LeetCode in Kotlin

3429. Paint House IV

Medium

You are given an even integer n representing the number of houses arranged in a straight line, and a 2D array cost of size n x 3, where cost[i][j] represents the cost of painting house i with color j + 1.

The houses will look beautiful if they satisfy the following conditions:

• No two adjacent houses are painted the same color.
• Houses equidistant from the ends of the row are not painted the same color. For example, if n = 6, houses at positions (0, 5), (1, 4), and (2, 3) are considered equidistant.

Return the minimum cost to paint the houses such that they look beautiful.

Example 1:

Input: n = 4, cost = [[3,5,7],[6,2,9],[4,8,1],[7,3,5]]

Output: 9

Explanation:

The optimal painting sequence is [1, 2, 3, 2] with corresponding costs [3, 2, 1, 3]. This satisfies the following conditions:

• No adjacent houses have the same color.
• Houses at positions 0 and 3 (equidistant from the ends) are not painted the same color (1 != 2).
• Houses at positions 1 and 2 (equidistant from the ends) are not painted the same color (2 != 3).

The minimum cost to paint the houses so that they look beautiful is 3 + 2 + 1 + 3 = 9.

Example 2:

Input: n = 6, cost = [[2,4,6],[5,3,8],[7,1,9],[4,6,2],[3,5,7],[8,2,4]]

Output: 18

Explanation:

The optimal painting sequence is [1, 3, 2, 3, 1, 2] with corresponding costs [2, 8, 1, 2, 3, 2]. This satisfies the following conditions:

• No adjacent houses have the same color.
• Houses at positions 0 and 5 (equidistant from the ends) are not painted the same color (1 != 2).
• Houses at positions 1 and 4 (equidistant from the ends) are not painted the same color (3 != 1).
• Houses at positions 2 and 3 (equidistant from the ends) are not painted the same color (2 != 3).

The minimum cost to paint the houses so that they look beautiful is 2 + 8 + 1 + 2 + 3 + 2 = 18.

Constraints:

• 2 <= n <= 105
• n is even.
• cost.length == n
• cost[i].length == 3
• 0 <= cost[i]\[j] <= 105

Solution

import kotlin.math.min

class Solution {
    fun minCost(n: Int, cost: Array<IntArray>): Long {
        var dp0: Long = 0
        var dp1: Long = 0
        var dp2: Long = 0
        var dp3: Long = 0
        var dp4: Long = 0
        var dp5: Long = 0
        for (i in 0..<n / 2) {
            val nextdp0: Long =
                min(min(dp2, dp3), dp4) + cost[i][0] + cost[n - i - 1][1]
            val nextdp1: Long =
                min(min(dp2, dp4), dp5) + cost[i][0] + cost[n - i - 1][2]
            val nextdp2: Long =
                min(min(dp0, dp1), dp5) + cost[i][1] + cost[n - i - 1][0]
            val nextdp3: Long =
                min(min(dp0, dp4), dp5) + cost[i][1] + cost[n - i - 1][2]
            val nextdp4: Long =
                min(min(dp0, dp1), dp3) + cost[i][2] + cost[n - i - 1][0]
            val nextdp5: Long =
                min(min(dp1, dp2), dp3) + cost[i][2] + cost[n - i - 1][1]
            dp0 = nextdp0
            dp1 = nextdp1
            dp2 = nextdp2
            dp3 = nextdp3
            dp4 = nextdp4
            dp5 = nextdp5
        }
        var ans = Long.Companion.MAX_VALUE
        for (x in longArrayOf(dp0, dp1, dp2, dp3, dp4, dp5)) {
            ans = min(ans, x).toLong()
        }
        return ans
    }
}

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