Hard
There is a row of m
houses in a small city, each house must be painted with one of the n
colors (labeled from 1
to n
), some houses that have been painted last summer should not be painted again.
A neighborhood is a maximal group of continuous houses that are painted with the same color.
houses = [1,2,2,3,3,2,1,1]
contains 5
neighborhoods [{1}, {2,2}, {3,3}, {2}, {1,1}]
.Given an array houses
, an m x n
matrix cost
and an integer target
where:
houses[i]
: is the color of the house i
, and 0
if the house is not painted yet.cost[i][j]
: is the cost of paint the house i
with the color j + 1
.Return the minimum cost of painting all the remaining houses in such a way that there are exactly target
neighborhoods. If it is not possible, return -1
.
Example 1:
Input: houses = [0,0,0,0,0], cost = [[1,10],[10,1],[10,1],[1,10],[5,1]], m = 5, n = 2, target = 3
Output: 9
Explanation: Paint houses of this way [1,2,2,1,1]
This array contains target = 3 neighborhoods, [{1}, {2,2}, {1,1}].
Cost of paint all houses (1 + 1 + 1 + 1 + 5) = 9.
Example 2:
Input: houses = [0,2,1,2,0], cost = [[1,10],[10,1],[10,1],[1,10],[5,1]], m = 5, n = 2, target = 3
Output: 11
Explanation: Some houses are already painted, Paint the houses of this way [2,2,1,2,2]
This array contains target = 3 neighborhoods, [{2,2}, {1}, {2,2}].
Cost of paint the first and last house (10 + 1) = 11.
Example 3:
Input: houses = [3,1,2,3], cost = [[1,1,1],[1,1,1],[1,1,1],[1,1,1]], m = 4, n = 3, target = 3
Output: -1
Explanation: Houses are already painted with a total of 4 neighborhoods [{3},{1},{2},{3}] different of target = 3.
Constraints:
m == houses.length == cost.length
n == cost[i].length
1 <= m <= 100
1 <= n <= 20
1 <= target <= m
0 <= houses[i] <= n
1 <= cost[i][j] <= 104
class Solution {
private lateinit var prev: Array<IntArray>
private lateinit var curr: Array<IntArray>
private lateinit var mins: Array<IntArray>
fun minCost(houses: IntArray, cost: Array<IntArray>, m: Int, n: Int, target: Int): Int {
prev = Array(n) { IntArray(target) }
curr = Array(n) { IntArray(target) }
mins = Array(2) { IntArray(target) }
for (i in 0 until m) calculate(i, houses, cost, n, target)
var min = Int.MAX_VALUE
for (i in 0 until n) min = Math.min(min, curr[i][target - 1])
return if (min == Int.MAX_VALUE) -1 else min
}
private fun calculate(house: Int, houses: IntArray, cost: Array<IntArray>, n: Int, target: Int) {
swap()
calculateMins(n, target)
if (houses[house] > 0) {
costInPaintedHouse(house, houses, cost, target)
} else {
costNotPaintedHouse(
house,
cost,
target,
)
}
}
private fun costInPaintedHouse(house: Int, houses: IntArray, cost: Array<IntArray>, target: Int) {
val color = houses[house] - 1
for (i in cost[house].indices) {
val group = Math.min(target - 1, house)
val newG = house == group
if (i == color) {
curr[i][0] = prev[i][0]
for (j in 1..group) {
curr[i][j] = if (mins[0][j - 1] == prev[i][j - 1]) mins[1][j - 1] else mins[0][j - 1]
curr[i][j] = if (newG && j == group) curr[i][j] else Math.min(curr[i][j], prev[i][j])
}
} else {
for (j in 0..group) curr[i][j] = Int.MAX_VALUE
}
}
}
private fun costNotPaintedHouse(house: Int, cost: Array<IntArray>, target: Int) {
for (i in cost[house].indices) {
val group = Math.min(target - 1, house)
val newG = house == group
curr[i][0] = if (prev[i][0] == Int.MAX_VALUE) prev[i][0] else prev[i][0] + cost[house][i]
for (j in 1..group) {
curr[i][j] = if (mins[0][j - 1] == prev[i][j - 1]) mins[1][j - 1] else mins[0][j - 1]
curr[i][j] = if (newG && j == group) curr[i][j] else Math.min(curr[i][j], prev[i][j])
curr[i][j] = if (curr[i][j] == Int.MAX_VALUE) Int.MAX_VALUE else curr[i][j] + cost[house][i]
}
}
}
private fun swap() {
val temp = prev
prev = curr
curr = temp
}
private fun calculateMins(n: Int, target: Int) {
for (i in 0 until target - 1) {
mins[0][i] = prev[0][i]
mins[1][i] = Int.MAX_VALUE
for (j in 1 until n) {
if (prev[j][i] <= mins[0][i]) {
mins[1][i] = mins[0][i]
mins[0][i] = prev[j][i]
} else if (prev[j][i] <= mins[1][i]) mins[1][i] = prev[j][i]
}
}
}
}