LeetCode in Kotlin

1463. Cherry Pickup II

Hard

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

class Solution {
    fun cherryPickup(grid: Array<IntArray>): Int {
        val m = grid.size
        val n = grid[0].size
        val dp = Array(n) { Array(n) { IntArray(m) } }
        dp[0][n - 1][0] = grid[0][0] + grid[0][n - 1]
        for (k in 1 until m) {
            for (i in 0..Math.min(n - 1, k)) {
                for (j in n - 1 downTo Math.max(0, n - 1 - k)) {
                    dp[i][j][k] = maxOfLast(dp, i, j, k) + grid[k][i] + if (i == j) 0 else grid[k][j]
                }
            }
        }
        var result = 0
        for (i in 0..Math.min(n - 1, m)) {
            for (j in n - 1 downTo Math.max(0, n - 1 - m)) {
                result = Math.max(result, dp[i][j][m - 1])
            }
        }
        return result
    }

    private fun maxOfLast(dp: Array<Array<IntArray>>, i: Int, j: Int, k: Int): Int {
        var result = 0
        for (x in -1..1) {
            for (y in -1..1) {
                val r = i + x
                val c = j + y
                if (r >= 0 && r < dp[0].size && c >= 0 && c < dp[0].size) {
                    result = Math.max(result, dp[r][c][k - 1])
                }
            }
        }
        return result
    }
}

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