There are a row of n houses, each house can be painted with one of the k colors. The cost of painting each house with a certain color is different. You have to paint all the houses such that no two adjacent houses have the same color.
The cost of painting each house with a certain color is represented by a n x k cost matrix. For example, costs[0][0] is the cost of painting house 0 with color 0; costs[1][2] is the cost of painting house 1 with color 2, and so on... Find the minimum cost to paint all houses.
Note: All costs are positive integers.
Example:
Input: [[1,5,3],[2,9,4]]
Output: 5
Explanation: Paint house 0 into color 0, paint house 1 into color 2. Minimum cost: 1 + 4 = 5;
Or paint house 0 into color 2, paint house 1 into color 0. Minimum cost: 3 + 2 = 5.
Follow up: Could you solve it in O(nk) runtime?
# Dynamic programming
class Solution(object):
def minCostII(self, costs):
"""
:type costs: List[List[int]]
:rtype: int
"""
if len(costs) == 0 or len(costs[0]) == 0:
return 0
n, k = len(costs), len(costs[0])
dp = [[0 for _ in range(k)] for _ in range(n)]
prev_min = float("inf")
prev_min_index = -1
prev_second_min = float("inf")
for i in range(n):
for j in range(k):
if i == 0:
dp[i][j] = costs[i][j]
elif j == prev_min_index:
dp[i][j] = prev_second_min + costs[i][j]
else:
dp[i][j] = prev_min + costs[i][j]
prev_min = float("inf")
prev_min_index = -1
prev_second_min = float("inf")
for j in range(k):
if dp[i][j] < prev_min:
prev_second_min = prev_min
prev_min = dp[i][j]
prev_min_index = j
elif dp[i][j] < prev_second_min:
prev_second_min = dp[i][j]
return min(dp[-1])