LeetCode: Paint House

Posted on February 14, 2018April 14, 2020 by braindenny

Paint House

Similar Problems:

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[0][0] is the cost of painting house 0 with color red; 
costs[1][2] is the cost of painting house 1 with color green, and so on...
Find the minimum cost to paint all houses.

Note:
All costs are positive integers.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

Solution: dp

## https://code.dennyzhang.com/paint-house
## Basic Ideas: dynamic programming
##
##  For each house, we need to choose one color
##  From left to right, loop the houses
##
## Complexity: Time O(n), Space O(1)
class Solution:
    def minCost(self, costs: List[List[int]]) -> int:
        r, b, g = 0, 0, 0
        for i in range(len(costs)):
            [rc, bc, gc] = costs[i]
            rc += min(b, g)
            bc += min(r, g)
            gc += min(r, b)
            r, b, g = rc, bc, gc
        return min(r, b, g)

## 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[0]

        # 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)

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