LeetCode in Kotlin

2304. Minimum Path Cost in a Grid

Medium

You are given a 0-indexed m x n integer matrix grid consisting of distinct integers from 0 to m * n - 1. You can move in this matrix from a cell to any other cell in the next row. That is, if you are in cell (x, y) such that x < m - 1, you can move to any of the cells (x + 1, 0), (x + 1, 1), …, (x + 1, n - 1). Note that it is not possible to move from cells in the last row.

Each possible move has a cost given by a 0-indexed 2D array moveCost of size (m * n) x n, where moveCost[i][j] is the cost of moving from a cell with value i to a cell in column j of the next row. The cost of moving from cells in the last row of grid can be ignored.

The cost of a path in grid is the sum of all values of cells visited plus the sum of costs of all the moves made. Return the minimum cost of a path that starts from any cell in the first row and ends at any cell in the last row.

Example 1:

Input: grid = [[5,3],[4,0],[2,1]], moveCost = [[9,8],[1,5],[10,12],[18,6],[2,4],[14,3]]

Output: 17

Explanation: The path with the minimum possible cost is the path 5 -> 0 -> 1.

• The sum of the values of cells visited is 5 + 0 + 1 = 6.

• The cost of moving from 5 to 0 is 3.

• The cost of moving from 0 to 1 is 8.

So the total cost of the path is 6 + 3 + 8 = 17.

Example 2:

Input: grid = [[5,1,2],[4,0,3]], moveCost = [[12,10,15],[20,23,8],[21,7,1],[8,1,13],[9,10,25],[5,3,2]]

Output: 6

Explanation: The path with the minimum possible cost is the path 2 -> 3.

• The sum of the values of cells visited is 2 + 3 = 5.

• The cost of moving from 2 to 3 is 1.

So the total cost of this path is 5 + 1 = 6.

Constraints:

• m == grid.length
• n == grid[i].length
• 2 <= m, n <= 50
• grid consists of distinct integers from 0 to m * n - 1.
• moveCost.length == m * n
• moveCost[i].length == n
• 1 <= moveCost[i][j] <= 100

Solution

class Solution {
    fun minPathCost(grid: Array<IntArray>, moveCost: Array<IntArray>): Int {
        val m = grid.size
        val n = grid[0].size
        val dp = Array(m) { IntArray(n) }
        System.arraycopy(grid[m - 1], 0, dp[m - 1], 0, n)
        for (i in m - 2 downTo 0) {
            for (j in 0 until n) {
                var min = Int.MAX_VALUE
                for (k in 0 until n) {
                    min = min.coerceAtMost(grid[i][j] + moveCost[grid[i][j]][k] + dp[i + 1][k])
                }
                dp[i][j] = min
            }
        }
        var min = Int.MAX_VALUE
        for (s in dp[0]) {
            min = min.coerceAtMost(s)
        }
        return min
    }
}

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