- Series: Paint Fence & Follow-up
- Paint Fence
- Review: Dynamic Programming Problems
- Tag: #dynamicprogramming, #paintfence
There are a row of n houses, each house can be painted with one of the three colors: red, blue or green. 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 3 cost matrix.
For example, costs is the cost of painting house 0 with color red; costs is the cost of painting house 1 with color green, and so on... Find the minimum cost to paint all houses.
All costs are positive integers.
Credits To: leetcode.com
Leave me comments, if you have better ways to solve.
## Blog link: https://code.dennyzhang.com/paint-house ## Basic Ideas: Dyanmic programming ## For house after house1, it can only have 3 possilities. ## ## Complexity: Time O(n), Space O(1) class Solution: def minCost(self, costs): """ :type costs: List[List[int]] :rtype: int """ length = len(costs) if length == 0: return 0 dp = [None]*3 # caculate house I dp = costs # caculate the following house for i in range(1, length): l = [None]*3 for j in range(0, 3): l[j] = costs[i][j] + min(dp[(j+1)%3], dp[(j+2)%3]) dp = l return min(dp)