[LeetCode] 256. Paint House

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.

Thought process:

1. Sub-problem: find the minimum cost to paint the houses up to current house in red, blue or green.
2. Function:
1. Red: min(f[i - 11][1], f[i - 1][2]) + costs[i][0].
2. Blue: min(f[i - 1][0], f[i - 1][2]) + costs[i][1].
3. Green: min(f[i - 1][0], f[i - 1][1]) + costs[i][2].
3. Initialization: f[0][i] = 0.
4. Answer: min(f[costs.length][i]).

Solution:

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class Solution { public int minCost(int[][] costs) { if (costs.length == 0) { return 0; } for (int i = 1; i < costs.length; i++) { costs[i][0] += Math.min(costs[i - 1][1], costs[i - 1][2]); costs[i][1] += Math.min(costs[i - 1][0], costs[i - 1][2]); costs[i][2] += Math.min(costs[i - 1][0], costs[i - 1][1]); } int last = costs.length - 1; return Math.min(costs[last][0], Math.min(costs[last][1], costs[last][2])); } }

Time complexity: O(n).