<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 {
private:
    int solve(vector<vector<int>>& grid, int i, int j1, int j2, vector<vector<vector<int>>>& dp){
        if(j1 < 0 || j2 < 0 || j1 > grid[0].size()-1 || j2 > grid[0].size()-1) return -1e8;
        if(i == grid.size()-1){
            if(j1 == j2) return grid[i][j1];
            return grid[i][j1] + grid[i][j2];
        }

        if(dp[i][j1][j2] != -1) return dp[i][j1][j2];

        int maxi = INT_MIN;
        for(int d1=-1; d1<=1; d1++){
            for(int d2=-1; d2<=1; d2++){
                int temp = INT_MIN;
                if(j1 == j2) temp = grid[i][j1] + solve(grid, i+1, j1+d1, j2+d2, dp);
                else temp = grid[i][j1] + grid[i][j2] + solve(grid, i+1, j1+d1, j2+d2, dp);
                maxi = max(maxi, temp);
            }
        }

        return dp[i][j1][j2] = maxi;
}


int solveTab(vector<vector<int>>& grid, int r, int c){
    vector<vector<vector<int>>> dp(r+1, vector<vector<int>>(c+1, vector<int>(c+1, 0)));
    
    for(int j1=0; j1<c; j1++){
        for(int j2=0; j2<c; j2++){
            if(j1 == j2) dp[r-1][j1][j2] = grid[r-1][j1];
            else dp[r-1][j1][j2] = grid[r-1][j1] + grid[r-1][j2];
        }
    }

    for(int i=r-2; i>=0; i--){
        for(int j1=0; j1<c; j1++){
            for(int j2=0; j2<c; j2++){
                for(int d1=-1; d1<=1; d1++){
                    for(int d2=-1; d2<=1; d2++){
                        int temp = INT_MIN;
                        if(j1+d1>c-1 || j1+d1<0 || j2+d2>c-1 || j2+d2<0) continue;
                        if(j1 == j2) temp = grid[i][j1] + dp[i+1][j1+d1][j2+d2];
                        else temp = grid[i][j1] + grid[i][j2] + dp[i+1][j1+d1][j2+d2];
                        dp[i][j1][j2] = max(dp[i][j1][j2], temp);
                    }
                }
            }
        }
    }
    
    return dp[0][0][c-1];
}

public:
    int cherryPickup(vector<vector<int>>& grid) {
        int r = grid.size();
        int c = grid[0].size();
        // vector<vector<vector<int>>> dp(r, vector<vector<int>>(c, vector<int>(c, -1)));
        // return solve(grid, 0, 0, c-1, dp);
        return solveTab(grid, r, c);
    }
};
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