LeetCode in Kotlin

3418. Maximum Amount of Money Robot Can Earn

Medium

You are given an m x n grid. A robot starts at the top-left corner of the grid (0, 0) and wants to reach the bottom-right corner (m - 1, n - 1). The robot can move either right or down at any point in time.

The grid contains a value coins[i][j] in each cell:

• If coins[i][j] >= 0, the robot gains that many coins.
• If coins[i][j] < 0, the robot encounters a robber, and the robber steals the absolute value of coins[i][j] coins.

The robot has a special ability to neutralize robbers in at most 2 cells on its path, preventing them from stealing coins in those cells.

Note: The robot’s total coins can be negative.

Return the maximum profit the robot can gain on the route.

Example 1:

Input: coins = [[0,1,-1],[1,-2,3],[2,-3,4]]

Output: 8

Explanation:

An optimal path for maximum coins is:

1. Start at (0, 0) with 0 coins (total coins = 0).
2. Move to (0, 1), gaining 1 coin (total coins = 0 + 1 = 1).
3. Move to (1, 1), where there’s a robber stealing 2 coins. The robot uses one neutralization here, avoiding the robbery (total coins = 1).
4. Move to (1, 2), gaining 3 coins (total coins = 1 + 3 = 4).
5. Move to (2, 2), gaining 4 coins (total coins = 4 + 4 = 8).

Example 2:

Input: coins = [[10,10,10],[10,10,10]]

Output: 40

Explanation:

An optimal path for maximum coins is:

1. Start at (0, 0) with 10 coins (total coins = 10).
2. Move to (0, 1), gaining 10 coins (total coins = 10 + 10 = 20).
3. Move to (0, 2), gaining another 10 coins (total coins = 20 + 10 = 30).
4. Move to (1, 2), gaining the final 10 coins (total coins = 30 + 10 = 40).

Constraints:

• m == coins.length
• n == coins[i].length
• 1 <= m, n <= 500
• -1000 <= coins[i][j] <= 1000

Solution

import kotlin.math.max

class Solution {
    fun maximumAmount(coins: Array<IntArray>): Int {
        val m = coins.size
        val n = coins[0].size
        val dp = Array<IntArray>(m) { IntArray(n) }
        val dp1 = Array<IntArray>(m) { IntArray(n) }
        val dp2 = Array<IntArray>(m) { IntArray(n) }
        dp[0][0] = coins[0][0]
        for (j in 1..<n) {
            dp[0][j] = dp[0][j - 1] + coins[0][j]
        }
        for (i in 1..<m) {
            dp[i][0] = dp[i - 1][0] + coins[i][0]
        }
        for (i in 1..<m) {
            for (j in 1..<n) {
                dp[i][j] = max(dp[i][j - 1], dp[i - 1][j]) + coins[i][j]
            }
        }
        dp1[0][0] = max(coins[0][0], 0)
        for (j in 1..<n) {
            dp1[0][j] = max(dp[0][j - 1], (dp1[0][j - 1] + coins[0][j]))
        }
        for (i in 1..<m) {
            dp1[i][0] = max(dp[i - 1][0], (dp1[i - 1][0] + coins[i][0]))
        }
        for (i in 1..<m) {
            for (j in 1..<n) {
                dp1[i][j] = max(
                    max(dp[i][j - 1], dp[i - 1][j]),
                    (max(dp1[i][j - 1], dp1[i - 1][j]) + coins[i][j]),
                )
            }
        }
        dp2[0][0] = max(coins[0][0], 0)
        for (j in 1..<n) {
            dp2[0][j] = max(dp1[0][j - 1], (dp2[0][j - 1] + coins[0][j]))
        }
        for (i in 1..<m) {
            dp2[i][0] = max(dp1[i - 1][0], (dp2[i - 1][0] + coins[i][0]))
        }
        for (i in 1..<m) {
            for (j in 1..<n) {
                dp2[i][j] = max(
                    max(dp1[i][j - 1], dp1[i - 1][j]),
                    (max(dp2[i][j - 1], dp2[i - 1][j]) + coins[i][j]),
                )
            }
        }
        return dp2[m - 1][n - 1]
    }
}

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