Leetcode - paint house I solution (Java)
Paint House I
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.
Analysis
We can use Dynamic programming to solve this problem.
Define DP[i][j] as the minimum cost to paint house until i with color j. where i ∈ [1, n] and j ∈ {0, 1, 2}.
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DP[i][0] += Math.min(DP[i-1][1], DP[i-1][2]); DP[i][1] += Math.min(DP[i-1][0], DP[i-1][2]); DP[i][2] += Math.min(DP[i-1][0], DP[i-1][1]); |
We can reuse the costs array. Here is the implementation:
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public int minCost(int[][] costs) { if(costs==null||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 n = costs.length-1; return Math.min(Math.min(costs[n][0], costs[n][1]), costs[n][2]); } |
