# Cherry Pickup II (Leetcode 1463) Solution Mani
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Given a `rows x cols` matrix `grid` representing a field of cherries. Each cell in `grid` represents the number of cherries that you can collect.

You have two robots that can collect cherries for you, Robot #1 is located at the top-left corner (0,0) , and Robot #2 is located at the top-right corner (0, cols-1) of the grid.

Return the maximum number of cherries collection using both robots  by following the rules below:

• From a cell (i,j), robots can move to cell (i+1, j-1) , (i+1, j) or (i+1, j+1).
• When any robot is passing through a cell, It picks it up all cherries, and the cell becomes an empty cell (0).
• When both robots stay on the same cell, only one of them takes the cherries.
• Both robots cannot move outside of the grid at any moment.
• Both robots should reach the bottom row in the `grid`.

Example 1: ```Input: grid = [[3,1,1],[2,5,1],[1,5,5],[2,1,1]]
Output: 24
Explanation: 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.
```

Example 2: ```Input: 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]]
Output: 28
Explanation: 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.
```

Example 3:

```Input: grid = [[1,0,0,3],[0,0,0,3],[0,0,3,3],[9,0,3,3]]
Output: 22
```

Example 4:

```Input: grid = [[1,1],[1,1]]
Output: 4
```

Constraints:

• `rows == grid.length`
• `cols == grid[i].length`
• `2 <= rows, cols <= 70`
• `0 <= grid[i][j] <= 100 `

Solution:

``````1
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/**
* Time Complexity: O(M * N ^ 2)
* Space Complexity: O(M * N ^ 2)
**/
class Solution {
public int cherryPickup(int[][] grid) {
dp = new int[grid.length][grid.length][grid.length];
for (int i = 0; i < dp.length; i++) {
for (int j = 0; j < dp[i].length; j++) {
Arrays.fill(dp[i][j], -1);
}
}

return solve(grid, 0, 0, grid.length - 1);
}

int[][][] dp;

int solve(int[][] grid, int x, int y1, int y2) {
if (x < 0 || x >= grid.length || y1 < 0 || y1 >= grid.length || y2 < 0 || y2 >= grid.length) {
return 0;
}

if (dp[x][y1][y2] != -1) {
return dp[x][y1][y2];
}

int res = grid[x][y1];
if (y1 != y2) {
res += grid[x][y2];
}

int max = Integer.MIN_VALUE;
for (int i = -1; i <= 1; i++) {
for (int j = -1; j <= 1; j++) {
max = Math.max(max, solve(grid, x + 1, y1 + i, y2 + j));
}
}

res = res + max;
dp[x][y1][y2] = res;

return res;
}

}
``````
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