LeetCode in Kotlin

3840. House Robber V

Medium

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed and is protected by a security system with a color code.

You are given two integer arrays nums and colors, both of length n, where nums[i] is the amount of money in the ith house and colors[i] is the color code of that house.

You cannot rob two adjacent houses if they share the same color code.

Return the maximum amount of money you can rob.

Example 1:

Input: nums = [1,4,3,5], colors = [1,1,2,2]

Output: 9

Explanation:

• Choose houses i = 1 with nums[1] = 4 and i = 3 with nums[3] = 5 because they are non-adjacent.
• Thus, the total amount robbed is 4 + 5 = 9.

Example 2:

Input: nums = [3,1,2,4], colors = [2,3,2,2]

Output: 8

Explanation:

• Choose houses i = 0 with nums[0] = 3, i = 1 with nums[1] = 1, and i = 3 with nums[3] = 4.
• This selection is valid because houses i = 0 and i = 1 have different colors, and house i = 3 is non-adjacent to i = 1.
• Thus, the total amount robbed is 3 + 1 + 4 = 8.

Example 3:

Input: nums = [10,1,3,9], colors = [1,1,1,2]

Output: 22

Explanation:

• Choose houses i = 0 with nums[0] = 10, i = 2 with nums[2] = 3, and i = 3 with nums[3] = 9.
• This selection is valid because houses i = 0 and i = 2 are non-adjacent, and houses i = 2 and i = 3 have different colors.
• Thus, the total amount robbed is 10 + 3 + 9 = 22.

Constraints:

• 1 <= n == nums.length == colors.length <= 105
• 1 <= nums[i], colors[i] <= 105

Solution

import kotlin.math.max

class Solution {
    fun rob(nums: IntArray, colors: IntArray): Long {
        val n = nums.size
        var dp0: Long = 0
        var dp1 = nums[0].toLong()
        for (i in 1..<n) {
            val new0: Long
            val new1: Long
            if (colors[i] == colors[i - 1]) {
                new1 = dp0 + nums[i]
                new0 = max(dp0, dp1)
            } else {
                new0 = max(dp1, dp0)
                new1 = max(dp0, dp1) + nums[i]
            }
            dp0 = new0
            dp1 = new1
        }
        return max(dp0, dp1)
    }
}

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