<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:
    long long dp[70][70][70]; // Adjust size according to constraints

    long long compute(int row, int col1, int col2, vector<vector<int>>& grid, int m, int n) {
        // base case
        if (col1 < 0 || col1 >= n || col2 < 0 || col2 >= n) return LLONG_MIN / 2;  // check bounds
        if (row == m) return 0;  // Changed this condition

        if (dp[row][col1][col2] != -1) return dp[row][col1][col2]; // Use memoized result

        // LORE
        // For every movement of Alice, Bob can have 3 movements
        // e.g.: Alice moves bottom left -> Bob can do bottom left, bottom down, bottom right
        // Similarly for 2 other movements of Alice, Bob will have 3 possible movements for each

        long long cherries = grid[row][col1];
        if (col1 != col2) cherries += grid[row][col2];

        long long maxi = LLONG_MIN / 2;  // initialize to a very small number

        for (int dAlice = -1; dAlice <= 1; dAlice++) { // movement of Alice
            int ncol1 = col1 + dAlice;
            for (int dBob = -1; dBob <= 1; dBob++) { // corresponding movement of Bob (3 possible)
                int ncol2 = col2 + dBob;
                // Ensure indices are valid before accessing grid
                if (ncol1 >= 0 && ncol1 < n && ncol2 >= 0 && ncol2 < n) { 
                    long long nextCherries = compute(row + 1, ncol1, ncol2, grid, m, n);
                    maxi = max(maxi, cherries + nextCherries);
                }
            }
        }
        return dp[row][col1][col2] = maxi;
    }

    int cherryPickup(vector<vector<int>>& grid) {
        // fixed point(start) to variable point(end)
        // whenever this always try starting with a fixed point

        // Alice (rob 1)
        // Bob (rob 2)
        // Row movement is the same for both of them 
        // Column will differ
        int m = grid.size();
        int n = grid[0].size();
        memset(dp, -1, sizeof(dp));  // Initialize dp array to -1
        long long result = compute(0, 0, n - 1, grid, m, n);
        return max(0LL, result);  // Ensure non-negative result
    }
};
14257