LeetCode in Kotlin
2742. Painting the Walls
Hard
You are given two 0-indexed integer arrays, cost and time, of size n representing the costs and the time taken to paint n different walls respectively. There are two painters available:
- A** paid painter** that paints the
ithwall intime[i]units of time and takescost[i]units of money. - A** free painter** that paints any wall in
1unit of time at a cost of0. But the free painter can only be used if the paid painter is already occupied.
Return the minimum amount of money required to paint the n walls.
Example 1:
Input: cost = [1,2,3,2], time = [1,2,3,2]
Output: 3
Explanation: The walls at index 0 and 1 will be painted by the paid painter, and it will take 3 units of time; meanwhile, the free painter will paint the walls at index 2 and 3, free of cost in 2 units of time. Thus, the total cost is 1 + 2 = 3.
Example 2:
Input: cost = [2,3,4,2], time = [1,1,1,1]
Output: 4
Explanation: The walls at index 0 and 3 will be painted by the paid painter, and it will take 2 units of time; meanwhile, the free painter will paint the walls at index 1 and 2, free of cost in 2 units of time. Thus, the total cost is 2 + 2 = 4.
Constraints:
1 <= cost.length <= 500cost.length == time.length1 <= cost[i] <= 1061 <= time[i] <= 500
Solution
class Solution {
fun paintWalls(cost: IntArray, time: IntArray): Int {
val n = cost.size
val dp = Array(n + 1) { IntArray(n + 1) }
return solve(n, cost, 0, time, dp)
}
private fun solve(wallsRem: Int, cost: IntArray, idx: Int, time: IntArray, dp: Array<IntArray>): Int {
if (wallsRem <= 0) return 0
if (idx >= cost.size) return 1000000000
if (dp[idx][wallsRem] != 0) {
return dp[idx][wallsRem]
}
val skip = solve(wallsRem, cost, idx + 1, time, dp)
val take = cost[idx] + solve(wallsRem - time[idx] - 1, cost, idx + 1, time, dp)
dp[idx][wallsRem] = skip.coerceAtMost(take)
return dp[idx][wallsRem]
}
}