Cherry Pickup II

Two robots Need to reach end by picking up cherries one present at top left and other at top right, return maximum cherries they can pick and if they pick from same cell, it should be counted as one

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

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). 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 passes through a cell, It picks up all cherries, and the cell becomes an empty cell. When both robots stay in the same cell, only one takes the cherries. Both robots cannot move outside of the grid at any moment. Both robots should reach the bottom row in 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.

Constraints:

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

Solution

we'll be iterating both robot together because there could be mutual cell and we don't want to add them to twice. While traversing row will be same for both robots.

For each robot there would be three path but combined there would be 9 path. So, we'll apply a nested loop to cover all these paths.

var cherryPickup = function (grid) {
    let m = grid.length;
    let n = grid[0].length;

    function recurr(i, j1, j2) {
        let maxi = -Infinity;

        if(j1<0 || j2<0 || j1>=n || j2>=n){
            return -1e8;
        }
        if (i == m - 1) {
            //base case

            if (j1 == j2) {
                return grid[i][j1]
            } else {
                return grid[i][j1] + grid[i][j2]
            }
        }

        for (let dj1 = -1; dj1 <= 1; dj1++) {

            for (let dj2 = -1; dj2 <= 1; dj2++) {
                let value = 0;

                if (j1 == j2) {
                    value += grid[i][j1]
                } else {
                    value += grid[i][j1] + grid[i][j2]
                }

                value += recurr(i + 1, j1 + dj1, j2 + dj2);

                maxi = Math.max(value, maxi);
            }
        }
        return maxi;
    }

   return recurr(0,0,n-1)

};

Memorization

var cherryPickup = function (grid) {
    let m = grid.length;
    let n = grid[0].length;
    let dp = Array.from({ length: m }, () =>
        Array.from({ length: n }, () => Array(n))
    );

    function recurr(i, j1, j2, dp) {
        let maxi = -Infinity;

        if (j1 < 0 || j2 < 0 || j1 >= n || j2 >= n) {
            return -1e8;
        }

        if (dp[i][j1][j2] !== undefined) {
            return dp[i][j1][j2]
        }
        if (i == m - 1) {
            //base case

            if (j1 == j2) {
                return grid[i][j1]
            } else {
                return grid[i][j1] + grid[i][j2]
            }
        }

        for (let dj1 = -1; dj1 <= 1; dj1++) {

            for (let dj2 = -1; dj2 <= 1; dj2++) {
                let value = 0;

                if (j1 == j2) {
                    value += grid[i][j1]
                } else {
                    value += grid[i][j1] + grid[i][j2]
                }

                value += recurr(i + 1, j1 + dj1, j2 + dj2, dp);

                maxi = Math.max(value, maxi);


            }
        }

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

    return recurr(0, 0, n - 1, dp)

};