Paint House II

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- LeetCode: Minimum Falling Path Sum 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)