<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 n = grid.size();
        int m = grid[0].size();
        vector<vector<vector<int>>> dp(n, vector<vector<int>> (m, vector<int> (m, -1)));    // {i, {j1, j2}}
        vector<vector<int>> nextMatrix(m, vector<int> (m, -1));    // {j1, j2}

        // Base case: Both robots reach bottom 
        for(int j1=0; j1<m; j1++) {
            for(int j2=0; j2<m; j2++) {
                if(j1 == j2) nextMatrix[j1][j2] = grid[n-1][j1];     // Both robots are standing at same cell, return cell only once
                else nextMatrix[j1][j2] = grid[n-1][j1] + grid[n-1][j2];    // Both robots are standing at different cell, return sum of both cells
            }
        }

        for(int i=n-2; i>=0; i--) {
            
            vector<vector<int>> currMatrix(m, vector<int> (m, INT_MIN));
            for(int j1=0; j1<m; j1++) {
                for(int j2=0; j2<m; j2++) {

                    int maxCherry = INT_MIN;    // To store maximum cheery picked while trying all movement combinations for both robots
                    // For each move of robot_1, robot_2 have 3 different move options
                    // Hence, in total 9 possible movement combinations
                    // Both robots move down simulatenously, but in columns they have separate choices 
                    // Hence, using dj1 & dj2 for depicting choice of robot_1 & robot_2 respectively
                    for(int dj1=-1; dj1<=1; dj1++) {
                        for(int dj2=-1; dj2<=1; dj2++) {
                
                            // Check if both robots are currently at same cell or not
                            if(j1 == j2) maxCherry = max(maxCherry, (( (i+1 < n) && (j1+dj1 >= 0) && (j1+dj1 < m) && (j2+dj2 >= 0) && (j2+dj2 < m) ) ? (grid[i][j1] + nextMatrix[j1+dj1][j2+dj2]) : INT_MIN));
                            else maxCherry = max(maxCherry, (( (i+1 < n) && (j1+dj1 >= 0) && (j1+dj1 < m) && (j2+dj2 >= 0) && (j2+dj2 < m) ) ? (grid[i][j1] + grid[i][j2] + nextMatrix[j1+dj1][j2+dj2]) : INT_MIN));
                        }
                    }

                    currMatrix[j1][j2] = maxCherry;
                }
            }

            nextMatrix = currMatrix;
        }

        return nextMatrix[0][m-1];    // Tells maxCherryPicked till both robots reach bottom, given they're initially at locations (0, 0) & (0, m-1) respectively
    }
};
25317