LeetCode: Paint House II

Paint House II

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• Tag: #paintfence

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?

Github: code.dennyzhang.com

Credits To: leetcode.com

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

• Solution:

## https://code.dennyzhang.com/paint-house-ii
## Basic Ideas: dynamic programming
##
##  From house to house, from color to color
##  From top to down, from left to right
##  dp(i, j)  = min(dp(i-1, k)) k != j
##
## Complexity: Time O(n*n*k), Space O(n*k)
class Solution(object):
    def minCostII(self, costs):
        """
        :type costs: List[List[int]]
        :rtype: int
        """
        if not costs: return 0
        n, m = len(costs), len(costs[0])
        if m == 0: return 0
        dp = copy.deepcopy(costs[0])
        for i in range(1, n):
            dp2 = copy.deepcopy(dp)
            for j in range(m):
                dp2[j], v = sys.maxsize, dp2[j]
                dp[j] = min(dp2) + costs[i][j]
                dp2[j] = v
        return min(dp)