>>> dp(rows, vector<vector<int>>(cols, vector<int>(cols, -1)));

        // Initial state: both robots start from the first row
        dp[0][0][cols - 1] = grid[0][0] + (cols > 1 ? grid[0][cols - 1] : 0);

        // Iterate through each row
        for (int r = 1; r < rows; ++r) {
            // Initialize the current DP state for this row
            vector<vector<int>> current(cols, vector<int>(cols, -1));
            for (int c1 = 0; c1 < cols; ++c1) {
                for (int c2 = 0; c2 < cols; ++c2) {
                    if (dp[r - 1][c1][c2] == -1) continue; // Skip invalid states

                    // Try all possible new positions for both robots
                    for (int newC1 = max(0, c1 - 1); newC1 <= min(cols - 1, c1 + 1); ++newC1) {
                        for (int newC2 = max(0, c2 - 1); newC2 <= min(cols - 1, c2 + 1); ++newC2) {
                            // Calculate cherries collected in the current row for both robots
                            int cherries = dp[r - 1][c1][c2] + grid[r][newC1];
                            if (newC1 != newC2) cherries += grid[r][newC2];
                            // Update the current DP state with the maximum cherries collected
                            current[newC1][newC2] = max(current[newC1][newC2], cherries);
                        }
                    }
                }
            }
            // Update DP table for the current row
            dp[r] = current;
        }

        // Find the maximum cherries collected in the last row
        int maxCherries = 0;
        for (int c1 = 0; c1 < cols; ++c1) {
            for (int c2 = 0; c2 < cols; ++c2) {
                maxCherries = max(maxCherries, dp[rows - 1][c1][c2]);
            }
        }
        return maxCherries;
    }
};

24527