DP-8. Paint House

Problem

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.

Example:

Input: [[17,2,17],[16,16,5],[14,3,19]]
Output: 10
Explanation: Paint house 0 into blue, paint house 1 into green, paint house 2 into blue.
Minimum cost: 2 + 5 + 3 = 10.

Explanation

This is an optimal problem, so we can naturally think of dp. What is the subproblem? The minimum cost of current house is dependent on the previous house. As there are only 3 colors and no two adjacent houses have the same color. So, we can calculate the three different cost of the current house and we can easily find the equation. If the current color is 0, what is the color of the previous house? Either 1 or 2. So we just need to get the minimum of those two cost plus the current cost[0].

Solution

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// 21:06
class Solution {
public int minCost(int[][] costs) {
int houses = costs.length;
int[][] dp = new int[houses + 1][3];

for (int i = 0; i < houses; i++) {
dp[i + 1][0] = Math.min(dp[i][1] + costs[i][0],
dp[i][2] + costs[i][0]);
dp[i + 1][1] = Math.min(dp[i][0] + costs[i][1],
dp[i][2] + costs[i][1]);
dp[i + 1][2] = Math.min(dp[i][0] + costs[i][2],
dp[i][1] + costs[i][2]);
}

return Math.min(Math.min(dp[houses][0], dp[houses][1]), dp[houses][2]);
}
}