LeetCode in Kotlin

3603. Minimum Cost Path with Alternating Directions II

Medium

You are given two integers m and n representing the number of rows and columns of a grid, respectively.

The cost to enter cell (i, j) is defined as (i + 1) * (j + 1).

You are also given a 2D integer array waitCost where waitCost[i][j] defines the cost to wait on that cell.

You start at cell (0, 0) at second 1.

At each step, you follow an alternating pattern:

• On odd-numbered seconds, you must move right or down to an adjacent cell, paying its entry cost.
• On even-numbered seconds, you must wait in place, paying waitCost[i][j].

Return the minimum total cost required to reach (m - 1, n - 1).

Example 1:

Input: m = 1, n = 2, waitCost = [[1,2]]

Output: 3

Explanation:

The optimal path is:

• Start at cell (0, 0) at second 1 with entry cost (0 + 1) * (0 + 1) = 1.
• Second 1: Move right to cell (0, 1) with entry cost (0 + 1) * (1 + 1) = 2.

Thus, the total cost is 1 + 2 = 3.

Example 2:

Input: m = 2, n = 2, waitCost = [[3,5],[2,4]]

Output: 9

Explanation:

The optimal path is:

• Start at cell (0, 0) at second 1 with entry cost (0 + 1) * (0 + 1) = 1.
• Second 1: Move down to cell (1, 0) with entry cost (1 + 1) * (0 + 1) = 2.
• Second 2: Wait at cell (1, 0), paying waitCost[1][0] = 2.
• Second 3: Move right to cell (1, 1) with entry cost (1 + 1) * (1 + 1) = 4.

Thus, the total cost is 1 + 2 + 2 + 4 = 9.

Example 3:

Input: m = 2, n = 3, waitCost = [[6,1,4],[3,2,5]]

Output: 16

Explanation:

The optimal path is:

• Start at cell (0, 0) at second 1 with entry cost (0 + 1) * (0 + 1) = 1.
• Second 1: Move right to cell (0, 1) with entry cost (0 + 1) * (1 + 1) = 2.
• Second 2: Wait at cell (0, 1), paying waitCost[0][1] = 1.
• Second 3: Move down to cell (1, 1) with entry cost (1 + 1) * (1 + 1) = 4.
• Second 4: Wait at cell (1, 1), paying waitCost[1][1] = 2.
• Second 5: Move right to cell (1, 2) with entry cost (1 + 1) * (2 + 1) = 6.

Thus, the total cost is 1 + 2 + 1 + 4 + 2 + 6 = 16.

Constraints:

• 1 <= m, n <= 105
• 2 <= m * n <= 105
• waitCost.length == m
• waitCost[0].length == n
• 0 <= waitCost[i][j] <= 105

Solution

import kotlin.math.min

class Solution {
    fun minCost(m: Int, n: Int, waitCost: Array<IntArray>): Long {
        val dp = LongArray(n)
        dp[0] = 1L
        for (j in 1..<n) {
            val entry = j + 1L
            val wait = waitCost[0][j].toLong()
            dp[j] = dp[j - 1] + entry + wait
        }
        for (i in 1..<m) {
            val entry00 = i + 1L
            val wait00 = waitCost[i][0].toLong()
            dp[0] = dp[0] + entry00 + wait00
            for (j in 1..<n) {
                val entry = (i + 1).toLong() * (j + 1)
                val wait = waitCost[i][j].toLong()
                val fromAbove = dp[j]
                val fromLeft = dp[j - 1]
                dp[j] = min(fromAbove, fromLeft) + entry + wait
            }
        }
        return dp[n - 1] - waitCost[m - 1][n - 1]
    }
}

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